@@Reachermordacai "hey dude I see on sundays, I gotta work the next 2794 days!" "Damn so you stop work in the middle of a week?" "How did you figure that one out?"
When I saw this video on TH-cam at a first glance, I thought I was looking at some old video from seven years ago or something, was surprised when I realized it was from today
For 40+ years I've been saying, "there is no rule for 7," meaning no way to check for divisibility as with 3, 5, 9 etc. Thank you for this. I now have my "rule for 7!"
Can't believe that 10 years later, I am still loving watch James Grime on Numberphile. I watched him when I was a nerdy high schooler and now I'm a nerdy adult. Thank you so much for all the videos over the years.
That's one of the great things about youtube that is not talked about enough. There are so many creators I have watched for a decade or longer and they were so important to me and continue to be so. In the media too much attention is paid to the drama and such. What about people like this? Or the green brothers? Vsauce? Rhett and link? I could go on. It's not an easy thing to do, to be able to take your fans with you as they pass into different eras of their lives. Most adults don't still watch what they watched as kids but so many of us do with creators like I mentioned above. These people are my teachers and actually helped guide me to being a better adult. Me and so many others. It's a beautiful thing. The creators, as well as fellow followers, feel like family and it's such a comfort.
When I was in the 7th grade, I was taught all the tests for divisibility except for 7. What we were told was "try dividing it by 7," which completely defeats the purpose of a divisibility test. Thank you for filling in this particular gap in my education.
@@BeBopScraBoo Yes, and the video is also completely missing the stated topic. I wanted to know why non-roundly dividing by 7 always creates the same decimal pattern.
@@Dowlphin decimal system is based on ten, which factors to 2 and 5. any fraction with a denominator that factors with any other number will have a repeating pattern.
The same algorithm (3:51) can be used to construct a formula for other primes: 11: x - y (k=10, j=99) 13: x + 4y or x - 9y (k=4, j=39) 17: x - 5y (k=12, j=119) 19: x + 2y (k=2, j=19) 23: x + 7y (k=7, j=69) 29: x + 3y (k=3, j=29) 31: x - 3y (k=28, j=279) 37: x - 11y (k=26, j=259) 41: x - 4y (k=37, j=369) 43: x + 13y or x - 30y (k=13, j=129) 47: x - 14y (k=33, j=329) 53: x + 16y (k=16, j=159) 59: x + 6y (k=6, j=59) 61: x - 6y (k=55, j=549) 67: x - 20y (k=47, j=469) 71: x - 7y (k=64, j=639) 73: x + 22y (k=22, j=219) 79: x + 8y (k=8, j=79) 83: x + 25y (k=25, j=249) 89: x + 9y (k=9, j=89) 97: x - 29y or x - 30x + x (k=68, j=679) where k is the multiplier of 10x + y: 10kx + ky (4:11) and j is a multiple of the prime that is subtracted: 10kx - jx + ky = x + ky (4:30) then subtract the prime if it helps And since the same algorithm is used, the results have the same properties, like iterability (1:24)
For the ones with a y coefficient greater than 10, we could split the number like so: 100a + b; where b is the last 2 digits and a is the rest: 37: a +10b (k=10, j=999) 43: a - 3b (k=6, j=559; originally 6b - 2a which is even and backwards, so divide by -2) 47: a + 8b (k=8, j=799) 53: a - 9b (k=18, j=1749; originally 18b - 2a -> divide by -2) 67: a - 2b (k=12, j=1139; orig. 12b - 6a -> div. -6) Couldn't find a decent formula for 73, 83, and 97 in a timely manner. I might revisit this later
@@leo848 I suppose, but for 2, 5 and therefore 10, for example, you don't need to iterate since you're just looking at the last digit (is it even?, is it 5 or 0?, and is it 0? respectively). For the alternating sum for 11, I have not taken the time to reverse engineer its algebraic derivation; it too may have the same property because it may be a similar algorithm.
@@leo848 not really. You can construct a finite state automaton (for instance, a 3-state automaton for divisibility by 3) that, after you run the digits through its arrows will straightaway give you a ‘yes’ or a ‘no’ without any need for iteration. I guess, we're just more used to addition and stuff, and devise these chevksum kind of tests, which fall into the trap of iteration.
@@nnaammuuss Of course. I meant that once you have a divisibility test that gives back a number that is divisible iff the original number is, one can always iterate.
Me and a friend of mine discovered a trick for 11 in middle school: Take 121, u split it in 1+21 = 22 so if the result is divisible by 11 the original is too. It works even for larger numbers like 35673 split in 3+56+73 = 132 132 split in 1+32 = 33 In case of even digits 1078 split in 10+78 = 88 Basically you split the number in sets of two digits starting from the end.
One of the great things about youtube that is not talked about enough. There are so many creators I have watched for a decade or longer and they were so important to me and continue to be so. In the media too much attention is paid to the drama and such. What about people like this? Or the green brothers? Vsauce? Rhett and link? I could go on. It's not an easy thing to do, to be able to take your fans with you as they pass into different eras of their lives. Most adults don't still watch what they watched as kids but so many of us do with creators like I mentioned above. These people are my teachers and actually helped guide me to being a better adult. Me and so many others. It's a beautiful thing. The creators, as well as fellow followers, feel like family and it's such a comfort.
One trick that might help for finding primes is that except for 2 and 3 all primes follow the pattern of a multiple of 6 plus or minus 1. This is true because +2 would be divisible by 2, +3 by 3, +4 by 2, and +5 is also -1. Unfortunately I discovered this long after I needed to write Basic computer programs to find primes in school.
Because of this video, I found out about the 1001 test, alternate sum of 3 digits: 123456788 is divisible by 7 because 123-456+788=455 is divisible by 7! Best part is that this test works with 11 and 13... 455 is also divisible by 13 so 123456788 is divisible by 13. This is golden when testing for primes!!
This video took me back in time, when I was a middle-schooler amazed by these tricks and properties of numbers. Even now in uni, I watch videos on this channel with the same flabbergasted look, enjoying every minute as an extraordinary discover. I really appreciate your content, you're doing great ✌️
I love the fact that everyone keeps saying this video feels like it was from years ago but no one is commenting how Dr. Grimes didn't age that much. He's a vampire who knows all the secrets of numbers
I always found multiples of 100, subtracted them and added on double the number of 100s you removed to what is remaining. Taking advantage of 98 being a multiple of 7. This works pretty well and is easy to do in your head
I seriously love watching Numberphile videos and you are my absolute favourite, James Grime Sir! I adore you James Grime Sir to a great extent. (Your smile is literally the best!) Thanks for making me love mathematics more than ever.
I’d always do it by comparing to other numbers. 7000 is divisible by 7, 7000-6468=532. If 532 is divisible by 7 then 6468 was. 700-532=168. 168-140=28. 28 is divisible by 7 so 6468 is. This method is much easier for me to do in my head because it’s only addition or subtraction.
@CarlFriedrichGauss1 easy 5194-4900=294, 294-280=14. 14=7*2. Therefore 5194 can be divided by 7. All you have to do is use is tens, hundreds, thousands of the times table.
When I was 13, I found another method for the divisibility by 7. It's in a way even more useful than the one in the video. The trick is to cut the number at the second digit. For instance, for 343, it's 3 and 43. After that, you double the rest (in the example, from 3, you get 6). If the sum is divisible by 7, then the original number is divisible by 7: 6+43 = 49. So, 343 is divisible by 7. What this method also provides is that it preserves modulo. So, if the number is not divisible by 7, the method also tells you how much is the remainder mod 7. For example, taking 716, you get 2×7 + 16 = 30, which is 28 + 2. So, 716 gives 2 as a remainder after dividing by 7.
Nice, this was actually the answer i expected from the video. For smaller numbers (under 6 digits) i think your solution is easier. For larger numbers i think the videos solution is better as each step requires multiplying 1digit by 5 and reduces the test number by. 1 digit, the mod 100 version each step requires multiplying an n-2 digit number by 2, it potentially reduces the number of digits by 2 but 3/10 operations will only reduce it by 1. Regarding finding the remainder the video requies 1 extra step, each iteration of the algorithm multiplies the remainder by 5, if you keep track of the number of iterations mod 6 you can multiply your final remainder by (1,3,2,6,4,5) mod 7.
This is very cool! I "discovered" the same method while staring at the digital clock and splitting the number at the colon when I was little. Thought I was one of the great mathematicians of the century.
This is such a great video.. i’ve shared with my kids and the “presentation” with special british commentary and voice intonations is perfect. Useful and just enough entertainment!
I poked around in excel to find out what happens with that seven trick in how other numbers terminate. 49 is in a loop of 49's as we saw in the video and all numbers 1-48 that aren't divisible by 7 are in a loop together and all of the numbers divisible by seven are in their own loop. 1,5,25,27,37,38,43,19,46,34,23,17,36,33,18,41,9,45,29,47,39,48,44,24,22,12,11,6,30,3,15,26,32,13,16,31,8,40,4,20,2,10 and 7,35,28,42,14,21
Cool! I've tried something like that as well, and it seems that powers of 7 all eventually reduce down to 49 as well: 7^2 = 49 --> 49 7^3 = 343 --> 49 7^4 = 2401 --> 245 --> 49 7^5 = 16807 --> 1715 --> 196 --> 49 And so on.
I just instantly see either 350+84 or 420+14 which makes it obviously divisible by 7. But I guess the point of the video is to showcase some more interesting methods. nice job.thanks.
There's another similar method which I use if the divisibility test is just too hard: Keep adding or subtracting the number which you want to know divisibility by to your number until it's a multiple of 10. If, say, your number which you want to know divisibility by is even, and your number is odd, there's no way of getting to a multiple of 10, so it isn't divisible. Divide your 10 multiple by 10 and try to see if that's divisible, if not, repeat the process until you see something you recognize. Example: divisibility by 13 for 832 832(i picked this number completely at random no joke)+13=845 845+(13*5)=845+65=910 910/10=91 91=13*7, but lets say i'm still unsure 91+13=104 104+13=117 117+13=130 No doubt about it, 130 is divisible by 13. 832 must be divisible by 13
I had learnt divisibility tests in school. But we hadn't learnt any for 7. I genuinely though it wasn't possible, (thinking they would've taught it if it was that way) But that 10x + y explanation completely blew me away. It's these very basic simplistic things in math which amaze you. Because it is so simple and 100% sure it was possible to come up with myself. Really encourages one to keep trying out stuff and working out things just for fun!
@@BeBopScraBoo You would be a crappy teacher. The WHOLE point of Mathematics is to recognize (and prove) beautiful patterns of set theory which can be inspiring for curiosity. I.e. The proof of Div by 7 is _Number Theory,_ specifically using *cyclic addition* which differs from *linear addition.* This _difference of perspective_ can inspire students to want to learn more.
Sadly I never learnt the Div by 7 mod trick in school either. When I was 30 I heard about the _subtract double the last digit from the remaining digits_ but didn’t know _why_ it worked so shortly after I set out to prove it. I independently derived some circular addition rules (modular arithmetic) and from that it was relatively straightforward: 10x + y ≡ 0 (mod 7) 50x + 5y ≡ 0*5 (mod 7) 49x + x + 5y ≡ 0 (mod 7) [49x ≡ 0 (mod 7)] + [x + 5y ≡ 0 (mod 7)] 0 + x + 5y ≡ 0 (mod 7) x + 5y - 7y ≡ 0 - 7y (mod 7) x - 2y ≡ 0 (mod 7) QED The sad part is kids do modular arithmetic (circular addition) when they learn how to tell analog time but no one every tells them they are doing Number Theory!
