@@michaelcolbourn6719 It can be both. According to merriam-webster a factiod is 1. an invented fact believed to be true because it appears in print 2. a briefly stated and usually trivial fact. :3~
@@shangerdanger not all numbers that end with 225 are square numbers, for example 2225. In other words, they could put multiple numbers that can look like possibly being the square, for example all of them ending with 225, so unless we can exclude the other numbers, this ending in itself is not enough. Even if it's the first candidate for an answer than many of us will have 🙂
A la vista se resuelve la opcion C 5x5=25, las otras opciones requieren digitos diferentes para reproducir esos numeros en la parte de sus unidades y decenas
This question is a great example of what I call "gifted child math." I looked at the answers and thought, "Oh, yeah. It's probably C. None of the others look like square numbers" with no further thought than that. For most of my time in school, I thought that this sort of intuition meant I was "good at math," and it wasn't until I got to college that I realized that I was actually just good at guessing, and *real* math required being able to prove my result.
Same here, A, B, and E can be ruled out right away just by looking, C and D are the only ones that seem possible, but went with C since the last few digits looked more like a perfect square.
A tried and true test taking method for me in college was, "when in doubt, C your way out." this video has confirmed that method, thanks! Also... C's get degrees ;)
If your classroom had an analog clock with a second hand, you could always divide the clock face into fractions, in this case 12 minutes, and let the clock decide.
@@lellab.8179 Arguably this zeroes at the end argument is right but not trivial. The trick is, if a number ends with an odd number of zeroes, it must either have an odd number of prime factors 2 or an odd number of prime factors 5. From that follows the result. It is a bit tricky to prove this but mainly tedious, not hard. I guess the easy way is to first show that a number ends with exactly as many zeroes as is the _minimum_ of the multiplicities of the 2 and 5 prime factors. Which is e.g. easy to show by induction. From there follows the result. Would be much easier in a prime base :)
that means he reaches a circle with the radius of 1,000. It's pretty fantastic that MILLIONS watch this, when there's censored MSM lying bizzarly about politics, wars and the economy!
Perhaps an illusion it may be, my first imagination of this problem is choosing a square number by at least looking the last 3 digits of these choices. And only 225 is a perfect square, i.e. 15² = 225 Hence, I directly just chose C as the answer. 😬
Multiples of odd powers of 10 like E can't be squares, since 1000 = 2³ · 5³. So, you have to multiply with 10 to get even powers of 2 and 5. So we can eliminate e) All squares of numbers ending with 5 end with 25, which can be proven with any of the binomial formulas: (10n + 5)² = 100n² + 100n + 25 (10n - 5)² = 100n² - 100 + 25 (10n + 5)² = 10n · 10(n+1) + 25 = 100n² + 100n + 25 Thus, we can eliminate d) All squares end on the digit 0, 1, 4, 5, 6 or 9. Thus we can eliminate b) 99.999.999 = 10⁸ - 1. According to the third binomial formula, the difference between two squares is the product of the sum and the difference of their roots: a² - b² = (a + b)(a - b). This product is only 1, when both factors are 1. This is only the case for 1 and 0. So, we can also rule out a) And so be got c) as the correct answer by elimination.
Fun! I did it pretty similarly, except b: It's divisible by 3, which gives 41111111, which can't be divisible by 3 again since it's digits don't add to 0 modulo 3.
@10:15 “All we need to do is choose the value of N that is closest to 10…” Actually, the choice which is closest to an integer MIGHT have the value of the log portion close to 1 (and N, therefore, close to 10), OR it might have the log portion close to 0 (and N, therefore, close to 0). In this particular case, it happens that all the log portions of the choices have a value greater than 0.5, and therefore we can truly look for the one closest to 10-but that was never actually assessed/proven, and it should have been.
Are you saying that he could have looked at the anwser choice closet to the nearest integer down instead of the the closest integer above? Like if it were 0.5 and it would be in between?
@@maxhagenauer24 ... Yes, because log[10] 1 = 0 which is an integer. so had there been numbers close to 1 more work would have been needed. And it can get even worse. Because of the number is 0.1 then the log is -1, and if its 0.01 it is -2, etc.. all of which are integers. But obviously the problem was set up to have all the arguments of the log clustered close to 10 to make it more approachable.
@@adb012 How does log_10 (1) = 0 make any difference? Why would you need to check more if that were the case? 0 is still an integer here. I guess you are right about the negative ones but we know there are not negatives in this problem because they are all greater than 1. The numbers don't even have to be close to 10 to make it work, just greater than 1.
A number ending in a 5 can actually only be a square if the number in the 10s place is a 2. Because all numbers ending in 5 are all 10n+5 where n is an integer. (10n+5)^2 expands to 100n^2+100n+25. 100n^2 and 100n are both multiples of 100. So, adding them together, you still have a multiple of 100 and then you add 25 to it. So, it's going add up in a number that ends 25.
Ok, I've been messing around with prime factorization and squares since I saw this video. Apparently you can solve if you know two rules. 1. A perfect square can't have an odd number in the 10s place unless there's a 6 in the 1s place. (eliminates a, b, and d) 2. A perfect square can't end in an odd number of 0s (eliminates e) Presh's way is probably more natural based on things people are likely to be aware of, but based on my messing around both of the above are true statements. 1 is provable because if you say x=the number you're squaring rounded down to a multiple of 10 and y=the final digit of the number you're squaring the the number is 10x+y (10x+y)^2=100x^2+20xy+y^2. 100x^2+20xy is going to be a multiple of 20, because you're adding two multiples of 20 together so 100x^2+20xy+y^2 is going to be 20n+y^2 for some number "n," all multiples of 20 end with a an even number in the 10s place and a 0 in the 1s place, and y^2 is a single digit number squared. So, it's a multiple of 20 plus either either 0, 1, 4, 9, 16, 25 (or 20*1+5), 36 (or 20*1+16), 49 (or 20*2+9), 64 (or 20*3+4), 81 (or 20*4+1). So, all perfect squares numbers are going to be either 0, 1, 4, 5, 9, or 16 plus a multiple of 20. 16 is the only possibility where you can get to an odd number in the 10s place and it will always leave you with 6 in the 1s place. 2 is provable because a number that ends in an all 0s you can write it as 10^x*y or 2^x*5^x*y, where x is the number of 0s at the end and y is the number with all of its trailing 0s lopped off. If there are an odd number of 0s, x is an odd number. So, you can't have a perfect square unless y is a multiple of both 2 and 5 because you need to multiply by both 2 and 5 at least one more time for the prime factorization to be all even exponents. But a multiple of both 2 and 5 is another way of saying a multiple of 10, which would mean y has to end in a 0 and we defined y as having had all of its trailing 0s lopped off. So, in the prime factorization of a number that ends in an odd number of 0s either 2 or 5 will necessarily be raised to an odd power.
ANY number ending in 5, its square ends in 25. You can also square it by multiplying the number we get by leaving the 5 outside,, with the next number. e.g. 365^2 = 133225. Or you can say that it ends in 25, and multiply 36x37 to get 13322. Put them together to get 133225. So in that case, answer (c) is instanty the correct.
Q2 is a good question. My approach was to start by computing 10 to the power of each one (one of the only things you can do with the info provided...) Then I considered what happens when you add integers to the exponent. If it was a proof question, you missed one detail, which is checking that all of the 'fractional parts' are greater than 0.5 (obviously true, but needs to be mentioned).
