I remember back in 4th or 5th grade we were doing multiplication and I couldn't solve it. My teacher approached me and shouted at me. She pinched my ear and shouted at me to the point where I just wanted to cry lol. She traumatized me making me hate math, now I gotta learn it again all by myself. It's fun 😀
You are one of the best teachers I have ever had with math. I am writing my GED math exam Wednesday. And you have helped me a lot. Thank you very much.
Brilliant. I love your tips, tricks and techniques and I find your explanations really easy to follow. I've recovered from my school maths experience enough to enjoy playing with numbers again and your channel is ace! I also use it to help some of my students.
Thank you, it was great! But you forgot an important point to say that if you come across a negative number at one time, subtract one from the previous digit in quotient and if it was negative again, subtract one to make it positive, then continue with that number. example: 50531 ÷ 43
Hi! Can I ask if you can solve using his method with this problem, 2345/8. It doesn't seem to work. I'm curious if you guys can using his style. If so, can help me out as to how?
@@Shintah07 So you have (+2) 1/2345 1) 1*2=2 so you have "2" with "0" for the remainder 2) 03 + 2*2= 7 so now you have "27" and also "0" for the remainder 3)04 +7*2 = 18 so now you have "27¹8" and also "0" for the remainder 4) 05 + 18*2 = 41 so finaly you have 288r41 5) but 41 is still divisible by 8 with a remainder so 41/8 = 5r1 So you do 41-1= 1 for the remainder and you had 5 so 288+5 = 293 finally you finish with 2345/8 = 293r1 so 293 +1/8
Your explanation for shinta's problem made perfect sense to me. But I don't understand what to do about a negative number. My problem is 6930 / 60. I did 6930 / 5 (-10) 1. 5 goes into 6 once, No remainder. 2. 1 × -10=-10+19=9, 4 remainder 3. 1×-10=-10+43=33, 3 remainder 4. 6×-10=-60+30=-30 Here I tried to subtract 4 from the previous digit, so 4. 2×-10=-20+30=10, 0 remainder I had written 1162, but after subtracting the 4 it should become 1122. The Answer is actually 115 with a remainder of 30. Where did I go wrong? Is this because I rounded 10? I'm not sure what to do when your number ends is zero either.
@@justabout6979 6930/60 = 6930/(6x10) = 6930/6/10 So you can do 6930/6 then you divide the result by 10 1) 6/6=1; No remainder 2) 9/6=1; 3 remainder 3) 33/6=5; 3 remainder 4) 30/6=5; No remainder 5) You found 1155 6) Now you can divide 1155 by 10. So, 1155/10=115.5
I just did my own sum and straight away got into a problem. I did 37648 / 43. Straight off the bat, 4 into 37 is 9 with r1. So my first part of the answer is 9++ and something. My calculator says 37648 / 43 = 875, not 900+. HELP? Whatever i do, right or wrong, with the rest of the sum, I can't impact on the 9 but it's wrong. It has to be 8! After that, the maths breaks down altogether. I put the 9 in and carry the r1 onto the 6. I now have a sum of 16 - 3 x 9, 27, which leaves 4 into - 11. What is that and how do i enter the next figure in the answer? What is -11/4? If it's -2 r-3 i could, as per 7:40 subtract the -2 from the first digit answer of 9 but that gives me a 7, still not an 8. Can anyone explain WTF is going on here?????
Doing division his way, it would go like this: (-3) 37648/4. The first problem you get is 37/4, equal to 9 r 1. The second problem you get is 16-27 = -11. Because this number is negative, you need to go back to the first problem and subtract one from your answer, so instead of 37/4 = 9 r 1, you'll get 37/4 = 8 r 5. Now, the second problem becomes 56+8*-3 = 32 and 32/4 = 8 r 0. The third you get is 4+8*-3 = -20. This is a negative number again, so you need to go back to the problem and subtract 1 from the answer you previously got, so instead of 32/4 = 8 r 0, you'll get 32/4 = 7 r 4. Now the third problem to solve is 44+7*-3 = 23 and 23/4 = 5 r 3. The fourth and last problem, finding your remainder, becomes: 38+5*-3 = 23. So the answer to 37648/43 is 875 r 23.
UPDATE: There is a problem if the remainder is still divisible by the divisor. Example: 5842/45. Using this method, we get 128 r82. This is wrong, as the correct answer should be 129 r37. To fix, if the remainder is still divisible by the original divisor, subtract. Thus, 82-45=37, but add 1 to 128=129.
Hi! Can I ask if you can solve using his method with this problem, 2345/8. It doesn't seem to work. I'm curious if you guys can using his style. If so, can help me out as to how?