@@MichaelPohoreski That's not the point of school though. The purpose of school is to learn skills you can apply in real life. The point of divisibility tests is that you can do them in your head without writing anything down. If you have to write it down, it's pointless and you can just divide in the first place. Yes, it's interesting how numbers converge into patterns, but there's not point worrying about number theory and patterns in elementary school
An alternative method (that I worked out for myself right now) you can do is *last 2 digits plus twice the rest* . Will shrink the number much faster, and the mental maths is barely harder. Proof is exactly the same as in the video. It's actually surprisingly easy to generate any number of these tricks.
@@leoirias3506 true, but if you're dealing with 3-4 digit number, you've only got to double 1-2 digits. Not very hard to do mentally I'd say? If you've got a much larger number... You won't. Nobody needs to know if a 5+ digit number is divisible by 7 in their head.
@@QuantumHistorian For a larger number, you can split it into 2 digit numbers starting from the end, and then multiply by increasing powers of 2. Eg: 12936 -> 36+29*2+1*4 -> 98 -> divisible by 7 To make it simpler, you can do a mod 7 for each of the individual elements: 36 -> 35+1->1 29*2->(28+1)*2->2 1*4->4 1+2+4=>7
This video is the basis for my favorite math subject: prime factorization. It's breaking something down to it's most elementary components lets you see how many ways you can manipulate and use a number.
I remember I used a USB stick duplicator at work once, with 32 slots - 1 for the master USB and 31 for the clones. It also had a display that counted the total of successful duplicates, and checking if the number of successfully made duplicates was divisible by 31 was a quick way to tell if the machine was working properly. So, I used the similar divisibility test for 31, as it felt simpler once I got used to the method. "Subtract 3 times the last digit from the rest".
Hmmm I think I get your trick: for 31, you could do -3*(10x+y) = -30x-3y = -31x - (x - 3y). So indeed the original number (10x+y) it can be divided by 31 iff x-3y is divisible by 31. It's a bit of fiddling to find the right factor (in this case -3), or at least it's not obvious to me immediately, but it's not too hard to derive these tricks for other numbers actually! For example I found 13 too: 4(10x+y)=39x+(x+4y) so you can check x+4y (because 39=13*3).
@@ikbintom Finding those fiddling factors comes in a couple ways if you have nice numbers: if you're looking for something divisible by XY, where X and Y are digits, then for a number ABC... you can multiply the last digit (ex: C) by X/Y, and subtract from the rest. Why? Every time you add a multiple of XY, you add Y to the last digit, and X/Y to the other digits. If you multiply Y by the ratio, and subtract that, you've essentially said 'ok, great, let's remove Y copies of XY from ABC, guaranteeing the last digit is now 0. Is the remainder left 10x a multiple we're aware of?. And if you repeat the process, you're just now doing the same operation, but with the 10s and 100s places. The second nice way is finding a convenient multiple near 100. For 7, 98 works great. Since that's 2 away from 100, we can just chop off all but the last 2 digits (would be 3 digits for a multiple near 1000), and add on 2 for every hundred we cut off - 343? 3*2+43 = 49. For 31, 93 is kinda close to 100, but 7 off. So let's try it : 568, 35+68=103, not a multiple. In fact, since it's 10 more than a multiple (103-10=93), you know 558 would be a multiple of 31, which it is (18x31).
The general divisibility trick for any prime p (other than 2 and 5) is as follows: Let our dividend be k and express it as 10x + y (x , y are non-negative integers) Find an integer z such that 10z when divided by p yields remainder 1. Now if x + (z)y is divisible by p, then 10x + y is divisible by p. This yields the required algorithm. Note: You can replace 10 with any natural number as long as it isn't a multiple of our chosen prime p.
I usually do it with a multiple of p that ends in 1. Call the multiple 10m+1. Then if x-my is divisible by p, then 10x+y is also divisible by p. For example, 17x3=51, so you can take off the last digit, multiply it by 5, and subtract that from the rest of the number. Repeat until you get a number that obviously is or isn't a multiple of 17. If it is, then the original number was divisible by 17, and if not, it wasn't.
Nice, yeah that yields the trick(s) for 11 of either adding 10N or subtracting N (for trailing digit N). And for 19 and 21 you get tricks involving ±2N.
This video actually changed my life. I was really ready to end it all, had a goodbye note written, had picked out which bridge I was going to jump off, and then I saw this video and now I feel the strength to carry on. Thank you, sir. Seven truly is amazing.
You see what happens now, when greatness is demanded of you? Honor your time wisely, henceforth. You now harness the powerful knowledge of dividing by 7!
Here's a trick that only requires you to know up to 10x7: take your number (let's use his example of 6468), and find the multiple of 7 that ends in the same digit. So, 6468 ends in 8, and 28 ends in 8, so subtract 28 and you have 6440. Drop the last 0, so you have 644. Repeat. 644 ends in 4, 14 ends in 4, 644-14=630, drop the 0, and you have 63, which is divisible by 7, so it's all divisible by 7. It's similar, but you don't have to deal with the 'multiply by 5' step, so it might be a bit mentally easier.
That is about the same as dividing by 7 normally, except you're going right-to-left instead of left-to-right. However, your method is what I would do to check divisibility by many larger prime numbers like 17 or 23, except that I may add or subtract and attack the number from both the left and right. For example, to test if 3893 is divisible by 17, I would find it easier to add 17 rather than subtract 153. 3893 + 17 = 3910 or 391 after dropping the 0. At that point, I would notice: 391 = 17 * 23 = (20 - 3)(20 + 3) = 20² - 3² However, I could have could have continued the process by subtracting 51 (or adding 119) to 391 to get 340 (or 510). As to how often this is necessary in day-to-day life, well...
I prefer this method as well, it's a 7 based multiplication + subtraction vs a 5 based multiplication + addition. So end of the day it's pretty much the same
I have an easier approach for this one. Take 6468 for example. Consider first two digits i.e. 64. 7*9=63. So 7*900= 6300. 6468-6300= 168. Now consider first two digits again i.e. 16. 7*2=14. 7*20=140. 168-140= 28. 28 is divisible by 7 so number is divisible by 7. Fun fact - this is nothing but doing elementary division by 7 in a more complicated way. The method given in the video and and in the comment section by people are are more complex than simply dividing by 7 and check. It's fun math but not practical at all
Isn't it amazing that the sponsor is placed at the _end_ ? I can't believe Numberphile is the only channel I've seen that doesn't intrude on its own content!
shitty advertisers (like Raid Shadow Legends and other mobile games, most VPN, etc.) force you to have the ad at the starts + link in pinned comment + link in description, while Brilliant, Kiwico, etc. don't.
Most educational channels (all of the SciShow channels, PBS channels, Stand-up Maths, etc) put the ad at the end if they have one. If you can see the viewership graph on the seek bar, viewer retention usually flatlines on those videos as soon as the ad starts.
Probably because Numberphile has more leverage in the situation than your average independent youtuber who may be more in need of the sponsorship money (and therefore much more able to say no to a sponsor demanding a more intrusive segment).
It's much easier to find the nearest obvious number that can be divided by 7, e.g. 7000. Minus 6468. It's 532 and check if it's can be divided. Btw, you can do the same for 532 and 700. or 560 (7*80) - 532 = 28. Easy even doing in mind.
James discussing Number Theory, a perfect start to my day. The idea that we see here as a ‘trick’ can be further generalised to derive an algorithm to check whether a number is divisible by a prime number that ends with 1,3,7,9 (i.e. all primes greater than 5). Very useful indeed.
This method seems more complicated than just using classic division tricks to check. For example, with 686, you could just divide 68 by 7 with a remainder, then you take that remainder, apply it to the remaining digit to get 56 and 56 is evenly divisible by 7. Plus, doing it this way also gives you the quotient, not just the yes/no on divisibility. And, just like the trick you showed, you can repeat the steps for chains longer than 3, _and_ it works for divisibility with _any_ number, not just 7.
One of the points of this division trick is to avoid division. I teach both methods, and people seem to prefer the one highlighted in the video for whatever reason. For larger numbers, say 675081 you have to keep dividing by 7 and adding the remainder, OR you could multiply and add, and just check the last number for divisibility. If you noticed, all of these tricks are just to find divisibility, not the quotient. But I agree getting the quotient is a nice bonus =) I personally do not see one more complicated than the other, but different people work differently, and have their preferences!
A method which I find easier to do in my head is to split the number up into groups and look at each group individually. For instance, 434 becomes 43 4 becomes 42 14 and both groups are divisible by 7, so the original number is also. As a bonus, the quotient is immediately given as 62. Thus 434 / 7 = 62. Do the same with 6468: 6468 becomes 63 168 becomes 63 14 28, so it is divisible by 7 and the quotient is 924.
You have to love James Grime. Really! His passion, energy, and love for mathematics are just splashing on me from the monitor. Truly fantastic.👏 I wish I had such a passionate teacher/lecturer for my math lessons. I'm glad I could enjoy on YT for the least. 🙂
I personally find 6300+140+28 easier to figure out in the head, works for arbitrary sized numbers quite quickly too if you know your x7 table well enough. Its a bit of doing the full division but in blocks of 2-3 digits... and in a way its related to this one...
Another interesting thing my friend and I figured out- if you divide a number which is not divisible by 7, by 7; you get repeating patterns after the decimal place- which is always 142857… in a cyclic order.
Yeah this one is completely nuts. But it's the best known cyclic number. Multiply 142857 by anything you get an anagram or 9's repeated if it's a power of 7.
basically, what i do is i add up all of the digits in the number, then i add 7 to the original number and add up the digits of that number. if the total of (original +7) is 2 less than the original sum, it’s a multiple of 7. it works *a lot* of the time but it isn’t foolproof because occasionally, the sun resets to a larger number e.g 7(7), 14(5), 21(3) 28(*10*) wait, so maybe it works if the next number is 2 less or 7 more? idek
True, but the decimal remainder starts with a different number in the sequence which is in numeric order in respect to the remainder. example: put the cycle in numeric order - 1,2,4,5,7,8 which correlates with remainders 1,2,3,4,5,6 If you divide 37 by 7 you get 5 with a remainder 2. So the decimal remainder is .285714 repeated. For 38/7, it would be 5.428571... (3 is the remainder. The third number in the sequence is 4.)
1:25 For numbers with 6 digits, the easy thing to do is to split it in the middle, so you get two numbers with three digits each. If the difference of those numbers is divisible by 7, then so is the original numbers. This works, because the method corresponds to subtracting multiples of 1001, which is a multiple of 7 (and of 11 and 13, so you can use this to check those divisibilities too). Using this method on given example of (00)6468, the first step gives 468-6=462, which is notably divisible by 7, because 462 equals 420 + 42 😎
7 is funky because it's the representation of the constant of -1. It is the mechanism that causes the offset nature of reality and is the reason we make sense of time the way we do
I'm delighted! This filled in the one remaining gap in my knowledge of divisibility by single digit numbers! I will use it often, because I often feel curious about whether a number is prime, and I like to see what I can work out before simply searching the web for the answer.
Well, it certainly makes sense that adding 5x and subtracting 2x would work the same when what you care about is divisibility by 7; when counting modulo 7, 5 and -2 are the same number.