Wow, I actually got it right! And I haven't sat in a classroom in decades! By the way, the way I eliminated option B was by adding up the digits. All multiples of three have digits that add up to a multiple of three, and all multiples of nine have digits that add up to a multiple of nine. Since the sum of the digits is 24, which is a multiple of 3, but not of 9, then there is a prime number that only appears as a factor once. (But as for the rest, I solved them exactly as shown!)
These really aren't that hard, so the real question is how much time do you get for each question to pass the Oxford admission test. I would probably waste half a minute just figuring out fastest way to solve, so I would need at least a minute each. If they only give you 30 seconds each, I would probably fail. Then again, these are the type of problems you could easily practice if you had prior tests to examine.
These are also only the first two questions, and definitely some of the easiest. You wouldn't want to spend more than 4 minutes on each off the multiple choice parts of question 1
For the 1st problem, 123,333,333; 713,291,035; and 987,654,000 are divisible by 3, 5, and 1,000 respectively, but aren't divisible by 3²=9, 5²=25, and 1,000²=1,000,000 respectively. Therefore they cannot be square number. And for 99,999,999; notice that the number right after that 100,000,000=10⁸ is a square number. Therefore 99,999,999 cannot be a square number since it's too big and too close to another square number. Therefore 649,485,225 is a square number and (c) is a correct answer. Took me about 10 minutes to solve.
10000 is a square number, and divisible by 1000, but not by 1000^2. 1000 is 5^3*2^3. This is only a rule with numbers which have all of their prime factors with a power of 1, which 1000 definetly isn't.
Whew.. I guessed right on instinct, without really knowing how to solve, except with a calculator. I wonder if could get into Oxford with instinctive guesses or with a 1 in 5 chance of randomly guessing right. Maybe now that I watched the video, I have slightly improved those odds.
Before watching the video: Q1: Square numbers only end in 00 (except 0), 1, 4, 9, 6, or 25. Therefore we can immediately eliminate b) and d). a) is 1 less than (10,000^2). It's not at least (2*10,000-1) less than (10,000^2) and therefore cannot be square. e) factor out (10*10), you're left with a number that ends in ...40, which no square number can end in.
the last 2 digits of every 50 consecutive integer squares follow a pattern which i memorized, so that eliminated A,B,D because the last 2 digits of those numbers are impossible to be squares since theyre not in the pattern. since E is divisible by 100, if E was a square, then E/100 would be a square too. so the last 2 digits of E/100 are 40, but it is impossible for the last 2 digits of a square to be 40, so that just leaves C
And had you memorized that 6494852 specifically was a square? Because your theory would have fallen down with 6494862. Just because the squares of all integers ending in 5 end in 25, it doesnt follow that all numbers ending in 25 are squares.
@coachRoome I was also doing some elimination based on my knowledge. The 25 ending number was the closest and hence I picked it up. I wasn't specifically going for power of 5.
This is the easiest question on this channel. 99, 33, 35, 000 can't be at the end of a square number. 25 can be at the end of a square number. . To prove just write the last two digits of square from 1 to 25. 01, 04, 09, 16, 25, 36, 49, 64, 81, 00, 21, 44, 69, 96, 25, 56, 89, 24, 61, 00, 41, 84, 29, 76, 25 from now on numbers will repeat backwards. So these are the only numbers which can be at the end of any square number.
i wouldve had no idea how to solve the first one but i recognized the answer from either a vsauce or veritasium video i watched recently and guessed correctly from that
For the first question since they didn't specify on which set of numbers the square operation is performed I assume it's the set of real numbers and then all of the answers are correct. Every number is the square of it's square root.
i think question 3 from your source is way more intresting, i was able to solve it and it was pretty hard but i couldn't solve without a calculator, you should make a video about it too
For a, factor out 9, we have 11111111, which is congruent to 8 mod 9 and no square number can have this property, so a is out; for b, no square can end in 3, so b is out; for d, any square that ends in 5 must be a multiple of 25, so d is out; and finally for e, any square that ends in 0 must end in an even amount of consecutive 0s, so e is out. So the answer must be c, even though I have absolutely 0 idea about its square root.
I went with the SWAG method (simple, wild a** guess) to solve. I looked at the list of answers, and chose 'C'. I then used my calculator to verify my answer, and found that I was correct. Unlike the author, I did not go through all of the permutations to calculate the answer. I simply realised that 225 was the square of 15. I was lucky this time.
I used a calculator. Without a calculator, it is possible to solve with a lot of arithmetic and time. For example, 27,000^2 = 729,000,000. To get to the next lowest square number, you subtract 2*27,000 - 1, meaning you subtract 53,999. 729,000,000 - 53,999 = 728,946,001, and the square root of that is 26,999. The square root of 713,291,035 is between 26,707 and 26,708, so it would have taken 27,000 - 26,707 = 293 times of going down one square number at a time until you get a square number less than 713,291,035 without getting 713,291,035 as a square number first. The same procedure could be used for other choices starting with big square numbers.
a) congruent to -1 mod 4, squares can only congruent to 0 or 1 mod 4. b) divisible by 3 but not by 9 d) divisible by 5 but not by 25 e) divisible by 100 but not by 10000
I noticed an overcomplication in the solution for the first question. When finding the perfect square you are able to reduce the possible answers to just (c) and (e) by consideration of the last two digits (the other 3 cannot be perfect squares based on the last two digits regardless of the remaining digits). After that (e) can be removed due to the odd number of trailing zeros (no perfect squares can have an odd number of trailing zeros). So really there are only 2 logic steps needed, and you only need to examine the last 2 digits to eliminate the first 3, and the final test only needs to know the last 4 digits (yet in the explanation shown digits beyond the last 4 were examined).
Yes, but that requires a "list of possible two-digit endings for square numbers". You can not help knowing this, most people will have to think about it a little bit. And THAT's the interesting video. 😁
1. I’ve seen in a veritasium vid that any power you give to five will add a digit onto 5 mutiplied by the previous power, so I knew it was c because it ended in 225. 2. I didn’t have a paper but if you plug it in I’m sure it’s quite easy to spot These questions aren’t that hard, they probably give them like a very short time for each question
So, I'm not entirely sure with US exams, I thought people were allowed to use calculators here in Australia. If that's the case then there is absolutely No difficulty solving that. If not, you can eliminate e) by try to divide it with 10,000; eliminate d) by divide it with 25, a) can be eliminated by (11,111,111) divide by 121, with b) you try divide it by 9. By the way, without really looking into any details, you'd probably have guessed c) is correct because 225 is 15*15,so that's going to be a good start point.
I knew immediately for the 1st question that it had to be c or e because of the last 2 digits. Then it was simple to guess c. There are only about 20 or so different 2 digit numbers that squares can end in
Thanks for sharing another interesting video! For the first problem: e) is a multiple of 1000, so it cannot be a perfect square. Only perfect squares mod 4 have remainder of 0 and 1. This eliminates all but b) and c). Once you divide b) by 3 and do the test again, you are left with c).
If there are an odd number of trailing zeroes, the number cannot be a perfect square. This is easily seen by dividing by the perfect square 100, eliminating two zeroes, and repeating the division by 100 until one zero is left. Hence, a perfect square cannot end in an odd number of zeroes. For a fun theorem I just thought up: In which number bases b (b would be an integer > 1) are there no perfect squares which have an odd number of trailing zeroes? This would be true if and only if b is square-free, for which the proof is left as an exercise for the reader. By this theorem, there are no perfect squares with an odd number of trailing zeroes in base 6, 10, or 15 (all have a square-free factorization). However, perfect squares with an odd numbers of trailing zeroes exist in base 4 (for example: 10), 8 (ex: 20), 9 (ex: 40), 12 (ex: 30), and 16 (ex: 40).