@@silverdustgirl In my example, 5842/45: the idea is "how many 45s are there inside 5842". Using the method in the video, the remainder is 82. However, 82 still "contains" one (1) 45. Thus, I propose subtract 45 from 82 to get 37 as the actual remainder. The "extra" 45 should then be added to 128. Therefore, there are 129 (45s) and a remainder of 37.
this is very effective for other people! i will have to use this in the future. ive just started middle school, so im going to need to learn a bit more about negatives before i do this, but thank you!
As to why they don't teach this in the lower grades: I think doing long division the long/traditional way shows the actual math processes with step by step clarity. I love this short cut, but it becomes harder to see the many steps that one glosses over performing tricks to get to the same answer. It becomes 'magic" without having to bother with the basic mathematical principles. It's fun to watch a magician, but you don't see how he did it. So I think, much like learning Newtonian physics, it's best to learn math the standard way in the lower grades, but if you can't to the mental shortcuts in college physics, you will be a very long time solving problems. This, I suspect is the "necessity" which is the mother of the "inventions." I think you see what I mean... Of course, learns shortcuts for doing things in many fields of endeavor. And I think it is quicker to teach the tricks knowing the basics, than to know the tricks then try to teach the basics to show how it works. However, for most of the shortcuts you have presented, it is truly simply a different pattern of doing the basics. But why this gimmicky manipulation of long division works... I suspect it would take a while to show how this works to someone without the basics (and many who did learn the basics but who don't have the fascination and facility with math that some of us do.) --"Who has not been amazed to find out that the function *y=e^x* , like a phoenix rising again from its own ashes, is its own derivative?" - François le Lyonnais Love the videos... (oh, and no, I'm not a math teacher... just a doctor with a hobby.)
I agree. The teaching of division seems to oscillate from one method to another with some places even removing it from the curriculum altogether (which I think is a crime). The problem seems to be trying to avoid confusion amongst students who start off learning single digit division with the 'bus stop' method writing the carry as they go from left to right. in some ways this method follows that logic but students can come unstuck with bigger numbers. Yet the traditional long division model has its limitations to requiring the correct divisor for each quotient decimal place. A more forgiving method is to subtract the divisor number or a multiple followed by a number of zeros and to the right write down the quotient number with the same number of zeros. One can then subtract and find the next divisor multiple that will fit followed by zeros then writing that quotient followed by the same number of zeros. Totting up all the right hand column quotients gives whole number result with remainder/divisor fraction left over from the subtraction column. This demonstrates that division is repeated subtraction and students can start with subtracting the divisor times 1x, 2x or 5x from left to right. That should be sufficient for high school use where a large number can easily be divided by a three or four digit number. Confidence & proficiency is improved if students are taught how to check their results using 'Sum of the Digits': Divisor x Quotient + remainder = Dividend
I can see why this method isn’t taught in schools. If we didn’t have Josh talking us through the adjustments he makes, the divisions by single digits he makes from the original problem just don’t look right. For example, in the final calculation, 308,379/9 is not 3351. Then if you gave a child a division by a single digit, would they get confused with the Tecmath long division method we see here? The idea about it being space saving is a red herring too. As a retired UK primary teacher, I know kids are required to show all the steps they take in their working out, thus they’d have to show all the adjustments that Josh talked us through and then erased. So there’d be little difference between the amount of paper used for the traditional method. The plain truth of the matter is that long division is a pain in the neck. It’s better for kids to have reliable methods for working with manageable numbers and use a calculator for larger numbers. Don’t get me wrong, I like a lot of Tecmath’s tips, esp for dividing by 5, 25 & 125, and long multiplication. I feel I could explain those methods to kids and how they work; this particular one though beats me, and it seems totally unnecessary in this post-Sir-Clive-Sinclair world to ever have to use it.
I'm not that sure (I'm not a mathematician nor English mother-tongue so sorry for typos and so on...). If I'm wrong let me know! It'll be a chance to learn!!
Thank you so much, this video helped me a lot. However, what if you need the answer with the decimal instead of the remainder? Could you help with that? Please? I loved this video and same with all the others, tysm!
You can just keep going with this method past the 1s column into the decimals. The math is exactly the same. (If you need 3 decimal places, it's effectively like taking the number you are dividing and multiplying it by 1000 and then taking your answer to the division and shifting the decimal over 3 places to "undo" the multiplication by 1000.) Just stop when you have enough decimal places and make sure your remainder isn't rolling over.
The main thing that sumped me here was the remainder business. I remember having to write a little R and a number after it, and never really understanding why they did that. This method can easily dispense with the whole remainder thing by appending a couple decimal places and performing a couple more sub-calculations until the R becomes a decimal I won't say this is less time consuming or takes up less paper, but it is more direct than the method I learned which was a sort of duck hunt where you end up with a lot of little side calcualtions to figure out how many times the small number goes into some portion of the big number. I don't think I would use this process unless I absolutely had no calculator handy. In that case, I would prefer this method over the longhand method I learned!