Ooo you're right. If the number is 10x+y, then the first method is about checking if x+5y is divisible by 7, and the second method is about checking if x-2y is divisible by 7. (I like to call them the x+5y & the x-2y methods). But modulo 7, they're the same thing! (For people not familiar with modular arithmetic, if x+5y is divisible by 7, so is x-2y. Because it is smaller by exactly 7y.) In fact I can make an infinite number of 7 divisibilty tests like these, like the 8x-9y divisibility test & the 12y-6x divisibilty tests by adding or subtracting 7x's & 7y's. (If you frame it as a number is divisible by 7, if 12 times the last digit minus 6 times the first digit is divisible by 7, you can perhaps use it as a cool party trick ;) )
I own a funnel cake stand and I used to sell them for $7 each (not anymore... inflation!). This would have been handy when counting up the money at the end of an event to make sure the drawer was right!
The method I found for determining divisibility by 7 was 3 times the ten's place plus the one's place. So 105 would be 10x3+5 or 35. 3x3+5 or 14. 1x3+4 or 7. For divisibility by 4, if the 10's place is odd the 1's place must be 2 or 6, or even but not divisible by 4. If the 10's place is even then the 1's place must be 0, 4, or 8, or divisible by 4.
For 434 I did 3X43 and added the 4. This is 129 + 4 =133, or 7 less than 140, so, a 7-mer. Or, using 133 as a rest stop, do it as 3 X 13, + 3, which is 39 + 3 = 42. I think our way beats his, although I've not analyzed just why it works.
@@davidcovington901 I believe it works because 1 is 1 more than a multiple of 7 while 10 is 3 away, so to compensate for dividing by 10 you multiply by 3. My Geometry teacher in high school told us that some POW had figured out a way but didn't tell us what it was. I guess I subconsciously kept working on the problem until I figured it, or rather apparently one, way.
There is a more smart way to see if it is divisible by 4. In three steps. 1: we look only in the last two digits. 2. It's even. 3: if the units are divisible by 4, which are 0,4,8 then the tens must be even, if it's not divisible by 4 which are 2,6 then the tens must be odds. So, for example with units as a 2, tens must be odds. 12,32,52,72,92
Very useful video. Lately I’ve been wanting to check Primality in my head, and this helps make it a lot easier. I knew the tricks for 2,3,&5, and memorized that 49, 77, and 91 were multiples of 7 to be able to do it for all two digit numbers. Knowing the tricks for 7 and 11 let me check all the way up to 169, so that’s pretty cool.
There's a maths based "latin square" puzzle that I love, and I'm usually trying to figure out if a given number is divisible by 7 somewhere in the puzzle. I'll try this method. Usually I'd break it up as 420 + 14, or 6300 + 140 + 28. For reference, the puzzle is called "Keen" and is on "Simon Tatham's Portable Puzzle Collection".
I've been waiting for this for 49 years actually. How fitting LOL. I use it for checking numbers for primailty in my head. Brilliant! Thank you! ( I did know all of the others already actually).
Knowing divisibility checks is really useful for prime factorizations too...7 and 11 in particular are terrific for this since they're on the bigger end of the early primes.
6468 is the polynomial 6x^3 + 4x^2 + 6x + 8 evaluated at x = 10. To check divisibility by 7, evaluate it instead at x = 3 by synthetic division, casting out 7’s along the way. 6*3+4 = 22 ~ 1 1*3+6 = 9 ~ 2 2*3+8 = 14 is divisible by 7. For 11, the alternating sum is equivalent to synthetically dividing by -1. For 4, synthetically divide the last two digits by 2. For 8, synthetically divide the last three digits by 2.
I actually have a secret method for determining whether a number is divisible by 7.. I just ask myself.. Is it 14? If so, then yes. Otherwise, I’m outta luck
I always felt the subtraction rule was a bit clumsy and I was faster at division (or finding the remainder, i should say) than doing that test. The addition rule is easier to do!
To check for a multiple of 8 i have an easier method (for me at least): I look at the last 3 digits, if the first is even then the last 2 must form a multiple of 8. if the first is odd, then the last 2 can't form a multiple of 8 but must form a multiple of 4. For example, 264 is a multiple of 8 because 2 is even and 64 is a multiple of 8. Another example, 128 is a multiple of 8 because 1 is odd and 28 is a multiple of 4 (but not of 8) I find this much easier than halving 3 times in a row. Checking whether the last 2 digits are a multiple of 8 is easy as long as you know your 8s table, checking whether they are multiple of 4 is a bit trickier, but you only have to do it half of the times anyways.
For 2-digit multiples of 4 there is a similar "trick": if the 10s digit is even, then the 1s digit should be one of 0,4,8 if the 10s digit is odd, then the 1s digit should be one of 2,6
The same trick "scaled down" works to find multiples of 4 too. If the 2nd digit is even, the last one must be 0, 4, or 8. If the 2nd digit is odd, the last one must be 2 or 6.
@@nekogod Another similar one is: take the last three digits multiply first by 4, second by 2, third by 1 and sum, check if divisible by 8, i.e. 264 = 2*4 + 6*2 + 4 = 24 The advantage is that it gives you a smaller number.
My favourite way to tackle divisibility by 7 is to exploit the fact that 1001 = 7 x 11 x 13. It's relatively easy to cast out lots of 1001 (though you can occasionally get some nasty carries) and it's guaranteed to bring you down to a number below 1001. Even though this might require a bit more memorisation (or mental arithmetic; just doing long division with numbers that small is surprisingly manageable, especially with a bit of practice/experience) from this point onwards than most of the methods discussed here, it does have the huge advantage of being able to attack 7, 11 and 13 simultaneously!
A similar process (as I’m sure others worked out) is to divide the number by 50, then add the whole number portion of the quotient to the remainder. So for 434 it would be 8 + 34 = 42, which is divisible by 7. This works because 50 - 1 = 49, as related to his explanation of how his method works. For very large numbers, you can repeat this process as needed 😸
I think the easier and more straight forward method would just be subtracting multiples of seven since it cannot change the remainder. This method is solid and works for any number, but for some small numbers there are better ways. 6468 - 6300 168 - 140 28 ✓ A nice bonus is that we get can easily perform the division right now, since 6468 = 6300 + 140 + 28 = 7 (900+20+4) = 7*924
Literally just division by hand as we learned in primary school, but with 0's instead of spaces. I wish we'd have learned why that worked back then. I think it might have helped quite a fair few to better understand, or might at least somewhat dispel the myth that maths is just magic.
I just want to say I haven’t been on TH-cam in a long time and rediscovering this account is helping bring back my love for math. Covid stopped me from loving school and learning and I lost my love for math and as a freshman in college taking calculus it’s been really hard. Thank you guys for always being so enthusiastic about math even if you never see this comment.
I worked out my own test for 7: split off the RH digit, multiply the rest by 3 and add the RH digit. The reason is as follows: 10x +y can be re-written as 7x+ ( 3x +y) . Now as 7 divides 7x, to divide 10x+y, it needs to divide 3x +y. A similar trick can be done for 13, but this time the test is (RH digit ) MINUS ( 3 times rest) ( ignore the minus sign if the result is negative). Edit: I've just worked it out for 17: similar to 13, but the test is (RH digit) minus (7 times rest). Didn't bother with 14, 15 or 16 as they are simply combinations of 2 tests ( respectively 2 and 7, 3 and 5, 2 and 8 ( or 4 twice)).
A game I've been playing since I was a kid is breaking down numbers I come across, like street addresses or telephone numbers, to find it's largest prime. I "win" if a number's largest divisor is less than 12. I've always known every other trick mentioned in this video but seven has ALWAYS been the bane of this game as I didn't have an easy method to quickly divide by it. Or at least, not until now. So glad to have learned this, this trick is going to speed up my games considerably!
Actually for the divisibility by 8, you consider the last 3 digits, and then only the last 2 digits, because second step is testing divisibility by 4. Nice video.
In school we were taught to subtract two times the last digit from the rest which is equivalent to this method but had the advantage that the resulting number is less than when you add five times the last digit, though subtracting-usually-tends to be a bit harder to perform without a pen and paper or using a calculator.
I like the double last and subtract because it can find out if the number is a multiple of 3 at the same time. May not be the most efficient a lot of the time, but it's nice seeing the subtraction of 21b rather than adding 49b.
It’s a lot easier to subtract the multiple of 7 that has the same last digit as the number we’re examining: 434 = 420 + 14 Remove the 0 from the first number in the sum. 42 is divisible by 7. It even gives you the quotient: 434/7 = (420/7+14/7) =60+2 =62
I sent a video to you guys a long time ago where I explore the significance of dividing by 7, and thought this may have been your response 😅 but my video was pointing out the 142857 loop of decimal points. I love all of these neat patterns that lie in mathematics !
I was hoping the video would be about that! I discovered that pattern when in school - how doubling 142857 shifts to 285714 and doubling again shifts to 571428.
@@thomasreichert2804 yess such a cool pattern! I was trying to find this pattern with other numbers too, but it only seemed to exist if the divisor was a multiple of 7, such as 1/14 or 1/28. I looked at the pattern as treating the numerator as the index of the decimal number from low to high, then the pattern continues on. For example 3/7, the 3rd smallest number from the 142857 pattern is 4, hence 3/7 = 0.428571 and so on.
@@mattcoulter7 yeh, it's great, you can work out any integer divided by 7 to as many decimal places as you want by just remembering that simple sequence
Oh, the difficulty of 7. I suffered a concussion in a bike accident in 1985. When I got to the hospital, the doctor asked me to count backwards from 100 by 7s as one of the mental acuity tests. My immediate reaction was something like "nobody can do that." I gather they look for the patient's reaction, the effort, and some relation to consistency; they know that nearly all patients will make mistakes. I did pass that portion of the tests and they released me about 2 hours after admission. The high point of the day was that math quiz. Really - it was either a truly bad day or I'm a nerd. Pick one or two.
Counting backwards in sevens is definitely quite traumatic. While checking if you were clear of a concussion injury, I think the doctors were trying to cause PTSD!
Divisibility of any number "a_0...a_n b" by 7 is by far easier when you consider the difference between 2*b and 7 if a_n is smaller than 2b for example: 364 ---> 2*4=8>6 then consider 2*8-7=1 gives 36-1 is divisibe by 7 266 ---> 2*6=12 too big then consider 14-12=-2 gives 26-(-2)=28 divisibe by 7 etc ...
Thank you so much for this ive needed this for so long 7 was killing me. Id figured out 2 to 11 on my own but 7 i had no idea this makes things so easy now and raises an interesting idea that im gonna test for 13
I've always known the rule about taking the last digit, multiplying it by 2, then subtracting the smaller number from the larger number. ... the x5 rule is new to me.
I use something that can also be used for 11* and 13. Add up every other trio of digits, both groups and subtract the totals. This is because 7*11*13 is 1,001. *: With 11, there's an even simpler rule: add up every other digit, both groups, and then subtract the totals.
Adding up every other digit into two groups and subtracting the totals is effectively the same as alternating addition and subtraction. Number: ABCDEF A - B + C - D + E - F = A + C + E - B - D - F = (A + C + E) - (B + D + F) Alternating will just keep the progressive total smaller.
@@SirRebrl true, and I could have added that to my original comment, but what I said works for all 3 can tell you, for instance, that 1,540 is divisible by both 7 and 11 (and therefore 77) because 540-1 = 539, the prime factorization of which is 7²*11.
Back in the days when applications started to be locked by a serial number it wasn't that uncommon that one of the checks was "divide by 7". Many times you needed to present a number say 12 digits long which seems impossible to guess but all that was needed was to look for one that could be divided by 7 (11 was another common).
As far as I know, the simplest is separate the last digit, multiply by 2, and reduce the rest. 343 => 34 | 3 , 3 x 2 = 6 => 34 - 6 = 28 (divisible by 7). But, it takes longer when reach more than 5 digits. And yes, this is for base-10 numbering system.