So high-school me wouldn't have gotten in to Oxford, but me after college would nail it. Boomers would've been better at question 2 since they learned on slider-rulers, and younger kids usually skip that.
I would just randomly guess that c) is the correct answer because it looks like a square of a number ending in 5 and the others don't look like squares, although I don't know enough number theory to know why. btw, given a concatenation operator (+) such that e.g. 12(+)34 = 1234, then x(+)5 squared is x(x+1)(+)25. e.g. 25 squared = (2 * 3)(+)25 = 625. I learned that as a kid and it's a fun party trick (for particularly nerdy parties), but never actually knew why it worked.
A trick I use for the first problem is just look at the first numbers in each option and check if they are perfect square. I don’t know if this method is reliable though…
Can you do three of these in a row, or do you need the step by step procedure that isn't going to confuse your outcome as an answer to the question that launched pass the ruler. Did you know you should be able to answer every math question by using a ruler. It's astronomical ability to use it with paper and pencil will help but I bet you've never experienced any of this as a help to an amazing outcome that delivers an accuracy.
I did the first problem from the top of my head: Any perfect square ends in 0, 1, 4, 5, 6 or 9, which immediately eliminates answer b. (Any number ending with any number of zeros)² ends with an even number of zeros, answer e eliminated. Answer a eliminated exactly like Presh explained. My feeling said (Any number ending with 5)² ends with 25, but 'my feeling' isn't a valid proof. Any number ending with 5 can be written as 10x+5. (10x+5)²=100x²+100x+25. 100(x²+x) always ends with 00. Adding 25 and the number always ends with 25. Feeling proven, answer d eliminated, answer c is the correct answer. Second problem: Oh, dear. Basic rules, almost forgot them, it's been a while. But as soon as I saw it, I knew where it was going. Nice solution.
Square numbers don`t end with 2, 3, 7 or 8 if they end with 5, it`s always 25 if the final digit is 9, next to is is even number or 0 if it ends with 0, it`s even sum of 0s done 2 seconds :)
At the end, to be sure that we have the right answer, we should also check that the largest number is closer to 10 than the smaller number is to 1, because otherwise we could have a number that rounds down closer. Now, 10/9.8 > 8.4/1, so yes, it is, but we should check that for completeness.
it's not _quite_ that simple. The log function is very far from linear, so you would have to actually look at the function's values if, say, you were comparing log (1.05) and log (9.9). The log function is much steeper close to 1 than it is close to 10, so the latter value is closer to an integer even though its argument is further away. log(1.05) = .021..., while log (9.9) = 0.995...
@@rickdesper Yes, it would be easy to argue that the other values were not going to round down, as they were much too near to the next multiple of 10. My point was just that some argument to this effect needs to be made in order for us to be sure we have found *all* of the solutions as asked for.
My approach to the first problem was to say that any squared number will always end in a string of digits that are also a square number, which rules out everything but C, I'm not sure if this is a correct approach though
for the first question, perfect squares are either 0 or 1 mod 4, so that leaves option C and E, and by your observations E cannot be a perfect square, so that leaves C as our perfect square
In the first question, C looked like the obvious choice due to that 25 ending. I ruled out E because none of the multiples of 1000 from 1000 to 9000 are square, i.e. it has to be a multiple of 10000 and thus end in 4 zeroes, not 3. A and B were definitely ruled out from the start. It came down to a choice between C and D, and C just looked more like a square number due to, once again, ending in 25, as 225 and 625 do.
From the thumbnail: The answer has to be C, because a square number cannot be congruent to 3 (mod 4), nor can it be equal to (a non-multiple of 5) * (5^3) (because a square would need an _even_ number of instances of the factor 5). EDIT: Oops, a little mistake! Option B is not congruent to 3 (mod 4), but a square number cannot end in a 3 .
1st question A is made of an even number of 9s. It cannot be, because 1 more is square. E has an odd number of zeros at its end. D ends in 35. B ends in 3. C is correct.
It's weird to me that nobody in the comments is talking about 2:47. How is it obvious that the last digits of the squares of 0-9 are the only possible last digits of any square number? I had to look up an actual proof of that to understand it
Think about the process of long multiplication. The last digit of the result is just the product of the last digits of the two numbers because every row except the first has no ones digit.
No it's pretty obvious for most people like me at least because I was taught that in my class by my maths teacher when we had just learnt about square for the first time
For the first problem I have 2 observations: 1. for c) and d) if X^2 ends in 5 => X ends in 5 => X^2 ends in 25 Demonstration is as follows X ends in 5 => X = Y + 5 where Y ends in 0 X^2 = (Y+5)^2 = Y^2 + 2*Y*5 + 25 = Y^2 + 10*Y + 25, but Y^2 ends in 00 and 10*Y ends in 00, because Y ends in 0 => X^2 ends in 25. So for a number that ends in 5 to be a perfect square it has to at least end in 25. ALWAYS. 2. For e) if X has n number of ZEROs at the end =>X^2 will have 2n number of ZEROs at the end, in any case an even number. Since e) ends 3 zeros it can't be a square.
I did it (mostly) by a different method. Firstly note that if a = b^2 then a = b^2 mod n for any n - so we can just list all the squares mod n and see if (a mod n) is on the list. Modulo 2 never helps (all values are possible squares) and modulo 3 doesn't help us in this case since all values are 0 or 1 mod 3. Modulo 4 the perfect squares are 0^2 = 0, (±1)^2 = 1 and 2^2 = 0. This rules out 99,999,999 since it is -1 mod 4 (for divisibility by 4 you can ignore the multiples of 100 since 100 is divisible by 4, so it's just 99 which is 100 - 1 which is -1 mod 4) It also rules out 713,291,035 by the same logic since 35 = -1 mod 4 Modulo 5, the perfect squares are 0^2 = 0, (±1)^2 = 1 and (±2)^2 = 4. This rules out 123,333,333 since it is 3 mod 5 I ruled out 987,654,000 by noting that any square must have an even number of zeros at the end (which is similar to the logic in the video but less rigorous). Technically you could use the same method we did above but you'd need to go as high as mod 13 but off the top of my head I don't know any tricks for computing values mod 13
All numbers that end in 5 square to something that ends in 25. The first 2 digits, 64 (8 × 8) were also a clue. I immediately guessed the answer without any significant calculations.
@@rubiks6 Huhwot? An even root? You're kidding me, right? So the square root of a large integer beginning with 81 may occasionally start with a 3 , because 3 is an "even root" of 81 ? (Note: the square root of a large integer beginning with 81 starts either with a 9 or with 28 ; never with 3 .)
First problem: If n is a perfect square there must be some integer k such that k is the square root of n k must have some prime factorization a*b*c*d… where it is possible that any pair of variables is equal, that is it’s possible a=b, etc but it’s also possible they’re not equal Since n=k^2, the prime factorization of n is a*a*b*b*c*c*d*d… that is a^2*b^2*c^2*d^2… that is if p is a prime factor of n, p^2 is a factor of n (d) is divisible by 5. Using the divisibility rule for 25 the answer is not d (b) is divisible by 3. Using the divisibility rule for 9 the answer is not b e is a bit more complicated but essentially if something ends in 0 it has 2 and 5 in its prime factorization, therefore if something ends in 0 and is a perfect square it must have 2 and 5 each an even number (other than zero) of times in its prime factorization. Since e ends in an odd number of zeroes this does not hold so the answer is not e The answer is not a because if we add 1 to a it becomes a perfect square as it will be an integer ending in an even number of 0s (same logic as above paragraph). Consecutive integers of this magnitude are obviously not both perfect squares as the distance between perfect squares is always increasing as the magnitude of the perfect squares increases. Therefore a is wrong This leaves only one option, c must be a perfect square assuming the statement in the question is correct
@@rubiks6unfortunately I am not well versed in number theory. Although it seems obvious now and is probably second nature to people with a math education, it was not apparent to me that “perfect square has every prime factor an even number of times” as was claimed in the video so I had to convince myself of this first. I hope to improve my intuition through solving problems presented in these videos so I can one day be as good at math as some of the other viewers!
www.youtube.com/@MindYourDecisions ALL 5 choices are square numbers, since EACH choice is the square of a not-necessarily integer SQUARE ROOT. So that the ROOT must be an INTEGER is an invalid ASSUMPTION. The issue is the word "exactly" in the question. So a prerequisite is determining under which conditions THAT word (in its meaning of "no less and NO MORE than") is valid in determining the answer.