How do you handle when subtracting results in a negative number? For example, I've been making up examples and am stuck with 53 (so round to 5) dividing into 31389. In my 2nd step, I get 13-18 = -5 which has put me at a standstill
@@craighunter9082 I agree with you but it's worse than that. I have a conundrum i just hit which I'd like you to look at. I tried 37648 / 43. Wether i use the method shown in this video or my old school taught method, I get the same problem. My first sum is 37/4 = 9. No matter what i do now i can't impact on the 9. The answer is going to be 900 and something, correct? My calculator tells me the answer is 875...... so my old school method is wrong as is this one in the video. Care to tell me what I'm doing wrong, please? Indeed as i proceed, i end up with -ve numbers. What is -11/4, which is the calculation for the second digit? Actually, my old method isn't wrong. 43 won't go into 37 so it has to go into 376 which does become an 8 something figure. Either way I'm confirming your point that this techmath method is flawed.
I realize it's like a year later and doesn't matter but to answer the OP: in your 2nd step you should have been adding the 18 since the +3 over 50 isn't negative. That said this method seems to be more work than it saves
@@craighunter9082 I'm running into alot of problems, i guess because I rounded 10 AND I'm using negatives. My problem is 6930 / 60. I did 6930 / 5 (-10) 1. 5 goes into 6 once, No remainder. 2. 1 × -10 = -10+19= 9, 4 remainder 3. 1× -10= -10+43= 33, 3 remainder 4. 6× -10= -60+30= -30 Here I tried to subtract 4 from the previous digit, so 4. 2× -10= -20+30= 10, 0 remainder I had written 1162, but after subtracting the 4 it should become 1122. The Answer is actually 115 with a remainder of 30. I'm not sure what to do when your number ends is zero either. You don't have to respond. This is just another example of a problem it doesn't work well.
Sir(/or anyone who knows)what should we do when the result is negative , and im having trouble with dividing large numbers with smaller numbers ,67961÷23 , first we have consider and -3, but initially we divide the first digit 6 by 2 and get 3 but the answer is in 2954.82 range ? What do we do?
In your case (dividing by 23) divide by 2, but *subtract* 3 times the previous answer digit. Take, for example, 27819/44 (his second example) which holds true for any divisor ending in (at least) anything from 1 to 4.
Hi sir wonderful method. Thank you for letting us know. @ 8.47 you have a reminder of 31. Why din’t you divide it further by 4 and how do we know where to stop the problem and write it as reminder
When your dividing by a number that ends in five is there a rule for that? I did one twice, adding five in one problem, and subtracting five in the other. The one that I added five to was the correct answer. So should you always add instead of subtract numbers that end with five?
What do you do if one of your digits in your answer is above 9? For example if you have a 5 for the first digit, but the second digit is something like 12, does the 5 become a 6, and the answer then becomes 62? And then, when you do that thing with the adding and multiplying, do you multiply by 12 or just 2?
what do you do if you have to divide by a negative number, it's happened to me a few times. when the difference is negative soemtimes I have to subtract a bigger number.
@@danibilling9435 Add decimal places to the dividend. For example, when he does 16283 / 38 he'll end up with 428. After the 428 you add a decimal point. You then take the last digit of 3 as he does, add the 8 x 2 = 16 to get the 19. But instead of just having the 19 as the remainder, you continue, by dividing the 19 by 4 to get 3 with 3 re.... wait..... it's supposed to be a 5 with no remaining...... I'm confused. Well forget this. Can @tecmath explain how to do this without just saying "continue with 0s" or "continue with more decimal points"?
That's simple! Just add two or three decimal places to the number (27819.000) and carry on with Josh's system! You'd do two more calculations to get 27819/44 = 632.25
There are three possibilities I can think of. 1. Do the same that the method tells you to, by taking the 512, rounding it down to 510, and then simply dividing by 51, and using the -2 you got to get from 512 to 510. 2. Do the same as above, but there's an extra step in which you then have to take the 51, and then change that to a 5, with an extra -1, but then you have to work out how to incorporate that part with the first part, which is just too mind-boggling to even imagine. 3. You round to the nearest 100, where 512 becomes 500, and you divide by 5, and not 51, and you take -12 as the modifier.
I've been practicing this but am running into an issue with examples such as 64925/23 Once you get to 4 - 9 what are you supposed to do? Borrow from the 3?
Can you explain to me how to divide 50,000/80,000? On the calculator it is .0625 but I don't understand because 50,000 only goes into 80,000 1 time with 30,000 left over?
I need your help for the question 64821/38. I've done your method and reach 1705.81, however, the answer is 1705.816. The last remainder is 30+14=44. So 44/4=11. I added 1 to the number before it but I don't understand how there are still numbers to follow. Please help.