For divisibility by 8: 1. Look at the hundreds place 2. If it's even, the last 2 digits need to be divisible by 8 3. If it's odd, the last 2 digits need to be divisible by 4, but not 8
Nice! Another thing you could do is 4 × Hundreds Place + 2 × Tens Place + Ones Place :) For example: 657,392 becomes 4 × 3 + 2 × 9 + 2 = 32 and since that is divisible by 8, so is the original number. This may not be easier, but the advantage is that you only have to deal with numbers up to 63, so most people wouldn't need to reiterate the test to find out their answer By the way, the reason why it works is because 100 is 4 more than a multiple of 8, and 10 is 2 more than 8 itself :)
In fact, for divisibility by 8 you don't need "the last three digits" but just "modulo 200", and for divisibility by 4 you don't need "the last two digits" but just "modulo 20". e.g. is XX68 divisible by 4? I don't know, but I can ignore XX60 so yes 8 is divisible by 4 Is YYY918 divisible by 8? I don't know but I can ignore YYY800, keep 118 and, well, 40-80-120-160 are easily divisible by 8, so 118 is not
An easier divisible by 8 test: Take the last three digits. If the hundreds digit is even, check if the last two digits are divisible by 8. If the hundreds digit is odd, check if the last two digits are divisible by 4 but not by 8. e.g. 6468 is not divisible by 8 because 68 is not; whereas 6568 is divisible by 8 because 68 is divisible by 4 but not by 8
useful for coding. in my math class we were doing some coding to assist in math proofs, more like proving something then making code to do that thing. being able to code to find if something is divisible by 7 is quite useful, but this method could be used for other numbers as well like 11, 13, 17, etc. trying to find if something is divisible by some uncommonly used prime number
Very interesting, and curious I have never heard of this before! It works, because 7*7 is one less than 5*10. Therefore it should work also as a trick for 13! 13*3 is one less than 4*10, so use the same trick with last digit times 4 to find if a number is divisible by 13. Same logic, same proof
I thought of the same too! I think it works for something like 17 then 23 or any number ending with 7 or 3 but the number which is to be multipled to last digit( like 4 for 13 and 5 for 7) will increase
19 is intersting: 19=20×10-1 so use twice last digit! All numbers either enter the 18 number loop of the numbers 1-18 if indivisible, or 19 if divisible.
10/p will give you a decimal for all primes. I don't know how that mod 7,11,13 etc will get you an integer. I think 1=10x mod p where x is the number to put in place of 5 may work. With p=11 I got x=10 and p=13 I got x=4. Those seem to work.
@@jasonwinters7560 Sorry, you're right, I've updated my comment to say 1/10 (which indicates the solution to 10x = 1 (mod p)) instead of 10/p, not sure why I made that error
Simplification for 8 is to calculate Last + 2*SecondFromLast + 4*ThirdFromLast and check if the result is divisible by 8, which is slightly faster than dividing by 2 thrice and matches with the other digitsum methods
I prefer using an equivalent test which is: if n=10m+r where r is a digit and m an integer, then 7|n iff 7|m-2*r (obviously this is the same as the one mentioned in the video, but it's better when n is non-negative and 7|n iff 7|(-n), so in case of a negative number I'd just negate the sign)
its way more easy if you just split your number in modulo groups. for example 6468 -> 64%7=1; then use the 1 as the next leading number before the next 6: 16%7 = 2; repeat the step: 28%7 = 0 . so if the last modulo is zero the whole number is divisible by 7. with this you dont need the addition and it works pretty well recursivly
I’ve been using those divisibility rules to gradually try to prove the collatz conjecture by induction. Basically, if I can prove that every number eventually goes to a number less than itself, regardless of how many steps, then every number goes to 1. Assuming that is true; All integers N eventually reach a number x such that x
I think the key to proving the collatz conjecture lies in the powers of 2. They're the only path to 1 available. I also think the key is to think of it as "can the even numbers 1 greater than a multiple of 3 eventually reach an odd number n that will satisfy '3n+1=2^x'"
I consider N = 2^n as the end of the series since it drops to 1 very quickly (exponentially). Note that N = 4^n - 1 is always an odd multiple of 3 [ say 3 M = 3 (2m+1) ]. This is provable by induction. As a result, there are infinite odd numbers N such that 3N+1 = 4^n which is 2^(2n) which is effectively another end of series. This addresses another subset of your last 25% odd positive integers. Perhaps suggestions from others can being your last 25% down to at least 15%.
11 is weird too, because substracting the last digit from the rest of the number and checking if the result is divisible by 11, will show if the number is divisible by 11.
@@divabhardwaj6381 Yeah, e.g. 143 is divisible by 11, because 14-3=11. Any divisibility rule can be found, when you multiple a number so that the last digit of the product is 1. For example, the rule for dividing by seven is found when 3*7=21.
i'm seing a pattern: 2 - last digit 4 - last 2 digits 8 - last 3 digits 16 - last 4 digits maybe? that would be a nice formula for: divisibility by n² can be checked by the last n digits.
@@MxIbatomik All numbers, even decimal numbers, can be divided by seven. However only whole numbers can be divided by seven such that there is no remainder.
When I was like 10yo I figured out by myself a very similar method to verify if a number is divisible by 7 and I was so proud of it: In case of 434, multiply by 3 every digit except for the last one and add it to the result (43x3)+4=133; Then again (13x3)+3=42 that is divisible by 7 so 343 is divisible by 7
2:59 What's more interesting is that any number that's divisible by 49, after doing enough times of this method will always reach 49. But others will always reach 7.
That's not exactly true. Consider 343, which is 49*7. Using the addition method (5*3+4+3), you get 22; using the subtraction method (2*3-4-3), you get -1; adding the results of both tests, you get 21.
Divisible by 7 can be useful in figuring out if there is a whole number of weeks in a number of days.
I came here to make that exact point :)
Could you demonstrate that please, thank you. :)
@@Reachermordacai "hey dude I see on sundays, I gotta work the next 2794 days!" "Damn so you stop work in the middle of a week?" "How did you figure that one out?"
Yeah but that isn't very useful at all unfortunately. We use days and months most of the time
@@YounesLayachi But we also use weekdays and weekends.
this feels like a very old-school numberphile video, love it
Yeaah! It really does! fun!
When I saw this video on TH-cam at a first glance, I thought I was looking at some old video from seven years ago or something, was surprised when I realized it was from today
James Grimes is classic!
What's changed
I literally started watching and was like "have I seen this already?"
For 40+ years I've been saying, "there is no rule for 7," meaning no way to check for divisibility as with 3, 5, 9 etc. Thank you for this. I now have my "rule for 7!"
My rule was doubling the last digit and subtracting it from the rest of the number.
yo man, will there be ant man 4?
my rule was imagine that number was base 3 and convert to denary so 343 3 + 4x3 + 3x9 = 42 divisible
@@darpanjain4250 i used the same. If we try to prove the method, the process remains same for both methods.
V0
Can't believe that 10 years later, I am still loving watch James Grime on Numberphile. I watched him when I was a nerdy high schooler and now I'm a nerdy adult. Thank you so much for all the videos over the years.
Wow I’m in the same boat exactly. Can’t believe it’s been 10 years
Exactly the same!! Finally getting myself to watch numberphile again after all these years. And it's just as brilliant.
I swear he hasn't aged a day in that time
It's awesome to watch him explain anything about math.
That's one of the great things about youtube that is not talked about enough. There are so many creators I have watched for a decade or longer and they were so important to me and continue to be so.
In the media too much attention is paid to the drama and such. What about people like this? Or the green brothers? Vsauce? Rhett and link? I could go on. It's not an easy thing to do, to be able to take your fans with you as they pass into different eras of their lives. Most adults don't still watch what they watched as kids but so many of us do with creators like I mentioned above.
These people are my teachers and actually helped guide me to being a better adult. Me and so many others. It's a beautiful thing.
The creators, as well as fellow followers, feel like family and it's such a comfort.
When I was in the 7th grade, I was taught all the tests for divisibility except for 7. What we were told was "try dividing it by 7," which completely defeats the purpose of a divisibility test. Thank you for filling in this particular gap in my education.
the test defeats the purpose of the test because it takes 5 times longer than just dividing by 7.
@@BeBopScraBoo But it doesn’t test your mental arithmetic so much (pardon the pun).
@@BeBopScraBoo this is exactly what i was thinking 😂😂
@@BeBopScraBoo Yes, and the video is also completely missing the stated topic. I wanted to know why non-roundly dividing by 7 always creates the same decimal pattern.
@@Dowlphin decimal system is based on ten, which factors to 2 and 5. any fraction with a denominator that factors with any other number will have a repeating pattern.
The same algorithm (3:51) can be used to construct a formula for other primes:
11: x - y (k=10, j=99)
13: x + 4y or x - 9y (k=4, j=39)
17: x - 5y (k=12, j=119)
19: x + 2y (k=2, j=19)
23: x + 7y (k=7, j=69)
29: x + 3y (k=3, j=29)
31: x - 3y (k=28, j=279)
37: x - 11y (k=26, j=259)
41: x - 4y (k=37, j=369)
43: x + 13y or x - 30y (k=13, j=129)
47: x - 14y (k=33, j=329)
53: x + 16y (k=16, j=159)
59: x + 6y (k=6, j=59)
61: x - 6y (k=55, j=549)
67: x - 20y (k=47, j=469)
71: x - 7y (k=64, j=639)
73: x + 22y (k=22, j=219)
79: x + 8y (k=8, j=79)
83: x + 25y (k=25, j=249)
89: x + 9y (k=9, j=89)
97: x - 29y or x - 30x + x (k=68, j=679)
where k is the multiplier of 10x + y: 10kx + ky (4:11)
and j is a multiple of the prime that is subtracted: 10kx - jx + ky = x + ky (4:30)
then subtract the prime if it helps
And since the same algorithm is used, the results have the same properties, like iterability (1:24)
For the ones with a y coefficient greater than 10, we could split the number like so: 100a + b; where b is the last 2 digits and a is the rest:
37: a +10b (k=10, j=999)
43: a - 3b (k=6, j=559; originally 6b - 2a which is even and backwards, so divide by -2)
47: a + 8b (k=8, j=799)
53: a - 9b (k=18, j=1749; originally 18b - 2a -> divide by -2)
67: a - 2b (k=12, j=1139; orig. 12b - 6a -> div. -6)
Couldn't find a decent formula for 73, 83, and 97 in a timely manner. I might revisit this later
Isn't iterability always necessarily given by the nature of these kinds of divisibility proofs?
@@leo848 I suppose, but for 2, 5 and therefore 10, for example, you don't need to iterate since you're just looking at the last digit (is it even?, is it 5 or 0?, and is it 0? respectively). For the alternating sum for 11, I have not taken the time to reverse engineer its algebraic derivation; it too may have the same property because it may be a similar algorithm.
@@leo848 not really. You can construct a finite state automaton (for instance, a 3-state automaton for divisibility by 3) that, after you run the digits through its arrows will straightaway give you a ‘yes’ or a ‘no’ without any need for iteration. I guess, we're just more used to addition and stuff, and devise these chevksum kind of tests, which fall into the trap of iteration.
@@nnaammuuss Of course. I meant that once you have a divisibility test that gives back a number that is divisible iff the original number is, one can always iterate.
Me and a friend of mine discovered a trick for 11 in middle school:
Take 121, u split it in 1+21 = 22 so if the result is divisible by 11 the original is too.
It works even for larger numbers like
35673 split in 3+56+73 = 132
132 split in 1+32 = 33
In case of even digits
1078 split in 10+78 = 88
Basically you split the number in sets of two digits starting from the end.