For the 2nd question, if you had log_10(1.2) & log_10(9.8), which would be closest to an integer? My guess is still 9.8 as the halfway point in log_10 is around 3.16 (sqrt(10)) but they are definitely close and I want an example that has this possibility.
Yes it was fun, but unfortunately Presh didn't mention, that the correct answer heavily depends on the metric. If you are in the discrete metric obviously all points have the same distance 1 to an integer. If d(x,y)=|arctan(x)-arctan(y)| is the distance, then d) is the right answer. But maybe the test says, which distance is to choose.
Hello Prash sir, can you help me to fined out the easiest and time saving solution for following problem? Three taps P, Q and R can fill the tank in 40 mins, 80 mins and 120 mins respectively. Initially all the taps are opened. After how much time (in mins) should the taps Q and R should closed so that the tank will completely filled in half an hour? Please help me to find out the easiest and time saving solution for above.
To eliminate the b choice you can also see that you can divide it by 3 but when you do, you can't divide it by 3 again. So it only has the prime factor 3 once and therefore is not a square number
c - Reasoning: a: 10^8-1 is one away from a square number and hence not square b: square numbers don't end in 3 c: correct answer d: if divided by 5 the resulting number ends in 7 and is therefore not divisible by 5 -> 5 is an unpaired divisor e: 1000 is not a square number, so any multiple of 1000 (aside from 1000 times another square number) cannot be a square number
What are you talking about? Is this about the second question with the logarithms? If so: how would you compare the mantissas between the five answer options if you don't have a calculator to calculate logarithms and you don't factor out the nearest power of 10?
@@yurenchu Yes, it's about the second question. You don't have to compare the mantissas (which the question prohibits anyway), just that the mantissa of log(96) = mantissa of log(9.6), so you can ignore where the decimal point is.
@@senshtatulo You'll somehow have to compare the mantissas (whether directly or indirectly), otherwise you can't tell which option is closest to an integer. And how would you compare the mantissas of, say, log(9), log(98) and log(972) against each other? How would you convince a viewer which option is closest to an integer?
@@senshtatulo But in order to be able to solve the puzzle, we have to compare 9 , 98 , and 972 (or their log_10 values) with _each other_ , not with (respectively) 10, 100 , and 1000. Moreover, if I'd apply your suggestion, I'd find that the difference between 9 and 10 is only 1 , while the difference between 972 and 1000 is 28 ; so would that mean that log_10(9) is closer to an integer than log_10(972) is?
Even without any knowledge of number theory or its existence at all, you can solve this problem in a second if you have looked at the two-digit “endings” of square numbers out of sheer curiosity. These are 00, *e*1, *e*4, *e*9, *o*6 and 25.
I eliminated b, d, and e immediately. Then I divided 99999999 by 9, but miscounted the digits. Besides 99999999 being one less than a square, 11111111≡8 (mod 9), and 8 is not a quadratic residue mod 9.
My most interesting math factoid is that if you square the number 111,111,111, the result is 12,345,678,987,654,321.
11² = 121, 121² = 14641
111² = 12321
1111² = 1234321
... and so on
edit:
22² = 484, 484² = 234(2)56
202² = 40804
222² = 48884 - 20²
33² = 989 + 100
88² =7788
... ▼to be continued below▼
cool
That's not a factoid it's just a true fact.
Mind blown 0_0
@@michaelcolbourn6719 It can be both. According to merriam-webster a factiod is 1. an invented fact believed to be true because it appears in print 2. a briefly stated and usually trivial fact. :3~
Btw, for anyone wondering, in the 1st question, 649,485,225 is a square number of 25,485.
And its prime factors are 3, 5 and 1699. (all of them squared)
shouldn't this be obvious by looking at it? because it ended in 225 and the others didn't end in a square type number
@@shangerdanger not all numbers that end with 225 are square numbers, for example 2225. In other words, they could put multiple numbers that can look like possibly being the square, for example all of them ending with 225, so unless we can exclude the other numbers, this ending in itself is not enough. Even if it's the first candidate for an answer than many of us will have 🙂
A la vista se resuelve la opcion C 5x5=25, las otras opciones requieren digitos diferentes para reproducir esos numeros en la parte de sus unidades y decenas
It must be (c).
a, b, & d cannot end with from a square. Any square ending in 0 must have an even number of 0s, so it leaves only (c).
This question is a great example of what I call "gifted child math." I looked at the answers and thought, "Oh, yeah. It's probably C. None of the others look like square numbers" with no further thought than that. For most of my time in school, I thought that this sort of intuition meant I was "good at math," and it wasn't until I got to college that I realized that I was actually just good at guessing, and *real* math required being able to prove my result.
Same. I legit said c bc it ended in 225 which is square. (15 sq). And thought that everything else would be unlikely to occur.
Me too. I thought c was the only one that looked like a square.
Same here, A, B, and E can be ruled out right away just by looking, C and D are the only ones that seem possible, but went with C since the last few digits looked more like a perfect square.
@@nolankryptonite3167 d can be ruled out immediately because if a square is divisible by 5 it must be by 25 so it must end in 00, 25, 50, 75
Same here... I could eliminate a and e. And c looked the most squary. ;)
A tried and true test taking method for me in college was, "when in doubt, C your way out." this video has confirmed that method, thanks!
Also... C's get degrees ;)
If your classroom had an analog clock with a second hand, you could always divide the clock face into fractions, in this case 12 minutes, and let the clock decide.
My son taught me that method when he was in high school. If you don't know, he said, the answer is always 'C'. Appears to work.
Always guess C for "correct" is also a thing where I live
For (e) the method I prefer is add the digits. They add up to 39, which is a multiple of 3 but not a multiple of 9.
This but for b)!
With all the 3s in there its already served on a silver tablet
Why go to the effort of adding the digits when you're staring at 3 0's at the end?
@@pietergeerkens6324 sure! But see the previous comment.
@@pietergeerkens6324Exactly my reasoning. Odd number of zeros at the end? Not a square.
@@lellab.8179 Arguably this zeroes at the end argument is right but not trivial.
The trick is, if a number ends with an odd number of zeroes, it must either have an odd number of prime factors 2 or an odd number of prime factors 5. From that follows the result. It is a bit tricky to prove this but mainly tedious, not hard.
I guess the easy way is to first show that a number ends with exactly as many zeroes as is the _minimum_ of the multiplicities of the 2 and 5 prime factors. Which is e.g. easy to show by induction. From there follows the result.
Would be much easier in a prime base :)
Congratulations on 3.14 million subscribers
Do you think the 3,141,592,653rd subscriber will get a prize?
@ yes
that means he reaches a circle with the radius of 1,000.
It's pretty fantastic that MILLIONS watch this, when there's censored MSM lying bizzarly about politics, wars and the economy!
@@Duck_Bidiyani A pie, perhaps?