The whole number remainder is 31 so this is the fraction 31/38. If you divide the fraction out you get a long decimal number 0.815789473. So the choice you have is to express the remainder as a fraction or decide to how many decimal places you want your answer to be. Some fractions give a finite answer like 1/16 = 0.0625 whilst others give a repeating recurring number like 1/12 = 0.083333 and others seem to go on and on and you don't know if they will repeat or ever end. It is an interesting thing to ponder why this might be? This is why when doing division of large numbers (by three or more digit divisors) I just use the multiples 2 & 5 (with the relevant zeros tagged on) as it quick & easy to do (using repetitive subtraction)
I’m not sure if this works, but I did the last problem slightly different: 308379/92 ... +8 308379/10 30/10=3 (8+12)/10=3r2 (23+24)/10=4r7 (77+32)/10=10r9 (99+80)/10=17r9
@@LJC4495 Hey ! not everyone speaks English. For some English is only needed when they come across videos like this. The world isn't centered around English. He probably has english as his 4th or 5th language. I too have at least 4 languages. Not counting the languages I am learning. otherwise I'll had to say 6 languages. Get a life kid.
@eugeniamuriambihu390 Hi This a shortened version of a series of posts (x4 in all) by Techmath. I suggest you watch them all and then have a go at last question. I just tried it and it worked a treat, as they build your knowledge gradually. I too struggled at first as I like to know ‘why’ rules/hacks work. However I just trusted the process and it worked.
I can see how this could make it easier for mental long division. You explain methods very clearly. Math is fun when you know what you are doing. 😀
I agree
Exactly, when you know what are you doing
I remember back in 4th or 5th grade we were doing multiplication and I couldn't solve it. My teacher approached me and shouted at me. She pinched my ear and shouted at me to the point where I just wanted to cry lol. She traumatized me making me hate math, now I gotta learn it again all by myself. It's fun 😀
Math is annoying asf x + y + a * B squared • blah blah = 🫠 annoying as hell too much complications to get to an answer
Proud to be 69th like, + I agree.
You should have been my maths teacher back on the day. 30+years of hating maths, and it needn't have been the case. Grateful to you mate. 👊👍
Tf is needn't
@@tyx168Need not
You are one of the best teachers I have ever had with math. I am writing my GED math exam Wednesday. And you have helped me a lot. Thank you very much.
0:27
You are the best math teacher in the world, thank you from the heart 💚
Awhy you choose 2 to multiply?@
Shkuntla devi is the best maths teacher in the world.
Thank you so much, I have a bad math teacher and your videos are helping me to get through math class without an F
I am a teacher and I learned a lot from you..I appreciate what you do ..thanks
Brilliant. I love your tips, tricks and techniques and I find your explanations really easy to follow. I've recovered from my school maths experience enough to enjoy playing with numbers again and your channel is ace! I also use it to help some of my students.
I loved it... very quick answers when you get used to it !
Absolutely fantastic..... Are you a maths teacher.... Your fantastic... Make it so understandable and easier to follow...
I am a math teacher. And thank you.
Wow, I was such a slow kid in math while in school, this would’ve made me a brainiac!
love how he's so happy
i like this method! Thanks for introducing this to us! Definitely uses much less space on paper. 👍
Try 37648/43. I bet you can't do it.
Thank you, it was great!
But you forgot an important point to say that if you come across a negative number at one time, subtract one from the previous digit in quotient and if it was negative again, subtract one to make it positive, then continue with that number.
example: 50531 ÷ 43
Hi! Can I ask if you can solve using his method with this problem, 2345/8. It doesn't seem to work. I'm curious if you guys can using his style. If so, can help me out as to how?
@@Shintah07 So you have (+2) 1/2345
1) 1*2=2 so you have "2" with "0" for the remainder
2) 03 + 2*2= 7 so now you have "27" and also "0" for the remainder
3)04 +7*2 = 18 so now you have "27¹8" and also "0" for the remainder
4) 05 + 18*2 = 41 so finaly you have 288r41
5) but 41 is still divisible by 8 with a remainder so 41/8 = 5r1
So you do 41-1= 1 for the remainder and you had 5 so 288+5 = 293
finally you finish with 2345/8 = 293r1 so 293 +1/8
Your explanation for shinta's problem made perfect sense to me. But I don't understand what to do about a negative number.
My problem is 6930 / 60. I did 6930 / 5 (-10)
1. 5 goes into 6 once, No remainder.
2. 1 × -10=-10+19=9, 4 remainder
3. 1×-10=-10+43=33, 3 remainder
4. 6×-10=-60+30=-30
Here I tried to subtract 4 from the previous digit, so
4. 2×-10=-20+30=10, 0 remainder
I had written 1162, but after subtracting the 4 it should become 1122. The Answer is actually 115 with a remainder of 30. Where did I go wrong? Is this because I rounded 10? I'm not sure what to do when your number ends is zero either.