:o that's given in my book
What about the magic 11s in pascal's triangle. Your algorithm reminded me of that.
I'd like to see a proof of this. This looks like coincidence so far to me.
@@michaelempeigne3519 no idea what the proof is but it worked with every number i tried
@@frasco_5518 Just because it works with numbers does not mean that it is true in general. This is the difference between math and science.
Seven is weird because it eight nine
🤓👍🏻
Most underrated comment ive seen
That's not weird, eating 3 square meals.
Darn it. I came here to say that.
This is colder than 7 degrees Fahrenheit
One of the great things about youtube that is not talked about enough. There are so many creators I have watched for a decade or longer and they were so important to me and continue to be so.
In the media too much attention is paid to the drama and such. What about people like this? Or the green brothers? Vsauce? Rhett and link? I could go on. It's not an easy thing to do, to be able to take your fans with you as they pass into different eras of their lives. Most adults don't still watch what they watched as kids but so many of us do with creators like I mentioned above.
These people are my teachers and actually helped guide me to being a better adult. Me and so many others. It's a beautiful thing.
The creators, as well as fellow followers, feel like family and it's such a comfort.
I use divisibility for finding primes and root approximations during tutoring. I lacked the 7 test so, thanks for that!
I usually find the closest number divisible by 70 then subtract/add from that. It’s not really a trick for 7 but it’ll always work.
7 has a lot of ways for divisibility determination
One trick that might help for finding primes is that except for 2 and 3 all primes follow the pattern of a multiple of 6 plus or minus 1. This is true because +2 would be divisible by 2, +3 by 3, +4 by 2, and +5 is also -1. Unfortunately I discovered this long after I needed to write Basic computer programs to find primes in school.
Because of this video, I found out about the 1001 test, alternate sum of 3 digits:
123456788 is divisible by 7 because 123-456+788=455 is divisible by 7!
Best part is that this test works with 11 and 13... 455 is also divisible by 13 so 123456788 is divisible by 13.
This is golden when testing for primes!!
@@thomaswilliams2273 it will catch all the primes but you also have to filter out the multiples of 5’s.
This video took me back in time, when I was a middle-schooler amazed by these tricks and properties of numbers.
Even now in uni, I watch videos on this channel with the same flabbergasted look, enjoying every minute as an extraordinary discover.
I really appreciate your content, you're doing great ✌️
Like using the missing fingers to do your nine times tables.
I love the fact that everyone keeps saying this video feels like it was from years ago but no one is commenting how Dr. Grimes didn't age that much. He's a vampire who knows all the secrets of numbers
It's the hidden beauty of numbers, and I love when I learn a new one.
@@ChrisM-qo1jc The Count! 5 fingers, 4 fingers, bwahaha...! 💚✌️😎
I always found multiples of 100, subtracted them and added on double the number of 100s you removed to what is remaining. Taking advantage of 98 being a multiple of 7. This works pretty well and is easy to do in your head
This sounds much easier than what was shown in the video.
can you elaborate step by step please
another way to work out if something is divisible by 7 is to just do all your math in base 7 and see if it ends in 0
Yeah, I think its easer way
The given method is pointless, much easier just to divide by 7.
I just subtract 7 over and over again. To check whether the number 70,000 was divisible by 7, I had to subtract 7 10,000 times. Turns out it is!
@@EebstertheGreat So easy, Eebster10010100.
@@christopherellis2663 it was joke :D
I seriously love watching Numberphile videos and you are my absolute favourite, James Grime Sir! I adore you James Grime Sir to a great extent. (Your smile is literally the best!) Thanks for making me love mathematics more than ever.
I’d always do it by comparing to other numbers. 7000 is divisible by 7, 7000-6468=532. If 532 is divisible by 7 then 6468 was. 700-532=168. 168-140=28. 28 is divisible by 7 so 6468 is. This method is much easier for me to do in my head because it’s only addition or subtraction.
6468=6300+168 =900*7+24*7=924×7
the method in the video can be programmed into a computer
@bornts8944
Use the modulo operator for computer and compare the result to zero.
Works for any number.
Do this - 5194
@CarlFriedrichGauss1 easy 5194-4900=294, 294-280=14. 14=7*2. Therefore 5194 can be divided by 7. All you have to do is use is tens, hundreds, thousands of the times table.
When I was 13, I found another method for the divisibility by 7. It's in a way even more useful than the one in the video. The trick is to cut the number at the second digit. For instance, for 343, it's 3 and 43. After that, you double the rest (in the example, from 3, you get 6). If the sum is divisible by 7, then the original number is divisible by 7: 6+43 = 49. So, 343 is divisible by 7.
What this method also provides is that it preserves modulo. So, if the number is not divisible by 7, the method also tells you how much is the remainder mod 7. For example, taking 716, you get 2×7 + 16 = 30, which is 28 + 2. So, 716 gives 2 as a remainder after dividing by 7.
100x + y
-98x (multiple of 7)
= 2x + y
Nice, this was actually the answer i expected from the video. For smaller numbers (under 6 digits) i think your solution is easier. For larger numbers i think the videos solution is better as each step requires multiplying 1digit by 5 and reduces the test number by. 1 digit, the mod 100 version each step requires multiplying an n-2 digit number by 2, it potentially reduces the number of digits by 2 but 3/10 operations will only reduce it by 1. Regarding finding the remainder the video requies 1 extra step, each iteration of the algorithm multiplies the remainder by 5, if you keep track of the number of iterations mod 6 you can multiply your final remainder by (1,3,2,6,4,5) mod 7.
Cool!
I think the method shown in the video also preserves mod.
@@joseville the method in the video multiplies the remainder by 5 for each iteration of the algorithm.
This is very cool! I "discovered" the same method while staring at the digital clock and splitting the number at the colon when I was little. Thought I was one of the great mathematicians of the century.
The most amazing thing about this video? James said "this weird trick" in the video, but that phrase doesn't appear in the title. Well done!
Mathematicians -hate- _love_ it!
The twinkle of mischief in his eyes was a nice touch too!
If it appeared in the title, I'd have passed on it. Everyone knows that "This one weird Trick!" is web-speak for clickbait! 🙂
I kept looking for the CONTINUE button
This weird trick
will allow you to anger people
in the comment section
*Read more*
This is such a great video.. i’ve shared with my kids and the “presentation” with special british commentary and voice intonations is perfect. Useful and just enough entertainment!
I poked around in excel to find out what happens with that seven trick in how other numbers terminate. 49 is in a loop of 49's as we saw in the video and all numbers 1-48 that aren't divisible by 7 are in a loop together and all of the numbers divisible by seven are in their own loop.
1,5,25,27,37,38,43,19,46,34,23,17,36,33,18,41,9,45,29,47,39,48,44,24,22,12,11,6,30,3,15,26,32,13,16,31,8,40,4,20,2,10
and
7,35,28,42,14,21
that's really odd!
Cool! I've tried something like that as well, and it seems that powers of 7 all eventually reduce down to 49 as well:
7^2 = 49 --> 49
7^3 = 343 --> 49
7^4 = 2401 --> 245 --> 49
7^5 = 16807 --> 1715 --> 196 --> 49
And so on.
Loops like this remind me of Goldbach’s conjecture… nice result!
One gets the next term in those sequences by multiplying by 5 modulo 49.
@@antonmiserez934 I think you mean the Collatz-conjecture
"Is it ever important to know if a number is divisible by seven?"
Peak question right there.
Just seeing Dr. Grime already tells me this is going to be sweet. Loved it.
Dr Grime looks so much older than the old days of the channel. Now I feel old 💀
@@iwatchwithnoads7480 And somehow he got even better at explaining
I just instantly see either 350+84 or 420+14 which makes it obviously divisible by 7. But I guess the point of the video is to showcase some more interesting methods. nice job.thanks.
I saw 700 - 14 for 686 :D
I think is more for students who struggle to immediately see that and how they can be supported :)
@@Matthew-bu7fg I do have to say that this method is indeed great, the examples could have been chosen better imo. But nevertheless great method.
There's another similar method which I use if the divisibility test is just too hard:
Keep adding or subtracting the number which you want to know divisibility by to your number until it's a multiple of 10. If, say, your number which you want to know divisibility by is even, and your number is odd, there's no way of getting to a multiple of 10, so it isn't divisible. Divide your 10 multiple by 10 and try to see if that's divisible, if not, repeat the process until you see something you recognize.
Example: divisibility by 13 for 832
832(i picked this number completely at random no joke)+13=845
845+(13*5)=845+65=910
910/10=91
91=13*7, but lets say i'm still unsure
91+13=104
104+13=117
117+13=130
No doubt about it, 130 is divisible by 13. 832 must be divisible by 13
I like your thoughts.
Even Numberhile now recognizes the elegance of the number 7
Thala for a reason
7 is the GOAT of multiple things
@@يوسفعمارنة-ش9ت HECK YEAH RONALDO SIUUUUU
I have a new trick.
numberhile
I had learnt divisibility tests in school. But we hadn't learnt any for 7. I genuinely though it wasn't possible, (thinking they would've taught it if it was that way)
But that 10x + y explanation completely blew me away. It's these very basic simplistic things in math which amaze you. Because it is so simple and 100% sure it was possible to come up with myself.
Really encourages one to keep trying out stuff and working out things just for fun!
if i was a teacher i'd refuse to teach the 'trick' because it's just friggin faster to divide by 7.
@@BeBopScraBoo You would be a crappy teacher. The WHOLE point of Mathematics is to recognize (and prove) beautiful patterns of set theory which can be inspiring for curiosity.
I.e. The proof of Div by 7 is _Number Theory,_ specifically using *cyclic addition* which differs from *linear addition.* This _difference of perspective_ can inspire students to want to learn more.
Sadly I never learnt the Div by 7 mod trick in school either. When I was 30 I heard about the _subtract double the last digit from the remaining digits_ but didn’t know _why_ it worked so shortly after I set out to prove it.
I independently derived some circular addition rules (modular arithmetic) and from that it was relatively straightforward:
10x + y ≡ 0 (mod 7)
50x + 5y ≡ 0*5 (mod 7)
49x + x + 5y ≡ 0 (mod 7)
[49x ≡ 0 (mod 7)] + [x + 5y ≡ 0 (mod 7)]
0 + x + 5y ≡ 0 (mod 7)
x + 5y - 7y ≡ 0 - 7y (mod 7)
x - 2y ≡ 0 (mod 7)
QED
The sad part is kids do modular arithmetic (circular addition) when they learn how to tell analog time but no one every tells them they are doing Number Theory!
@@MichaelPohoreski That's not the point of school though. The purpose of school is to learn skills you can apply in real life. The point of divisibility tests is that you can do them in your head without writing anything down. If you have to write it down, it's pointless and you can just divide in the first place. Yes, it's interesting how numbers converge into patterns, but there's not point worrying about number theory and patterns in elementary school
An alternative method (that I worked out for myself right now) you can do is *last 2 digits plus twice the rest* . Will shrink the number much faster, and the mental maths is barely harder. Proof is exactly the same as in the video. It's actually surprisingly easy to generate any number of these tricks.
I guess the hard part, for some people, would be to double the rest if its a big number, at least mentally.
Nice find, way easier to remember than the one Grime suggested
@@leoirias3506 true, but if you're dealing with 3-4 digit number, you've only got to double 1-2 digits. Not very hard to do mentally I'd say? If you've got a much larger number... You won't. Nobody needs to know if a 5+ digit number is divisible by 7 in their head.