@@davidspencer3726 Nevermind, he probably won't ever reach that many subcribers LOL
Perhaps an illusion it may be, my first imagination of this problem is choosing a square number by at least looking the last 3 digits of these choices.
And only 225 is a perfect square,
i.e. 15² = 225
Hence, I directly just chose C as the answer. 😬
I also did though the same thing
me too.
It's a little lucky, but it is the right answer, the last 3 numbers themselves don't need to for a square number.
All perfect squares that end in 5 end in 25. That is correct. Not 225 though, e.g. 25^2 = 625.
Multiples of odd powers of 10 like E can't be squares, since 1000 = 2³ · 5³. So, you have to multiply with 10 to get even powers of 2 and 5. So we can eliminate e)
All squares of numbers ending with 5 end with 25, which can be proven with any of the binomial formulas:
(10n + 5)² = 100n² + 100n + 25
(10n - 5)² = 100n² - 100 + 25
(10n + 5)² = 10n · 10(n+1) + 25 = 100n² + 100n + 25
Thus, we can eliminate d)
All squares end on the digit 0, 1, 4, 5, 6 or 9. Thus we can eliminate b)
99.999.999 = 10⁸ - 1. According to the third binomial formula, the difference between two squares is the product of the sum and the difference of their roots: a² - b² = (a + b)(a - b). This product is only 1, when both factors are 1. This is only the case for 1 and 0. So, we can also rule out a)
And so be got c) as the correct answer by elimination.
Fun! I did it pretty similarly, except b:
It's divisible by 3, which gives 41111111, which can't be divisible by 3 again since it's digits don't add to 0 modulo 3.
@10:15 “All we need to do is choose the value of N that is closest to 10…”
Actually, the choice which is closest to an integer MIGHT have the value of the log portion close to 1 (and N, therefore, close to 10), OR it might have the log portion close to 0 (and N, therefore, close to 0). In this particular case, it happens that all the log portions of the choices have a value greater than 0.5, and therefore we can truly look for the one closest to 10-but that was never actually assessed/proven, and it should have been.
Was just about to say this, you should first say that 4 ^ 2 = 16 > 10, hence 10 ^ (1 / 2)
Are you saying that he could have looked at the anwser choice closet to the nearest integer down instead of the the closest integer above? Like if it were 0.5 and it would be in between?
@@maxhagenauer24 ... Yes, because log[10] 1 = 0 which is an integer. so had there been numbers close to 1 more work would have been needed. And it can get even worse. Because of the number is 0.1 then the log is -1, and if its 0.01 it is -2, etc.. all of which are integers. But obviously the problem was set up to have all the arguments of the log clustered close to 10 to make it more approachable.
@@adb012 How does log_10 (1) = 0 make any difference? Why would you need to check more if that were the case? 0 is still an integer here. I guess you are right about the negative ones but we know there are not negatives in this problem because they are all greater than 1. The numbers don't even have to be close to 10 to make it work, just greater than 1.
@@maxhagenauer24 Well, what is closer to an integer? log_10 (9.8) or log_10 (1.02)?
A number ending in a 5 can actually only be a square if the number in the 10s place is a 2. Because all numbers ending in 5 are all 10n+5 where n is an integer. (10n+5)^2 expands to 100n^2+100n+25. 100n^2 and 100n are both multiples of 100. So, adding them together, you still have a multiple of 100 and then you add 25 to it. So, it's going add up in a number that ends 25.
Also, the number in the 10s place can't be odd unless number in the 1s place is 6, So, a, b, and d could all be eliminated with that 1 rule.
Ok, I've been messing around with prime factorization and squares since I saw this video. Apparently you can solve if you know two rules.
1. A perfect square can't have an odd number in the 10s place unless there's a 6 in the 1s place. (eliminates a, b, and d)
2. A perfect square can't end in an odd number of 0s (eliminates e)
Presh's way is probably more natural based on things people are likely to be aware of, but based on my messing around both of the above are true statements.
1 is provable because if you say x=the number you're squaring rounded down to a multiple of 10 and y=the final digit of the number you're squaring the the number is 10x+y (10x+y)^2=100x^2+20xy+y^2. 100x^2+20xy is going to be a multiple of 20, because you're adding two multiples of 20 together so 100x^2+20xy+y^2 is going to be 20n+y^2 for some number "n," all multiples of 20 end with a an even number in the 10s place and a 0 in the 1s place, and y^2 is a single digit number squared. So, it's a multiple of 20 plus either either 0, 1, 4, 9, 16, 25 (or 20*1+5), 36 (or 20*1+16), 49 (or 20*2+9), 64 (or 20*3+4), 81 (or 20*4+1). So, all perfect squares numbers are going to be either 0, 1, 4, 5, 9, or 16 plus a multiple of 20. 16 is the only possibility where you can get to an odd number in the 10s place and it will always leave you with 6 in the 1s place.
2 is provable because a number that ends in an all 0s you can write it as 10^x*y or 2^x*5^x*y, where x is the number of 0s at the end and y is the number with all of its trailing 0s lopped off. If there are an odd number of 0s, x is an odd number. So, you can't have a perfect square unless y is a multiple of both 2 and 5 because you need to multiply by both 2 and 5 at least one more time for the prime factorization to be all even exponents. But a multiple of both 2 and 5 is another way of saying a multiple of 10, which would mean y has to end in a 0 and we defined y as having had all of its trailing 0s lopped off. So, in the prime factorization of a number that ends in an odd number of 0s either 2 or 5 will necessarily be raised to an odd power.
In problem 1 I got the right answer only from vibes alone.
May the multiple guess be ever in your favor
ANY number ending in 5, its square ends in 25. You can also square it by multiplying the number we get by leaving the 5 outside,, with the next number.
e.g. 365^2 = 133225. Or you can say that it ends in 25, and multiply 36x37 to get 13322. Put them together to get 133225. So in that case, answer (c) is instanty the correct.
Q2 is a good question. My approach was to start by computing 10 to the power of each one (one of the only things you can do with the info provided...) Then I considered what happens when you add integers to the exponent.
If it was a proof question, you missed one detail, which is checking that all of the 'fractional parts' are greater than 0.5 (obviously true, but needs to be mentioned).
For (b), I actually noticed that it was divisible by 3, and not by 9, thus not a square. It's essentially the same method as Presh used for (d).
Wow, I actually got it right! And I haven't sat in a classroom in decades!
By the way, the way I eliminated option B was by adding up the digits. All multiples of three have digits that add up to a multiple of three, and all multiples of nine have digits that add up to a multiple of nine. Since the sum of the digits is 24, which is a multiple of 3, but not of 9, then there is a prime number that only appears as a factor once. (But as for the rest, I solved them exactly as shown!)
As for the second question, I didn't even attempt it. I don't even remember if I ever learned logs.
I didn’t know any of the other reasonings beforehand, but I still sussed out the correct answer-because I knew that one rule about squares of five
I didn't think to check the possible endings of a perfect square for option b. If you sum the digits you can see that it is divisible by 3 but not 9.
Nice and clean solutions. Just one tiny thing: at 0:48 you misspoke, transposing the 45 and 54.
The number theory about perfect squares have to have prime-number squares was new to me. Makes eliminating those ending in a five pretty quick.
These really aren't that hard, so the real question is how much time do you get for each question to pass the Oxford admission test. I would probably waste half a minute just figuring out fastest way to solve, so I would need at least a minute each. If they only give you 30 seconds each, I would probably fail. Then again, these are the type of problems you could easily practice if you had prior tests to examine.