@@justabout6979 6930/60 = 6930/(6x10) = 6930/6/10
So you can do 6930/6 then you divide the result by 10
1) 6/6=1;
No remainder
2) 9/6=1;
3 remainder
3) 33/6=5;
3 remainder
4) 30/6=5;
No remainder
5) You found 1155
6) Now you can divide 1155 by 10. So, 1155/10=115.5
I never knew that division was so easy in it own way
THANK YOU
Love your short tricks sir 🙆
Love from India 🇮🇳
I just did my own sum and straight away got into a problem. I did 37648 / 43. Straight off the bat, 4 into 37 is 9 with r1. So my first part of the answer is 9++ and something. My calculator says 37648 / 43 = 875, not 900+. HELP? Whatever i do, right or wrong, with the rest of the sum, I can't impact on the 9 but it's wrong. It has to be 8! After that, the maths breaks down altogether. I put the 9 in and carry the r1 onto the 6. I now have a sum of 16 - 3 x 9, 27, which leaves 4 into - 11. What is that and how do i enter the next figure in the answer? What is -11/4? If it's -2 r-3 i could, as per 7:40 subtract the -2 from the first digit answer of 9 but that gives me a 7, still not an 8. Can anyone explain WTF is going on here?????
Doing division his way, it would go like this: (-3) 37648/4. The first problem you get is 37/4, equal to 9 r 1. The second problem you get is 16-27 = -11. Because this number is negative, you need to go back to the first problem and subtract one from your answer, so instead of 37/4 = 9 r 1, you'll get 37/4 = 8 r 5. Now, the second problem becomes 56+8*-3 = 32 and 32/4 = 8 r 0. The third you get is 4+8*-3 = -20. This is a negative number again, so you need to go back to the problem and subtract 1 from the answer you previously got, so instead of 32/4 = 8 r 0, you'll get 32/4 = 7 r 4. Now the third problem to solve is 44+7*-3 = 23 and 23/4 = 5 r 3. The fourth and last problem, finding your remainder, becomes: 38+5*-3 = 23. So the answer to 37648/43 is 875 r 23.
Good method
Explanation is very good
You are the best. Thanks for this tricks. Now I can calculate much faster
Thanks, Josh
Thanks again for another great lesson, you are brilliant 🙂
I like the method thank you
Love your vids dude
Easy explanation: 12:49
wow you are a legend! you sound like thor mate!
I like your channel very much
Haiii just wondering, is there also an easier way to divide decimal numbers in long division? Arigato~
Nice teacher
I will like your channel very much
UPDATE: There is a problem if the remainder is still divisible by the divisor. Example: 5842/45. Using this method, we get 128 r82. This is wrong, as the correct answer should be 129 r37.
To fix, if the remainder is still divisible by the original divisor, subtract.
Thus, 82-45=37, but add 1 to 128=129.
thank you for the update 💖
It is work good!
Hi! Can I ask if you can solve using his method with this problem, 2345/8. It doesn't seem to work. I'm curious if you guys can using his style. If so, can help me out as to how?
why do you add 1 to 128?
thank you :)
@@silverdustgirl In my example, 5842/45: the idea is "how many 45s are there inside 5842". Using the method in the video, the remainder is 82. However, 82 still "contains" one (1) 45.
Thus, I propose subtract 45 from 82 to get 37 as the actual remainder. The "extra" 45 should then be added to 128.
Therefore, there are 129 (45s) and a remainder of 37.
Thank you so much! I learned plenty. You are great!
Effective method and I enjoy pretty much watching tue entire video 😅
Any calculus tricks Josh?
this is very effective for other people! i will have to use this in the future. ive just started middle school, so im going to need to learn a bit more about negatives before i do this, but thank you!
Yeah
I’m also going into middle school but I do Kumon, so I have to do like 80 3 by 2 division probs in 30 minutes (or you should do it in that time).
Thank you sir
Amazing it really improved my calculations
Thanks man😁
As to why they don't teach this in the lower grades:
I think doing long division the long/traditional way shows the actual math processes with step by step clarity. I love this short cut, but it becomes harder to see the many steps that one glosses over performing tricks to get to the same answer. It becomes 'magic" without having to bother with the basic mathematical principles. It's fun to watch a magician, but you don't see how he did it. So I think, much like learning Newtonian physics, it's best to learn math the standard way in the lower grades, but if you can't to the mental shortcuts in college physics, you will be a very long time solving problems. This, I suspect is the "necessity" which is the mother of the "inventions." I think you see what I mean... Of course, learns shortcuts for doing things in many fields of endeavor. And I think it is quicker to teach the tricks knowing the basics, than to know the tricks then try to teach the basics to show how it works. However, for most of the shortcuts you have presented, it is truly simply a different pattern of doing the basics. But why this gimmicky manipulation of long division works... I suspect it would take a while to show how this works to someone without the basics (and many who did learn the basics but who don't have the fascination and facility with math that some of us do.)