@@QuantumHistorian For a larger number, you can split it into 2 digit numbers starting from the end, and then multiply by increasing powers of 2. Eg: 12936 -> 36+29*2+1*4 -> 98 -> divisible by 7
To make it simpler, you can do a mod 7 for each of the individual elements:
36 -> 35+1->1
29*2->(28+1)*2->2
1*4->4
1+2+4=>7
@@rohitraghunathanIngenious... thank you!!
This video is the basis for my favorite math subject: prime factorization. It's breaking something down to it's most elementary components lets you see how many ways you can manipulate and use a number.
I remember I used a USB stick duplicator at work once, with 32 slots - 1 for the master USB and 31 for the clones. It also had a display that counted the total of successful duplicates, and checking if the number of successfully made duplicates was divisible by 31 was a quick way to tell if the machine was working properly. So, I used the similar divisibility test for 31, as it felt simpler once I got used to the method. "Subtract 3 times the last digit from the rest".
Hmmm I think I get your trick: for 31, you could do -3*(10x+y) = -30x-3y = -31x - (x - 3y). So indeed the original number (10x+y) it can be divided by 31 iff x-3y is divisible by 31.
It's a bit of fiddling to find the right factor (in this case -3), or at least it's not obvious to me immediately, but it's not too hard to derive these tricks for other numbers actually! For example I found 13 too: 4(10x+y)=39x+(x+4y) so you can check x+4y (because 39=13*3).
alternative to thisss ssubtract 3 times method you indicate, you can also do " rest + 28 times the unit digit"
@@michaelempeigne3519 that's true! I just thought 3 times a number was a bit easier than 28 times 😀
@@ikbintom Well, you thought wrong. It’s a lot easier to multiply by 28 then it is by 3, OBVIOUSLY.
@@ikbintom Finding those fiddling factors comes in a couple ways if you have nice numbers: if you're looking for something divisible by XY, where X and Y are digits, then for a number ABC... you can multiply the last digit (ex: C) by X/Y, and subtract from the rest. Why? Every time you add a multiple of XY, you add Y to the last digit, and X/Y to the other digits. If you multiply Y by the ratio, and subtract that, you've essentially said 'ok, great, let's remove Y copies of XY from ABC, guaranteeing the last digit is now 0. Is the remainder left 10x a multiple we're aware of?. And if you repeat the process, you're just now doing the same operation, but with the 10s and 100s places. The second nice way is finding a convenient multiple near 100. For 7, 98 works great. Since that's 2 away from 100, we can just chop off all but the last 2 digits (would be 3 digits for a multiple near 1000), and add on 2 for every hundred we cut off - 343? 3*2+43 = 49. For 31, 93 is kinda close to 100, but 7 off. So let's try it : 568, 35+68=103, not a multiple. In fact, since it's 10 more than a multiple (103-10=93), you know 558 would be a multiple of 31, which it is (18x31).
The general divisibility trick for any prime p (other than 2 and 5) is as follows:
Let our dividend be k and express it as 10x + y (x , y are non-negative integers)
Find an integer z such that 10z when divided by p yields remainder 1.
Now if x + (z)y is divisible by p, then 10x + y is divisible by p. This yields the required algorithm.
Note: You can replace 10 with any natural number as long as it isn't a multiple of our chosen prime p.
I usually do it with a multiple of p that ends in 1. Call the multiple 10m+1. Then if x-my is divisible by p, then 10x+y is also divisible by p.
For example, 17x3=51, so you can take off the last digit, multiply it by 5, and subtract that from the rest of the number. Repeat until you get a number that obviously is or isn't a multiple of 17. If it is, then the original number was divisible by 17, and if not, it wasn't.
Nice, yeah that yields the trick(s) for 11 of either adding 10N or subtracting N (for trailing digit N). And for 19 and 21 you get tricks involving ±2N.
This video actually changed my life. I was really ready to end it all, had a goodbye note written, had picked out which bridge I was going to jump off, and then I saw this video and now I feel the strength to carry on. Thank you, sir. Seven truly is amazing.
You see what happens now, when greatness is demanded of you?
Honor your time wisely, henceforth. You now harness the powerful knowledge of dividing by 7!
I’m not sure if you’re being serious or not, my friend.
@name i guess ok
Here's a trick that only requires you to know up to 10x7: take your number (let's use his example of 6468), and find the multiple of 7 that ends in the same digit. So, 6468 ends in 8, and 28 ends in 8, so subtract 28 and you have 6440. Drop the last 0, so you have 644. Repeat. 644 ends in 4, 14 ends in 4, 644-14=630, drop the 0, and you have 63, which is divisible by 7, so it's all divisible by 7. It's similar, but you don't have to deal with the 'multiply by 5' step, so it might be a bit mentally easier.
That is about the same as dividing by 7 normally, except you're going right-to-left instead of left-to-right.
However, your method is what I would do to check divisibility by many larger prime numbers like 17 or 23, except that I may add or subtract and attack the number from both the left and right.
For example, to test if 3893 is divisible by 17, I would find it easier to add 17 rather than subtract 153.
3893 + 17 = 3910 or 391 after dropping the 0.
At that point, I would notice:
391 = 17 * 23 = (20 - 3)(20 + 3) = 20² - 3²
However, I could have could have continued the process by subtracting 51 (or adding 119) to 391 to get 340 (or 510).
As to how often this is necessary in day-to-day life, well...
I prefer this method as well, it's a 7 based multiplication + subtraction vs a 5 based multiplication + addition. So end of the day it's pretty much the same
I have an easier approach for this one. Take 6468 for example. Consider first two digits i.e. 64. 7*9=63. So 7*900= 6300. 6468-6300= 168. Now consider first two digits again i.e. 16. 7*2=14. 7*20=140. 168-140= 28. 28 is divisible by 7 so number is divisible by 7. Fun fact - this is nothing but doing elementary division by 7 in a more complicated way. The method given in the video and and in the comment section by people are are more complex than simply dividing by 7 and check. It's fun math but not practical at all
Isn't it amazing that the sponsor is placed at the _end_ ? I can't believe Numberphile is the only channel I've seen that doesn't intrude on its own content!
shitty advertisers (like Raid Shadow Legends and other mobile games, most VPN, etc.) force you to have the ad at the starts + link in pinned comment + link in description, while Brilliant, Kiwico, etc. don't.
Yet another reason to love Numberphile
Most educational channels (all of the SciShow channels, PBS channels, Stand-up Maths, etc) put the ad at the end if they have one. If you can see the viewership graph on the seek bar, viewer retention usually flatlines on those videos as soon as the ad starts.
I know old Jacksfilms videos and Oddheader do that as well
Probably because Numberphile has more leverage in the situation than your average independent youtuber who may be more in need of the sponsorship money (and therefore much more able to say no to a sponsor demanding a more intrusive segment).
It's much easier to find the nearest obvious number that can be divided by 7, e.g. 7000. Minus 6468. It's 532 and check if it's can be divided. Btw, you can do the same for 532 and 700. or 560 (7*80) - 532 = 28. Easy even doing in mind.
When you have Dr. Grime on, you know it's gonna be great.
James discussing Number Theory, a perfect start to my day.
The idea that we see here as a ‘trick’ can be further generalised to derive an algorithm to check whether a number is divisible by a prime number that ends with 1,3,7,9 (i.e. all primes greater than 5). Very useful indeed.
Wow. I love this. I really love when the concept is abstracted out. Fantastic.
Please dont let it be a 7-8-9 joke
I'm angry and happy to see this comment came through.
It turns out that 7 was a 6 offender
The answer is no! 7 !8 9
This method seems more complicated than just using classic division tricks to check. For example, with 686, you could just divide 68 by 7 with a remainder, then you take that remainder, apply it to the remaining digit to get 56 and 56 is evenly divisible by 7. Plus, doing it this way also gives you the quotient, not just the yes/no on divisibility. And, just like the trick you showed, you can repeat the steps for chains longer than 3, _and_ it works for divisibility with _any_ number, not just 7.
You are the smartest person in the comments section.
One of the points of this division trick is to avoid division. I teach both methods, and people seem to prefer the one highlighted in the video for whatever reason. For larger numbers, say 675081 you have to keep dividing by 7 and adding the remainder, OR you could multiply and add, and just check the last number for divisibility. If you noticed, all of these tricks are just to find divisibility, not the quotient. But I agree getting the quotient is a nice bonus =)
I personally do not see one more complicated than the other, but different people work differently, and have their preferences!
isn't this just a long division though
So to figure out if a number is divisible by seven you have to figure out if it's divisible by seven? That's not very helpful
1:09
I do it by head instead, decomposing the multiples.
434 -14 = 420
7x60 = 420
7x2= 14
434 = 7x62
Man, I love that seven one, I feel like I must have learned it once but completely forgot it, as I knew all the rest. Awesome job!
I'm really happy to learn the 7 one because it completes the set. Most of them I learned in elementary school, but 7 was always troublesome.
A method which I find easier to do in my head is to split the number up into groups and look at each group individually. For instance, 434 becomes 43 4 becomes 42 14 and both groups are divisible by 7, so the original number is also. As a bonus, the quotient is immediately given as 62. Thus 434 / 7 = 62. Do the same with 6468: 6468 becomes 63 168 becomes 63 14 28, so it is divisible by 7 and the quotient is 924.
I agree. I would guess that this is not more work than the method presented in the video, in count of number of arithmetic operations ...
Yes, this is exactly what I do. And really nothing to remember as the method in the video describes. And, for me at least, it’s faster!
Yeah I immediately noticed 434 was divisible by 7 this way
That's just long division with fewer steps.
@@badrunna-im Exactly. :-)
You have to love James Grime. Really! His passion, energy, and love for mathematics are just splashing on me from the monitor. Truly fantastic.👏
I wish I had such a passionate teacher/lecturer for my math lessons. I'm glad I could enjoy on YT for the least. 🙂
Man, what a nice guy professor Grime seems. I've known him for so long on this platform, I feel he's part of my family now.
parasocial relationship moment
I always love the James Grime videos!
I personally find 6300+140+28 easier to figure out in the head, works for arbitrary sized numbers quite quickly too if you know your x7 table well enough. Its a bit of doing the full division but in blocks of 2-3 digits... and in a way its related to this one...
Another interesting thing my friend and I figured out- if you divide a number which is not divisible by 7, by 7; you get repeating patterns after the decimal place- which is always 142857… in a cyclic order.
Yeah this one is completely nuts. But it's the best known cyclic number. Multiply 142857 by anything you get an anagram or 9's repeated if it's a power of 7.
basically, what i do is i add up all of the digits in the number, then i add 7 to the original number and add up the digits of that number. if the total of (original +7) is 2 less than the original sum, it’s a multiple of 7. it works *a lot* of the time but it isn’t foolproof because occasionally, the sun resets to a larger number
e.g 7(7), 14(5), 21(3) 28(*10*)
wait, so maybe it works if the next number is 2 less or 7 more? idek
@@aaronmarchand999that is typology, i think you meant anagram
@@xaryop7950 No look it up, comes from Gurdjieff, unfortunately has been used for typology in recent years but that's not the original meaning
True, but the decimal remainder starts with a different number in the sequence which is in numeric order in respect to the remainder.
example: put the cycle in numeric order - 1,2,4,5,7,8 which correlates with remainders 1,2,3,4,5,6
If you divide 37 by 7 you get 5 with a remainder 2. So the decimal remainder is .285714 repeated.
For 38/7, it would be 5.428571... (3 is the remainder. The third number in the sequence is 4.)