Eh you have 2 and a half hours for the MAT
These are also only the first two questions, and definitely some of the easiest. You wouldn't want to spend more than 4 minutes on each off the multiple choice parts of question 1
You just passed pi million subs!
Congratz!
I'm Glad I managed to get both of them correct :)
Very clever thinking on question 1. I was easily able to get question 2, but I would have never found the non square number.
For 123,333,333, we can add the digits to get 24, so the large number is divisible by 3 but not by 9, and also not a perfect square.
Good exercises.
The second question reminded me of working with slide rules.
0:52 Reading error?
For the 1st problem, 123,333,333; 713,291,035; and 987,654,000 are divisible by 3, 5, and 1,000 respectively, but aren't divisible by 3²=9, 5²=25, and 1,000²=1,000,000 respectively. Therefore they cannot be square number. And for 99,999,999; notice that the number right after that 100,000,000=10⁸ is a square number. Therefore 99,999,999 cannot be a square number since it's too big and too close to another square number. Therefore 649,485,225 is a square number and (c) is a correct answer.
Took me about 10 minutes to solve.
No square number ends in 3, no square numbers end in a 5 without 2 in front, no square number ends with an odd amount of zeros
10000 is a square number, and divisible by 1000, but not by 1000^2. 1000 is 5^3*2^3. This is only a rule with numbers which have all of their prime factors with a power of 1, which 1000 definetly isn't.
@willzhao5889 yeah, I did notice some of that afterward, but that's my first solution of solving this
Minutes or seconds?
How does that mean they aren't perfect squares though?
bro right now has 3.14 millon subs which is like pi times million . Remark for a math youtuber
Whew.. I guessed right on instinct, without really knowing how to solve, except with a calculator. I wonder if could get into Oxford with instinctive guesses or with a 1 in 5 chance of randomly guessing right. Maybe now that I watched the video, I have slightly improved those odds.
Before watching the video:
Q1:
Square numbers only end in 00 (except 0), 1, 4, 9, 6, or 25. Therefore we can immediately eliminate b) and d).
a) is 1 less than (10,000^2). It's not at least (2*10,000-1) less than (10,000^2) and therefore cannot be square.
e) factor out (10*10), you're left with a number that ends in ...40, which no square number can end in.
These were fun!
I like your channel and I have watched all of your videos. I wonder where is the ending theme song?
Beautiful Maths ❤ thx😊
the last 2 digits of every 50 consecutive integer squares follow a pattern which i memorized, so that eliminated A,B,D because the last 2 digits of those numbers are impossible to be squares since theyre not in the pattern. since E is divisible by 100, if E was a square, then E/100 would be a square too. so the last 2 digits of E/100 are 40, but it is impossible for the last 2 digits of a square to be 40, so that just leaves C
During my Olympiad test, i had noticed that all the powers of 5 have 25 at the end so I instantly recognised it.
And had you memorized that 6494852 specifically was a square? Because your theory would have fallen down with 6494862.
Just because the squares of all integers ending in 5 end in 25, it doesnt follow that all numbers ending in 25 are squares.
@coachRoome I was also doing some elimination based on my knowledge. The 25 ending number was the closest and hence I picked it up. I wasn't specifically going for power of 5.
This is the easiest question on this channel.
99, 33, 35, 000 can't be at the end of a square number.
25 can be at the end of a square number.
.
To prove just write the last two digits of square from 1 to 25.
01, 04, 09, 16, 25, 36, 49, 64, 81, 00, 21, 44, 69, 96, 25, 56, 89, 24, 61, 00, 41, 84, 29, 76, 25 from now on numbers will repeat backwards. So these are the only numbers which can be at the end of any square number.
i wouldve had no idea how to solve the first one but i recognized the answer from either a vsauce or veritasium video i watched recently and guessed correctly from that
For the first question since they didn't specify on which set of numbers the square operation is performed I assume it's the set of real numbers and then all of the answers are correct. Every number is the square of it's square root.
i think question 3 from your source is way more intresting, i was able to solve it and it was pretty hard but i couldn't solve without a calculator, you should make a video about it too
For a, factor out 9, we have 11111111, which is congruent to 8 mod 9 and no square number can have this property, so a is out; for b, no square can end in 3, so b is out; for d, any square that ends in 5 must be a multiple of 25, so d is out; and finally for e, any square that ends in 0 must end in an even amount of consecutive 0s, so e is out. So the answer must be c, even though I have absolutely 0 idea about its square root.
At the time I saw the video I was impressed that youtube showed that you had 3.14M subscribers :D
I went with the SWAG method (simple, wild a** guess) to solve. I looked at the list of answers, and chose 'C'. I then used my calculator to verify my answer, and found that I was correct. Unlike the author, I did not go through all of the permutations to calculate the answer. I simply realised that 225 was the square of 15. I was lucky this time.
I used a calculator. Without a calculator, it is possible to solve with a lot of arithmetic and time. For example, 27,000^2 = 729,000,000. To get to the next lowest square number, you subtract 2*27,000 - 1, meaning you subtract 53,999. 729,000,000 - 53,999 = 728,946,001, and the square root of that is 26,999. The square root of 713,291,035 is between 26,707 and 26,708, so it would have taken 27,000 - 26,707 = 293 times of going down one square number at a time until you get a square number less than 713,291,035 without getting 713,291,035 as a square number first. The same procedure could be used for other choices starting with big square numbers.
a) congruent to -1 mod 4, squares can only congruent to 0 or 1 mod 4. b) divisible by 3 but not by 9 d) divisible by 5 but not by 25 e) divisible by 100 but not by 10000
That 225 in option C jumped out at me like a Five Nights at Freddy’s jumpscare. I knows me a 15 squared when I sees it.
I noticed an overcomplication in the solution for the first question. When finding the perfect square you are able to reduce the possible answers to just (c) and (e) by consideration of the last two digits (the other 3 cannot be perfect squares based on the last two digits regardless of the remaining digits). After that (e) can be removed due to the odd number of trailing zeros (no perfect squares can have an odd number of trailing zeros).
So really there are only 2 logic steps needed, and you only need to examine the last 2 digits to eliminate the first 3, and the final test only needs to know the last 4 digits (yet in the explanation shown digits beyond the last 4 were examined).
Yes, but that requires a "list of possible two-digit endings for square numbers". You can not help knowing this, most people will have to think about it a little bit. And THAT's the interesting video. 😁
Interesting !
1. I’ve seen in a veritasium vid that any power you give to five will add a digit onto 5 mutiplied by the previous power, so I knew it was c because it ended in 225.
2. I didn’t have a paper but if you plug it in I’m sure it’s quite easy to spot
These questions aren’t that hard, they probably give them like a very short time for each question
So, I'm not entirely sure with US exams, I thought people were allowed to use calculators here in Australia. If that's the case then there is absolutely No difficulty solving that. If not, you can eliminate e) by try to divide it with 10,000; eliminate d) by divide it with 25, a) can be eliminated by (11,111,111) divide by 121, with b) you try divide it by 9. By the way, without really looking into any details, you'd probably have guessed c) is correct because 225 is 15*15,so that's going to be a good start point.
I knew immediately for the 1st question that it had to be c or e because of the last 2 digits. Then it was simple to guess c. There are only about 20 or so different 2 digit numbers that squares can end in
Thanks for sharing another interesting video!
For the first problem: e) is a multiple of 1000, so it cannot be a perfect square.
Only perfect squares mod 4 have remainder of 0 and 1. This eliminates all but b) and c). Once you divide b) by 3 and do the test again, you are left with c).
They're all square numbers.
I don't make the rules.