--"Who has not been amazed to find out that the function *y=e^x* , like a phoenix rising again from its own ashes, is its own derivative?" - François le Lyonnais
Love the videos... (oh, and no, I'm not a math teacher... just a doctor with a hobby.)
I agree. The teaching of division seems to oscillate from one method to another with some places even removing it from the curriculum altogether (which I think is a crime). The problem seems to be trying to avoid confusion amongst students who start off learning single digit division with the 'bus stop' method writing the carry as they go from left to right. in some ways this method follows that logic but students can come unstuck with bigger numbers. Yet the traditional long division model has its limitations to requiring the correct divisor for each quotient decimal place. A more forgiving method is to subtract the divisor number or a multiple followed by a number of zeros and to the right write down the quotient number with the same number of zeros. One can then subtract and find the next divisor multiple that will fit followed by zeros then writing that quotient followed by the same number of zeros. Totting up all the right hand column quotients gives whole number result with remainder/divisor fraction left over from the subtraction column. This demonstrates that division is repeated subtraction and students can start with subtracting the divisor times 1x, 2x or 5x from left to right. That should be sufficient for high school use where a large number can easily be divided by a three or four digit number. Confidence & proficiency is improved if students are taught how to check their results using 'Sum of the Digits': Divisor x Quotient + remainder = Dividend
It is strange, but really cool. I am getting it, but I will need to practice. Thanks
Can you place the old way side by side with your new way? I want to show this to my daughter.
I can see why this method isn’t taught in schools. If we didn’t have Josh talking us through the adjustments he makes, the divisions by single digits he makes from the original problem just don’t look right. For example, in the final calculation, 308,379/9 is not 3351. Then if you gave a child a division by a single digit, would they get confused with the Tecmath long division method we see here? The idea about it being space saving is a red herring too. As a retired UK primary teacher, I know kids are required to show all the steps they take in their working out, thus they’d have to show all the adjustments that Josh talked us through and then erased. So there’d be little difference between the amount of paper used for the traditional method. The plain truth of the matter is that long division is a pain in the neck. It’s better for kids to have reliable methods for working with manageable numbers and use a calculator for larger numbers. Don’t get me wrong, I like a lot of Tecmath’s tips, esp for dividing by 5, 25 & 125, and long multiplication. I feel I could explain those methods to kids and how they work; this particular one though beats me, and it seems totally unnecessary in this post-Sir-Clive-Sinclair world to ever have to use it.
What happens if when you do the subtraction the result is negative. Or can this never happen for some reason?
I'm not that sure (I'm not a mathematician nor English mother-tongue so sorry for typos and so on...). If I'm wrong let me know! It'll be a chance to learn!!
?
What the heck sorry i don’t speak math like crazy
@@connergamerguy7523 Neither I do!! I don't know how to explain it better -_-
Lol
Wow, that's Vedic Math a way of learning-earning
Vedic math is awesome
@@tecmath agreed with you 100%
It's arithmetic trick
thanks for doing your thing josh ❤
This is brilliant. Can't wait to impress my family.
Thank you so much, this video helped me a lot. However, what if you need the answer with the decimal instead of the remainder? Could you help with that? Please? I loved this video and same with all the others, tysm!
You can just keep going with this method past the 1s column into the decimals. The math is exactly the same. (If you need 3 decimal places, it's effectively like taking the number you are dividing and multiplying it by 1000 and then taking your answer to the division and shifting the decimal over 3 places to "undo" the multiplication by 1000.) Just stop when you have enough decimal places and make sure your remainder isn't rolling over.
@@patrickpablo217 Ohk tysm!
As messy as the normal method is I probably still prefer it over this but each to they're own I guess. Thanks for teaching it anyway.
The main thing that sumped me here was the remainder business. I remember having to write a little R and a number after it, and never really understanding why they did that.
This method can easily dispense with the whole remainder thing by appending a couple decimal places and performing a couple more sub-calculations until the R becomes a decimal
I won't say this is less time consuming or takes up less paper, but it is more direct than the method I learned which was a sort of duck hunt where you end up with a lot of little side calcualtions to figure out how many times the small number goes into some portion of the big number. I don't think I would use this process unless I absolutely had no calculator handy. In that case, I would prefer this method over the longhand method I learned!
Big tumbs upp
Nice.