1:25 For numbers with 6 digits, the easy thing to do is to split it in the middle, so you get two numbers with three digits each. If the difference of those numbers is divisible by 7, then so is the original numbers. This works, because the method corresponds to subtracting multiples of 1001, which is a multiple of 7 (and of 11 and 13, so you can use this to check those divisibilities too). Using this method on given example of (00)6468, the first step gives 468-6=462, which is notably divisible by 7, because 462 equals 420 + 42 😎
7 is funky because it's the representation of the constant of -1. It is the mechanism that causes the offset nature of reality and is the reason we make sense of time the way we do
I'm delighted! This filled in the one remaining gap in my knowledge of divisibility by single digit numbers! I will use it often, because I often feel curious about whether a number is prime, and I like to see what I can work out before simply searching the web for the answer.
Well, it certainly makes sense that adding 5x and subtracting 2x would work the same when what you care about is divisibility by 7; when counting modulo 7, 5 and -2 are the same number.
Ooo you're right. If the number is 10x+y, then the first method is about checking if x+5y is divisible by 7, and the second method is about checking if x-2y is divisible by 7. (I like to call them the x+5y & the x-2y methods). But modulo 7, they're the same thing!
(For people not familiar with modular arithmetic, if x+5y is divisible by 7, so is x-2y. Because it is smaller by exactly 7y.)
In fact I can make an infinite number of 7 divisibilty tests like these, like the 8x-9y divisibility test & the 12y-6x divisibilty tests by adding or subtracting 7x's & 7y's. (If you frame it as a number is divisible by 7, if 12 times the last digit minus 6 times the first digit is divisible by 7, you can perhaps use it as a cool party trick ;) )
Hey -2 is my method! I was wondering why that worked! Thanks!
What’s neat is that the proof for x+5y mod 7 is contained in the proof for x-2y mod 7:
10x + y ≡ 0 (mod 7)
50x + 5y ≡ 0*5 (mod 7)
49x + x + 5y ≡ 0 (mod 7)
[49x ≡ 0 (mod 7)] + [x + 5y ≡ 0 (mod 7)]
0 + x + 5y ≡ 0 (mod 7)
x + 5y - 7y ≡ 0 - 7y (mod 7)
[x - 2y ≡ 0 (mod 7)] - [7y ≡ 0 (mod 7)]
x - 2y ≡ 0 (mod 7) + 0
x - 2y ≡ 0 (mod 7)
QED
I completely forgot about this channel & how much math(s) makes my brain happy… Just subscribed.
I own a funnel cake stand and I used to sell them for $7 each (not anymore... inflation!). This would have been handy when counting up the money at the end of an event to make sure the drawer was right!
And if the drawer wasn't right, what next? I'm not sure I'd want to count.
The method I found for determining divisibility by 7 was 3 times the ten's place plus the one's place. So 105 would be 10x3+5 or 35. 3x3+5 or 14. 1x3+4 or 7. For divisibility by 4, if the 10's place is odd the 1's place must be 2 or 6, or even but not divisible by 4. If the 10's place is even then the 1's place must be 0, 4, or 8, or divisible by 4.
For 434 I did 3X43 and added the 4.
This is 129 + 4 =133, or 7 less than 140, so, a 7-mer.
Or, using 133 as a rest stop,
do it as 3 X 13, + 3, which is 39 + 3 = 42.
I think our way beats his, although I've not analyzed just why it works.
@@davidcovington901 I believe it works because 1 is 1 more than a multiple of 7 while 10 is 3 away, so to compensate for dividing by 10 you multiply by 3. My Geometry teacher in high school told us that some POW had figured out a way but didn't tell us what it was. I guess I subconsciously kept working on the problem until I figured it, or rather apparently one, way.
@@thomaswilliams2273 Thanks!!!
@@thomaswilliams2273 Just divide the number by 7 ffs. Why would you do all these other steps as a “test”?
Just do the division.🤦♂️🤡
This only works if the number has 3 digits. Consider 1105
There is a more smart way to see if it is divisible by 4.
In three steps.
1: we look only in the last two digits.
2. It's even.
3: if the units are divisible by 4, which are 0,4,8 then the tens must be even, if it's not divisible by 4 which are 2,6 then the tens must be odds.
So, for example with units as a 2, tens must be odds.
12,32,52,72,92
Nice!
Just wanted to say that I'm especially excited when I see Dr. James Grime on Numberphile! Thank you for sharing your love for Math and the numbers! ❤️
Very useful video. Lately I’ve been wanting to check Primality in my head, and this helps make it a lot easier. I knew the tricks for 2,3,&5, and memorized that 49, 77, and 91 were multiples of 7 to be able to do it for all two digit numbers. Knowing the tricks for 7 and 11 let me check all the way up to 169, so that’s pretty cool.
There's a maths based "latin square" puzzle that I love, and I'm usually trying to figure out if a given number is divisible by 7 somewhere in the puzzle. I'll try this method.
Usually I'd break it up as 420 + 14, or 6300 + 140 + 28.
For reference, the puzzle is called "Keen" and is on "Simon Tatham's Portable Puzzle Collection".
I've been waiting for this for 49 years actually. How fitting LOL. I use it for checking numbers for primailty in my head. Brilliant! Thank you! ( I did know all of the others already actually).
Knowing divisibility checks is really useful for prime factorizations too...7 and 11 in particular are terrific for this since they're on the bigger end of the early primes.
6468 is the polynomial 6x^3 + 4x^2 + 6x + 8 evaluated at x = 10. To check divisibility by 7, evaluate it instead at x = 3 by synthetic division, casting out 7’s along the way.
6*3+4 = 22 ~ 1
1*3+6 = 9 ~ 2
2*3+8 = 14 is divisible by 7.
For 11, the alternating sum is equivalent to synthetically dividing by -1.
For 4, synthetically divide the last two digits by 2.
For 8, synthetically divide the last three digits by 2.
I love seeing someone so excited about divisibility by 7!
I actually have a secret method for determining whether a number is divisible by 7.. I just ask myself..
Is it 14? If so, then yes.
Otherwise, I’m outta luck
Hey your method works for 0% of the natural numbers
🤣
What if you're trying to figure out if 7 is divisible by 7?
@@imveryangryitsnotbutter lol also out of luck
49 is only number I subconsciously know is divisible by 7. Rest is outta my reach
I always felt the subtraction rule was a bit clumsy and I was faster at division (or finding the remainder, i should say) than doing that test. The addition rule is easier to do!
To check for a multiple of 8 i have an easier method (for me at least):
I look at the last 3 digits, if the first is even then the last 2 must form a multiple of 8.
if the first is odd, then the last 2 can't form a multiple of 8 but must form a multiple of 4.
For example, 264 is a multiple of 8 because 2 is even and 64 is a multiple of 8.
Another example, 128 is a multiple of 8 because 1 is odd and 28 is a multiple of 4 (but not of 8)
I find this much easier than halving 3 times in a row. Checking whether the last 2 digits are a multiple of 8 is easy as long as you know your 8s table, checking whether they are multiple of 4 is a bit trickier, but you only have to do it half of the times anyways.
For 2-digit multiples of 4 there is a similar "trick":
if the 10s digit is even, then the 1s digit should be one of 0,4,8
if the 10s digit is odd, then the 1s digit should be one of 2,6
It does work everytime? So cool!
The same trick "scaled down" works to find multiples of 4 too. If the 2nd digit is even, the last one must be 0, 4, or 8. If the 2nd digit is odd, the last one must be 2 or 6.
Another way to do 8's is take the first 2 digits, multiply by 2 and add the 3rd digit. So for 264, 26*2 = 52+4 = 56 = 7*8 for 128, 12*2=24+8=32=4*8
@@nekogod Another similar one is: take the last three digits multiply first by 4, second by 2, third by 1 and sum, check if divisible by 8, i.e. 264 = 2*4 + 6*2 + 4 = 24
The advantage is that it gives you a smaller number.
My favourite way to tackle divisibility by 7 is to exploit the fact that 1001 = 7 x 11 x 13. It's relatively easy to cast out lots of 1001 (though you can occasionally get some nasty carries) and it's guaranteed to bring you down to a number below 1001. Even though this might require a bit more memorisation (or mental arithmetic; just doing long division with numbers that small is surprisingly manageable, especially with a bit of practice/experience) from this point onwards than most of the methods discussed here, it does have the huge advantage of being able to attack 7, 11 and 13 simultaneously!
its guaranteed to get a 3 digit number, as 1-001 is 0 and 1-000 is 1.
A similar process (as I’m sure others worked out) is to divide the number by 50, then add the whole number portion of the quotient to the remainder. So for 434 it would be 8 + 34 = 42, which is divisible by 7. This works because 50 - 1 = 49, as related to his explanation of how his method works. For very large numbers, you can repeat this process as needed 😸
That's quick and easy!
If you're interested in the proofs for these divisibility tricks, look up "divisibility and modular arithmetic"
I think the easier and more straight forward method would just be subtracting multiples of seven since it cannot change the remainder. This method is solid and works for any number, but for some small numbers there are better ways.
6468 - 6300
168 - 140
28 ✓
A nice bonus is that we get can easily perform the division right now, since
6468 = 6300 + 140 + 28 =
7 (900+20+4) = 7*924
434 = 420 + 14, 6468 = 6300 + 140 + 28. This is the skill I got from taking math tests without a calculator. This will always be faster
Literally just division by hand as we learned in primary school, but with 0's instead of spaces. I wish we'd have learned why that worked back then. I think it might have helped quite a fair few to better understand, or might at least somewhat dispel the myth that maths is just magic.
I just want to say I haven’t been on TH-cam in a long time and rediscovering this account is helping bring back my love for math. Covid stopped me from loving school and learning and I lost my love for math and as a freshman in college taking calculus it’s been really hard. Thank you guys for always being so enthusiastic about math even if you never see this comment.
I am from India🇮🇳
Never heard of that "7" test, it's genius!
I worked out my own test for 7: split off the RH digit, multiply the rest by 3 and add the RH digit. The reason is as follows: 10x +y can be re-written as 7x+ ( 3x +y) . Now as 7 divides 7x, to divide 10x+y, it needs to divide 3x +y.
A similar trick can be done for 13, but this time the test is (RH digit ) MINUS ( 3 times rest) ( ignore the minus sign if the result is negative).
Edit: I've just worked it out for 17: similar to 13, but the test is (RH digit) minus (7 times rest).
Didn't bother with 14, 15 or 16 as they are simply combinations of 2 tests ( respectively 2 and 7, 3 and 5, 2 and 8 ( or 4 twice)).
A game I've been playing since I was a kid is breaking down numbers I come across, like street addresses or telephone numbers, to find it's largest prime. I "win" if a number's largest divisor is less than 12. I've always known every other trick mentioned in this video but seven has ALWAYS been the bane of this game as I didn't have an easy method to quickly divide by it. Or at least, not until now. So glad to have learned this, this trick is going to speed up my games considerably!
I love this. Thank you for sharing with us. :)
Loved it, thanks so much
Actually for the divisibility by 8, you consider the last 3 digits, and then only the last 2 digits, because second step is testing divisibility by 4. Nice video.
In school we were taught to subtract two times the last digit from the rest which is equivalent to this method but had the advantage that the resulting number is less than when you add five times the last digit, though subtracting-usually-tends to be a bit harder to perform without a pen and paper or using a calculator.
They showed that :D
@@DepFromDiscord Yeah I kinda hurried to the comments section🤭🙈
I like the double last and subtract because it can find out if the number is a multiple of 3 at the same time. May not be the most efficient a lot of the time, but it's nice seeing the subtraction of 21b rather than adding 49b.