"is a multiple of 1000, so it cannot be a perfect square" this, as you wrote it, isn't always true, given that 10 ^ 4 = 10 * 10 ^ 3.
If there are an odd number of trailing zeroes, the number cannot be a perfect square. This is easily seen by dividing by the perfect square 100, eliminating two zeroes, and repeating the division by 100 until one zero is left. Hence, a perfect square cannot end in an odd number of zeroes.
For a fun theorem I just thought up: In which number bases b (b would be an integer > 1) are there no perfect squares which have an odd number of trailing zeroes? This would be true if and only if b is square-free, for which the proof is left as an exercise for the reader. By this theorem, there are no perfect squares with an odd number of trailing zeroes in base 6, 10, or 15 (all have a square-free factorization). However, perfect squares with an odd numbers of trailing zeroes exist in base 4 (for example: 10), 8 (ex: 20), 9 (ex: 40), 12 (ex: 30), and 16 (ex: 40).
So high-school me wouldn't have gotten in to Oxford, but me after college would nail it. Boomers would've been better at question 2 since they learned on slider-rulers, and younger kids usually skip that.
Also, if a number divisible by 3 but not by 9 then it's not a perfect square (it eliminates e and b in P1)
I would just randomly guess that c) is the correct answer because it looks like a square of a number ending in 5 and the others don't look like squares, although I don't know enough number theory to know why.
btw, given a concatenation operator (+) such that e.g. 12(+)34 = 1234, then x(+)5 squared is x(x+1)(+)25. e.g. 25 squared = (2 * 3)(+)25 = 625. I learned that as a kid and it's a fun party trick (for particularly nerdy parties), but never actually knew why it worked.
I just calculated the square roots directly to get the answer. Fortunately I could stop at C so it didn't take too long.
hooray solved both questions :D
A is not divisible by 11^2, B isn't by 3^2, D isn't by 5^2 and E/10^2 isn't by 10^2, even though they're all divisible by the bases.
E is divisible by 10^2.
@@johnosullivan675 E/10^2 is divisible by 10 but not 100, so the point stands.
A trick I use for the first problem is just look at the first numbers in each option and check if they are perfect square. I don’t know if this method is reliable though…
Can you do three of these in a row, or do you need the step by step procedure that isn't going to confuse your outcome as an answer to the question that launched pass the ruler. Did you know you should be able to answer every math question by using a ruler. It's astronomical ability to use it with paper and pencil will help but I bet you've never experienced any of this as a help to an amazing outcome that delivers an accuracy.
For part 2 you would also have to make sure that log10 9.8 is closer to 1 than log10 8.4 is to 0 by checking what log10 x = 0.5 solves for.
log_10(x) = 0.5 means x = sqrt(10) = 3.16...
But yes, it's sloppy that he left that out.
8.4 >> sqrt(10) so this is a step that you could skip over (as long as you mention that fact).
I did the first problem from the top of my head:
Any perfect square ends in 0, 1, 4, 5, 6 or 9, which immediately eliminates answer b.
(Any number ending with any number of zeros)² ends with an even number of zeros, answer e eliminated.
Answer a eliminated exactly like Presh explained.
My feeling said (Any number ending with 5)² ends with 25, but 'my feeling' isn't a valid proof.
Any number ending with 5 can be written as 10x+5. (10x+5)²=100x²+100x+25. 100(x²+x) always ends with 00. Adding 25 and the number always ends with 25. Feeling proven, answer d eliminated, answer c is the correct answer.
Second problem: Oh, dear. Basic rules, almost forgot them, it's been a while. But as soon as I saw it, I knew where it was going. Nice solution.
Square numbers don`t end with 2, 3, 7 or 8
if they end with 5, it`s always 25
if the final digit is 9, next to is is even number or 0
if it ends with 0, it`s even sum of 0s
done 2 seconds :)
At the end, to be sure that we have the right answer, we should also check that the largest number is closer to 10 than the smaller number is to 1, because otherwise we could have a number that rounds down closer. Now, 10/9.8 > 8.4/1, so yes, it is, but we should check that for completeness.
it's not _quite_ that simple. The log function is very far from linear, so you would have to actually look at the function's values if, say, you were comparing log (1.05) and log (9.9). The log function is much steeper close to 1 than it is close to 10, so the latter value is closer to an integer even though its argument is further away.
log(1.05) = .021..., while log (9.9) = 0.995...
@@rickdesper Yes, it would be easy to argue that the other values were not going to round down, as they were much too near to the next multiple of 10. My point was just that some argument to this effect needs to be made in order for us to be sure we have found *all* of the solutions as asked for.
I eliminated (b) with a different reasoning but othwise had the same thought process.
Second question is really interesting
What happened with the music at the end? It's missing in the last videos
My approach to the first problem was to say that any squared number will always end in a string of digits that are also a square number, which rules out everything but C, I'm not sure if this is a correct approach though
for the first question, perfect squares are either 0 or 1 mod 4, so that leaves option C and E, and by your observations E cannot be a perfect square, so that leaves C as our perfect square
In the first question, C looked like the obvious choice due to that 25 ending. I ruled out E because none of the multiples of 1000 from 1000 to 9000 are square, i.e. it has to be a multiple of 10000 and thus end in 4 zeroes, not 3. A and B were definitely ruled out from the start. It came down to a choice between C and D, and C just looked more like a square number due to, once again, ending in 25, as 225 and 625 do.
From the thumbnail:
The answer has to be C, because a square number cannot be congruent to 3 (mod 4), nor can it be equal to (a non-multiple of 5) * (5^3) (because a square would need an _even_ number of instances of the factor 5).
EDIT: Oops, a little mistake! Option B is not congruent to 3 (mod 4), but a square number cannot end in a 3 .
Nb : Every integer's square will finish by 0, 1, 4, 5, 6 or 9 (because the first multiplication will always be 1x1, 2x2, 3x3,...).
Why the reupload?
1st question A is made of an even number of 9s. It cannot be, because 1 more is square. E has an odd number of zeros at its end. D ends in 35. B ends in 3. C is correct.
It's weird to me that nobody in the comments is talking about 2:47. How is it obvious that the last digits of the squares of 0-9 are the only possible last digits of any square number? I had to look up an actual proof of that to understand it
Think about the process of long multiplication. The last digit of the result is just the product of the last digits of the two numbers because every row except the first has no ones digit.
No it's pretty obvious for most people like me at least because I was taught that in my class by my maths teacher when we had just learnt about square for the first time
It’s because the last digit is going to give you the last digit, like 1000 is not going to give you the last digit because it’s too big
For the first problem I have 2 observations:
1. for c) and d) if X^2 ends in 5 => X ends in 5 => X^2 ends in 25
Demonstration is as follows X ends in 5 => X = Y + 5 where Y ends in 0
X^2 = (Y+5)^2 = Y^2 + 2*Y*5 + 25 = Y^2 + 10*Y + 25, but Y^2 ends in 00 and 10*Y ends in 00, because Y ends in 0 => X^2 ends in 25. So for a number that ends in 5 to be a perfect square it has to at least end in 25. ALWAYS.
2. For e) if X has n number of ZEROs at the end =>X^2 will have 2n number of ZEROs at the end, in any case an even number. Since e) ends 3 zeros it can't be a square.
Just noticed! 😲😲 You have almost Pi = 3.14M subscribers!!!
Very easy question...sloved in 30 sec
I did it (mostly) by a different method. Firstly note that if a = b^2 then a = b^2 mod n for any n - so we can just list all the squares mod n and see if (a mod n) is on the list.
Modulo 2 never helps (all values are possible squares) and modulo 3 doesn't help us in this case since all values are 0 or 1 mod 3.