You really helped me thanks
How do you handle when subtracting results in a negative number? For example, I've been making up examples and am stuck with 53 (so round to 5) dividing into 31389. In my 2nd step, I get 13-18 = -5 which has put me at a standstill
@@craighunter9082 I agree with you but it's worse than that. I have a conundrum i just hit which I'd like you to look at. I tried 37648 / 43. Wether i use the method shown in this video or my old school taught method, I get the same problem. My first sum is 37/4 = 9. No matter what i do now i can't impact on the 9. The answer is going to be 900 and something, correct? My calculator tells me the answer is 875...... so my old school method is wrong as is this one in the video. Care to tell me what I'm doing wrong, please? Indeed as i proceed, i end up with -ve numbers. What is -11/4, which is the calculation for the second digit? Actually, my old method isn't wrong. 43 won't go into 37 so it has to go into 376 which does become an 8 something figure. Either way I'm confirming your point that this techmath method is flawed.
@@fedup3449 just traditional way...easy n simple
I realize it's like a year later and doesn't matter but to answer the OP: in your 2nd step you should have been adding the 18 since the +3 over 50 isn't negative. That said this method seems to be more work than it saves
It looks like Fonze said you're trying to divide by 47 with the multiplier for 53.
@@craighunter9082 I'm running into alot of problems, i guess because I rounded 10 AND I'm using negatives.
My problem is 6930 / 60. I did 6930 / 5 (-10)
1. 5 goes into 6 once, No remainder.
2. 1 × -10 = -10+19= 9, 4 remainder
3. 1× -10= -10+43= 33, 3 remainder
4. 6× -10= -60+30= -30
Here I tried to subtract 4 from the previous digit, so
4. 2× -10= -20+30= 10, 0 remainder
I had written 1162, but after subtracting the 4 it should become 1122. The Answer is actually 115 with a remainder of 30. I'm not sure what to do when your number ends is zero either.
You don't have to respond. This is just another example of a problem it doesn't work well.
This method seems quicker but it seems harder to remember - and I can see learners struggling with it.
Absolutely genius
I tried to use this for 848/16 and it just won't work
Sir(/or anyone who knows)what should we do when the result is negative , and im having trouble with dividing large numbers with smaller numbers ,67961÷23 , first we have consider and -3, but initially we divide the first digit 6 by 2 and get 3 but the answer is in 2954.82 range ? What do we do?
In your case (dividing by 23) divide by 2, but *subtract* 3 times the previous answer digit. Take, for example, 27819/44 (his second example) which holds true for any divisor ending in (at least) anything from 1 to 4.
Good method of long division if you understand it 😊thanks.
ah! good old arithmetic stuff
Hi Josh, Does it matter is you got up or down? For example 780/15. So 15 would make it 2 or 1?
Very nice!!
Its 12:30 am and im here to rewire my brain
Hi sir wonderful method. Thank you for letting us know. @ 8.47 you have a reminder of 31. Why din’t you divide it further by 4 and how do we know where to stop the problem and write it as reminder
When your dividing by a number that ends in five is there a rule for that? I did one twice, adding five in one problem, and subtracting five in the other. The one that I added five to was the correct answer. So should you always add instead of subtract numbers that end with five?
DO YOU REALLY EXPECT ME TO DO MATH AT 12 AM?!
im est btw
Yes. That’s when insomniacs work the best. 💪🏼
@@WellOfEchoes of course! i didnt realize. thanks for the info!!
خیلی عالی بودی 💯
Faster than "that way they teach you in school"
I wish you where my teacher in school
I am watching this with context because I don’t even remember how too divide… I will like to Learn this way But I am lost completely
Game changer
interesting example: 314159 / 63 ;)
Make the tech physics for numericals too , we will support you
What about 9108/23?
Can u divide 7735/17, with this trick?
Good
What should i choose if i get a divisor of 55,65,75,etc?
5 & above round up
How would I do 715 divided by 8
I got the whole number but can’t figure out how to get the decimal.
What do you do if one of your digits in your answer is above 9?
For example if you have a 5 for the first digit, but the second digit is something like 12, does the 5 become a 6, and the answer then becomes 62?
And then, when you do that thing with the adding and multiplying, do you multiply by 12 or just 2?
It’s way easier for sure I’ve been using your method but what happens wen u can’t have remainders what do I do?
With the subtraction
What if you get something like 22-45
You add the whole divisor to 22 and take 1 off the previous answer.
@@gboldero1 oh I see thanks
I tried to divide 4567 by 25. Why am I not getting the correct answer. How to proceed?
I literally can't do maths, but this should help.
what do you do if you have to divide by a negative number, it's happened to me a few times. when the difference is negative soemtimes I have to subtract a bigger number.
how would you do 66772 / 32
I had it set up as :
3 (-2) | 66772
but eventually I get to 7-18 and I don't know what to do
how many figures do have to get before using this method
Kindly teach euclid's division Lemma
When it's 38 and I make it 4 put the 2 to the side to multiply with what if the number is 8 cuz you put 4 into 8 2 times no reminder
how do you keep diving to get the decimals? thanks
Great question!!
@@danibilling9435 Add decimal places to the dividend.