What’s neat is that the proof for x+5y mod 7 is contained in the proof for x-2y mod 7:
10x + y ≡ 0 (mod 7)
50x + 5y ≡ 0*5 (mod 7)
49x + x + 5y ≡ 0 (mod 7)
[49x ≡ 0 (mod 7)] + [x + 5y ≡ 0 (mod 7)]
0 + x + 5y ≡ 0 (mod 7)
x + 5y - 7y ≡ 0 - 7y (mod 7)
[x - 2y ≡ 0 (mod 7)] - [7y ≡ 0 (mod 7)]
x - 2y ≡ 0 (mod 7) + 0
x - 2y ≡ 0 (mod 7)
QED
It’s a lot easier to subtract the multiple of 7 that has the same last digit as the number we’re examining:
434 = 420 + 14
Remove the 0 from the first number in the sum. 42 is divisible by 7. It even gives you the quotient:
434/7 = (420/7+14/7)
=60+2
=62
I sent a video to you guys a long time ago where I explore the significance of dividing by 7, and thought this may have been your response 😅 but my video was pointing out the 142857 loop of decimal points. I love all of these neat patterns that lie in mathematics !
I was hoping the video would be about that! I discovered that pattern when in school - how doubling 142857 shifts to 285714 and doubling again shifts to 571428.
@@thomasreichert2804 yess such a cool pattern! I was trying to find this pattern with other numbers too, but it only seemed to exist if the divisor was a multiple of 7, such as 1/14 or 1/28. I looked at the pattern as treating the numerator as the index of the decimal number from low to high, then the pattern continues on. For example 3/7, the 3rd smallest number from the 142857 pattern is 4, hence 3/7 = 0.428571 and so on.
@@mattcoulter7 yeh, it's great, you can work out any integer divided by 7 to as many decimal places as you want by just remembering that simple sequence
Oh, the difficulty of 7. I suffered a concussion in a bike accident in 1985. When I got to the hospital, the doctor asked me to count backwards from 100 by 7s as one of the mental acuity tests. My immediate reaction was something like "nobody can do that." I gather they look for the patient's reaction, the effort, and some relation to consistency; they know that nearly all patients will make mistakes. I did pass that portion of the tests and they released me about 2 hours after admission. The high point of the day was that math quiz. Really - it was either a truly bad day or I'm a nerd. Pick one or two.
Counting backwards in sevens is definitely quite traumatic. While checking if you were clear of a concussion injury, I think the doctors were trying to cause PTSD!
Divisibility of any number "a_0...a_n b" by 7 is by far easier when you consider the difference between 2*b and 7 if a_n is smaller than 2b for example:
364 ---> 2*4=8>6 then consider 2*8-7=1 gives 36-1 is divisibe by 7
266 ---> 2*6=12 too big then consider 14-12=-2 gives 26-(-2)=28 divisibe by 7
etc ...
Thank you so much for this ive needed this for so long 7 was killing me. Id figured out 2 to 11 on my own but 7 i had no idea this makes things so easy now and raises an interesting idea that im gonna test for 13
I've always known the rule about taking the last digit, multiplying it by 2, then subtracting the smaller number from the larger number. ... the x5 rule is new to me.
Same. The one you said is my go-to 7 divisibility rule
It's essentially the same because +5 and -2 are the same modulo 7
Same!
His blank stare when hearing football references is so completely relatable to me lol.
I use something that can also be used for 11* and 13. Add up every other trio of digits, both groups and subtract the totals. This is because 7*11*13 is 1,001.
*: With 11, there's an even simpler rule: add up every other digit, both groups, and then subtract the totals.
Adding up every other digit into two groups and subtracting the totals is effectively the same as alternating addition and subtraction.
Number: ABCDEF
A - B + C - D + E - F
= A + C + E - B - D - F
= (A + C + E) - (B + D + F)
Alternating will just keep the progressive total smaller.
@@SirRebrl true, and I could have added that to my original comment, but what I said works for all 3 can tell you, for instance, that 1,540 is divisible by both 7 and 11 (and therefore 77) because 540-1 = 539, the prime factorization of which is 7²*11.
Back in the days when applications started to be locked by a serial number it wasn't that uncommon that one of the checks was "divide by 7".
Many times you needed to present a number say 12 digits long which seems impossible to guess but all that was needed was to look for one that could be divided by 7 (11 was another common).
Fun fact: the powers of 11 up to 11^9 (or one less than the radix) are sequential palindromes stopping at the nth power.
Very neat..
@@JamesDavy2009 Huh! Does this translate across bases? (So for example 11 in base 6 (7), would that be a palindrome all the way to 11^5?
@@qamarat8366 yep - that's what the "radix" in his message is referring to!
As far as I know, the simplest is separate the last digit, multiply by 2, and reduce the rest. 343 => 34 | 3 , 3 x 2 = 6 => 34 - 6 = 28 (divisible by 7). But, it takes longer when reach more than 5 digits. And yes, this is for base-10 numbering system.
For divisibility by 8:
1. Look at the hundreds place
2. If it's even, the last 2 digits need to be divisible by 8
3. If it's odd, the last 2 digits need to be divisible by 4, but not 8
Nice! Another thing you could do is 4 × Hundreds Place + 2 × Tens Place + Ones Place :)
For example:
657,392 becomes 4 × 3 + 2 × 9 + 2 = 32 and since that is divisible by 8, so is the original number.
This may not be easier, but the advantage is that you only have to deal with numbers up to 63, so most people wouldn't need to reiterate the test to find out their answer
By the way, the reason why it works is because 100 is 4 more than a multiple of 8, and 10 is 2 more than 8 itself :)
In fact, for divisibility by 8 you don't need "the last three digits" but just "modulo 200", and for divisibility by 4 you don't need "the last two digits" but just "modulo 20".
e.g. is XX68 divisible by 4? I don't know, but I can ignore XX60 so yes 8 is divisible by 4
Is YYY918 divisible by 8? I don't know but I can ignore YYY800, keep 118 and, well, 40-80-120-160 are easily divisible by 8, so 118 is not
@@NicholasTheKing
For any power of 2 (2^n), the last n digits must be divisible by 2^n.
An easier divisible by 8 test: Take the last three digits. If the hundreds digit is even, check if the last two digits are divisible by 8. If the hundreds digit is odd, check if the last two digits are divisible by 4 but not by 8.
e.g. 6468 is not divisible by 8 because 68 is not; whereas 6568 is divisible by 8 because 68 is divisible by 4 but not by 8
Similar rule works for division by 4:
0,4,8 if the tens are even.
2,6 if the tends are odd.
useful for coding. in my math class we were doing some coding to assist in math proofs, more like proving something then making code to do that thing. being able to code to find if something is divisible by 7 is quite useful, but this method could be used for other numbers as well like 11, 13, 17, etc. trying to find if something is divisible by some uncommonly used prime number
Very interesting, and curious I have never heard of this before! It works, because 7*7 is one less than 5*10. Therefore it should work also as a trick for 13! 13*3 is one less than 4*10, so use the same trick with last digit times 4 to find if a number is divisible by 13. Same logic, same proof
I thought of the same too! I think it works for something like 17 then 23 or any number ending with 7 or 3 but the number which is to be multipled to last digit( like 4 for 13 and 5 for 7) will increase
Wow!!!! It works.
For 17: 3*17=5*10+1.
So multiply last digit by 5 and subtract.
19 is intersting: 19=20×10-1 so use twice last digit!
All numbers either enter the 18 number loop of the numbers 1-18 if indivisible, or 19 if divisible.
This can be easily extended to all primes, simply find 1/10 (mod p) then use this in place of 5 with the same algorithm
10/p will give you a decimal for all primes. I don't know how that mod 7,11,13 etc will get you an integer. I think 1=10x mod p where x is the number to put in place of 5 may work. With p=11 I got x=10 and p=13 I got x=4. Those seem to work.
@@jasonwinters7560 Sorry, you're right, I've updated my comment to say 1/10 (which indicates the solution to 10x = 1 (mod p)) instead of 10/p, not sure why I made that error
Simplification for 8 is to calculate Last + 2*SecondFromLast + 4*ThirdFromLast and check if the result is divisible by 8, which is slightly faster than dividing by 2 thrice and matches with the other digitsum methods
The thing about this "trick" for divisibility by 7 is that it's less efficient than just doing the long division algorithm.
Just what I was thinking
I prefer using an equivalent test which is: if n=10m+r where r is a digit and m an integer, then 7|n iff 7|m-2*r (obviously this is the same as the one mentioned in the video, but it's better when n is non-negative and 7|n iff 7|(-n), so in case of a negative number I'd just negate the sign)
its way more easy if you just split your number in modulo groups. for example 6468 -> 64%7=1; then use the 1 as the next leading number before the next 6: 16%7 = 2; repeat the step: 28%7 = 0 . so if the last modulo is zero the whole number is divisible by 7. with this you dont need the addition and it works pretty well recursivly
I’ve been using those divisibility rules to gradually try to prove the collatz conjecture by induction.
Basically, if I can prove that every number eventually goes to a number less than itself, regardless of how many steps, then every number goes to 1.
Assuming that is true; All integers N eventually reach a number x such that x
I think the key to proving the collatz conjecture lies in the powers of 2. They're the only path to 1 available. I also think the key is to think of it as "can the even numbers 1 greater than a multiple of 3 eventually reach an odd number n that will satisfy '3n+1=2^x'"
I see 3n + 1 strikes again...
I consider N = 2^n as the end of the series since it drops to 1 very quickly (exponentially).
Note that N = 4^n - 1 is always an odd multiple of 3 [ say 3 M = 3 (2m+1) ]. This is provable by induction.
As a result, there are infinite odd numbers N such that 3N+1 = 4^n which is 2^(2n) which is effectively another end of series.
This addresses another subset of your last 25% odd positive integers. Perhaps suggestions from others can being your last 25% down to at least 15%.
11 is weird too, because substracting the last digit from the rest of the number and checking if the result is divisible by 11, will show if the number is divisible by 11.
Care to explain?
@@divabhardwaj6381 Yeah, e.g. 143 is divisible by 11, because 14-3=11. Any divisibility rule can be found, when you multiple a number so that the last digit of the product is 1. For example, the rule for dividing by seven is found when 3*7=21.
This was way more helpful that you would think. Thank you so much
ya there was some weird odd even rule for it
Also 37 is weird, because multiplying it by a multiple of 3 gives a triple digit number. E. g. 6*37=222 and 9*37=333.
i'm seing a pattern:
2 - last digit
4 - last 2 digits
8 - last 3 digits
16 - last 4 digits maybe?
that would be a nice formula for: divisibility by n² can be checked by the last n digits.
5:45 James’s reaction really goes to show that Maths-nerds aren’t big on sports 😅.
Another fun fact about the number 7: only 14.28% of numbers are divisive by 7.
Is it all numbers or only non-decimals numbers?
14 and 28 are both divisible by 7. Huzzah!
@oogaBooga This is because every seventh number is divisible by seven. 1/7 = .142857143... ≈ 14.29%
@@MxIbatomik All numbers, even decimal numbers, can be divided by seven. However only whole numbers can be divided by seven such that there is no remainder.
14.285714 is a better approximation. The 6 decimal places repeat.
When I was like 10yo I figured out by myself a very similar method to verify if a number is divisible by 7 and I was so proud of it:
In case of 434, multiply by 3 every digit except for the last one and add it to the result (43x3)+4=133;
Then again (13x3)+3=42 that is divisible by 7 so 343 is divisible by 7
2:59 What's more interesting is that any number that's divisible by 49, after doing enough times of this method will always reach 49. But others will always reach 7.
That's not exactly true. Consider 343, which is 49*7. Using the addition method (5*3+4+3), you get 22; using the subtraction method (2*3-4-3), you get -1; adding the results of both tests, you get 21.
The addition method gets you 49: 3×5 + 34.