Modulo 4 the perfect squares are 0^2 = 0, (±1)^2 = 1 and 2^2 = 0. This rules out 99,999,999 since it is -1 mod 4 (for divisibility by 4 you can ignore the multiples of 100 since 100 is divisible by 4, so it's just 99 which is 100 - 1 which is -1 mod 4)
It also rules out 713,291,035 by the same logic since 35 = -1 mod 4
Modulo 5, the perfect squares are 0^2 = 0, (±1)^2 = 1 and (±2)^2 = 4. This rules out 123,333,333 since it is 3 mod 5
I ruled out 987,654,000 by noting that any square must have an even number of zeros at the end (which is similar to the logic in the video but less rigorous). Technically you could use the same method we did above but you'd need to go as high as mod 13 but off the top of my head I don't know any tricks for computing values mod 13
987654000 = 3 (mod 9) and hence cannot be a square.
@@yurenchugood point; I think I was only looking at prime moduli
All numbers that end in 5 square to something that ends in 25. The first 2 digits, 64 (8 × 8) were also a clue. I immediately guessed the answer without any significant calculations.
"The first 2 digits, 64 (8 × 8) were also a clue."
The square root of option C does *not* start with the digit 8 though.
@@yurenchu - No, but it begins with 2, which is an even root of 64.
@@rubiks6 Huhwot? An even root? You're kidding me, right?
So the square root of a large integer beginning with 81 may occasionally start with a 3 , because 3 is an "even root" of 81 ?
(Note: the square root of a large integer beginning with 81 starts either with a 9 or with 28 ; never with 3 .)
First problem:
If n is a perfect square there must be some integer k such that k is the square root of n
k must have some prime factorization
a*b*c*d… where it is possible that any pair of variables is equal, that is it’s possible a=b, etc but it’s also possible they’re not equal
Since n=k^2, the prime factorization of n is a*a*b*b*c*c*d*d…
that is a^2*b^2*c^2*d^2…
that is if p is a prime factor of n, p^2 is a factor of n
(d) is divisible by 5. Using the divisibility rule for 25 the answer is not d
(b) is divisible by 3. Using the divisibility rule for 9 the answer is not b
e is a bit more complicated but essentially if something ends in 0 it has 2 and 5 in its prime factorization, therefore if something ends in 0 and is a perfect square it must have 2 and 5 each an even number (other than zero) of times in its prime factorization. Since e ends in an odd number of zeroes this does not hold so the answer is not e
The answer is not a because if we add 1 to a it becomes a perfect square as it will be an integer ending in an even number of 0s (same logic as above paragraph). Consecutive integers of this magnitude are obviously not both perfect squares as the distance between perfect squares is always increasing as the magnitude of the perfect squares increases. Therefore a is wrong
This leaves only one option, c must be a perfect square assuming the statement in the question is correct
Wow. You did way too much thinking. The answer was obvious if you know a tiny bit of number theory. (Way less than you used.)
@@rubiks6unfortunately I am not well versed in number theory. Although it seems obvious now and is probably second nature to people with a math education, it was not apparent to me that “perfect square has every prime factor an even number of times” as was claimed in the video so I had to convince myself of this first. I hope to improve my intuition through solving problems presented in these videos so I can one day be as good at math as some of the other viewers!
www.youtube.com/@MindYourDecisions ALL 5 choices are square numbers, since EACH choice is the square of a not-necessarily integer SQUARE ROOT. So that the ROOT must be an INTEGER is an invalid ASSUMPTION.
The issue is the word "exactly" in the question. So a prerequisite is determining under which conditions THAT word (in its meaning of "no less and NO MORE than") is valid in determining the answer.
I assume that when he created the slide he accidentally left out the word "perfect". He clearly said "perfect square" many times..
For the 2nd question, if you had log_10(1.2) & log_10(9.8), which would be closest to an integer? My guess is still 9.8 as the halfway point in log_10 is around 3.16 (sqrt(10)) but they are definitely close and I want an example that has this possibility.
My mind is weird. I knew 100% that it was C based in the number configurations, but could have never explained to you why.
immediately knew 1st one, any square that ends with 5 must have 25 as the first 2 digits
The second problem was fun.
Yes it was fun, but unfortunately Presh didn't mention, that the correct answer heavily depends on the metric. If you are in the discrete metric obviously all points have the same distance 1 to an integer. If d(x,y)=|arctan(x)-arctan(y)| is the distance, then d) is the right answer. But maybe the test says, which distance is to choose.
Hello Prash sir, can you help me to fined out the easiest and time saving solution for following problem?
Three taps P, Q and R can fill the tank in 40 mins, 80 mins and 120 mins respectively. Initially all the taps are opened. After how much time (in mins) should the taps Q and R should closed so that the tank will completely filled in half an hour?
Please help me to find out the easiest and time saving solution for above.
0:39 My answer is C
Now to see if the root number starts with 8
I did 1st one because i studied about perfect square and its rules but i can't do the 2nd one because i'm not in that standard
To eliminate the b choice you can also see that you can divide it by 3 but when you do, you can't divide it by 3 again. So it only has the prime factor 3 once and therefore is not a square number
congrats on the pi! 3.14 million
c - Reasoning:
a: 10^8-1 is one away from a square number and hence not square
b: square numbers don't end in 3
c: correct answer
d: if divided by 5 the resulting number ends in 7 and is therefore not divisible by 5 -> 5 is an unpaired divisor
e: 1000 is not a square number, so any multiple of 1000 (aside from 1000 times another square number) cannot be a square number
d: without dividing, any square ending in five must end in 25
Why did you factor out 10 for most of the options? The mantissa is the same even if you don't do that.
What are you talking about? Is this about the second question with the logarithms?
If so: how would you compare the mantissas between the five answer options if you don't have a calculator to calculate logarithms and you don't factor out the nearest power of 10?
@@yurenchu Yes, it's about the second question.
You don't have to compare the mantissas (which the question prohibits anyway), just that the mantissa of log(96) = mantissa of log(9.6), so you can ignore where the decimal point is.
@@senshtatulo You'll somehow have to compare the mantissas (whether directly or indirectly), otherwise you can't tell which option is closest to an integer. And how would you compare the mantissas of, say, log(9), log(98) and log(972) against each other?
How would you convince a viewer which option is closest to an integer?
@@yurenchu I would compare 9 with 10, 98 with 100, and 972 with 1000.
@@senshtatulo But in order to be able to solve the puzzle, we have to compare 9 , 98 , and 972 (or their log_10 values) with _each other_ , not with (respectively) 10, 100 , and 1000.
Moreover, if I'd apply your suggestion, I'd find that the difference between 9 and 10 is only 1 , while the difference between 972 and 1000 is 28 ; so would that mean that log_10(9) is closer to an integer than log_10(972) is?
C because the square of any number that ends in 5 has 25 at the end.
Even without any knowledge of number theory or its existence at all, you can solve this problem in a second if you have looked at the two-digit “endings” of square numbers out of sheer curiosity.
These are 00, *e*1, *e*4, *e*9, *o*6 and 25.
Remember similar problem on your channel. Thus didn't have much problem solving it
I got as far as eliminating option B using modular arithmetic, then stopped.
i still curious bout the 1st question. Without havin elimination choice, can how we determine exactly the root square of 649,485,225 is 25,485?
I eliminated b, d, and e immediately. Then I divided 99999999 by 9, but miscounted the digits. Besides 99999999 being one less than a square, 11111111≡8 (mod 9), and 8 is not a quadratic residue mod 9.
Option d is correct as we know that digital sum of s perfect square number can only be 1,4,7 and 9. Only option d satisfies