For example, when he does 16283 / 38 he'll end up with 428.
After the 428 you add a decimal point.
You then take the last digit of 3 as he does, add the 8 x 2 = 16 to get the 19.
But instead of just having the 19 as the remainder, you continue, by dividing the 19 by 4 to get 3 with 3 re.... wait..... it's supposed to be a 5 with no remaining......
I'm confused. Well forget this.
Can @tecmath explain how to do this without just saying "continue with 0s" or "continue with more decimal points"?
I love your videos. I have a question though, what if I have to put my answer into a decimal number what should I do?
Keep going with zeros after a decimal.
Thank you.😊
I actually love maths more now because I actually know how to do it.
How would you solve 937/33? Since 3x-3=-9, would you write the first digit as a 2?
Teachers: "This is not my way so it's wrong!"
You are certainty a good math teacher; but, the old way is good as long as one knows the time tables WELL.
me: *using calculator if tecmath is correct*
me: NO WAY
tecmath: its the eazziest way to divide than trash caclualtor
me: yessir
how did u get 4
does this work with 100s?
Good question!
I hate running into the next question with long division. this is faster, compact on paper, and easier. Brilliant trick
I get it yes
This works pretty well, but actually carrying it on to work out the decimal value rather than the remainder would be useful
That's simple! Just add two or three decimal places to the number (27819.000) and carry on with Josh's system! You'd do two more calculations to get 27819/44 = 632.25
I tapped in 6034/43 into my calculator and the result was 140.33 to 2dp.
You worked it out as 140 r14...
which answer is correct ?
Well you get back with 140*43 = 6020 + 14 =6034 so the answers ist not false..
The remainder is the fraction out of the divisor that is left.
140 r 14 means 140 + 14/43 = 140.32558139534
when i divide 262,144 by 512 which I know is 512 I can't get this method to work. Does it only work for two digit numbers?
Agree... I don’t know if I like this 🤷♂️ call me a hater
There are three possibilities I can think of.
1. Do the same that the method tells you to, by taking the 512, rounding it down to 510, and then simply dividing by 51, and using the -2 you got to get from 512 to 510.
2. Do the same as above, but there's an extra step in which you then have to take the 51, and then change that to a 5, with an extra -1, but then you have to work out how to incorporate that part with the first part, which is just too mind-boggling to even imagine.
3. You round to the nearest 100, where 512 becomes 500, and you divide by 5, and not 51, and you take -12 as the modifier.
I've been practicing this but am running into an issue with examples such as 64925/23
Once you get to 4 - 9 what are you supposed to do? Borrow from the 3?
Can you explain to me how to divide 50,000/80,000? On the calculator it is .0625 but I don't understand because 50,000 only goes into 80,000 1 time with 30,000 left over?
Do you mean 80,000 into 50,000
50000/80000 is 0 with 50,000 left over. Which is 0.0625.
80000/50000 is 1 with 30,000 left over, which is 1.6
I need your help for the question 64821/38. I've done your method and reach 1705.81, however, the answer is 1705.816. The last remainder is 30+14=44. So 44/4=11. I added 1 to the number before it but I don't understand how there are still numbers to follow. Please help.
The whole number remainder is 31 so this is the fraction 31/38. If you divide the fraction out you get a long decimal number 0.815789473. So the choice you have is to express the remainder as a fraction or decide to how many decimal places you want your answer to be. Some fractions give a finite answer like 1/16 = 0.0625 whilst others give a repeating recurring number like 1/12 = 0.083333 and others seem to go on and on and you don't know if they will repeat or ever end. It is an interesting thing to ponder why this might be? This is why when doing division of large numbers (by three or more digit divisors) I just use the multiples 2 & 5 (with the relevant zeros tagged on) as it quick & easy to do (using repetitive subtraction)
It is not applicable for 6763 divided by 26. This trick gives me 258, answer is 260.?,?
I’m not sure if this works, but I did the last problem slightly different:
308379/92 ...
+8 308379/10
30/10=3
(8+12)/10=3r2
(23+24)/10=4r7
(77+32)/10=10r9
(99+80)/10=17r9
i sittil dont anderstan
And apparently not understanding how to spell also .
Hes simart, spelling by pronouncing
@@LJC4495 Hey ! not everyone speaks English. For some English is only needed when they come across videos like this.
The world isn't centered around English.
He probably has english as his 4th or 5th language.
I too have at least 4 languages. Not counting the languages I am learning. otherwise I'll had to say 6 languages.
Get a life kid.
@eugeniamuriambihu390 Hi This a shortened version of a series of posts (x4 in all) by Techmath. I suggest you watch them all and then have a go at last question. I just tried it and it worked a treat, as they build your knowledge gradually.
I too struggled at first as I like to know ‘why’ rules/hacks work. However I just trusted the process and it worked.
@@RealGigaMind🤓