Can you solve these Oxford admissions questions?

[@Brain81505]

Btw, for anyone wondering, in the 1st question, 649,485,225 is a square number of 25,485.

[@pegasoltaeclair0611]

And its prime factors are 3, 5 and 1699. (all of them squared)

[@shangerdanger]

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

[@verkuilb]

@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.

[@erikkonstas]

Was just about to say this, you should first say that 4 ^ 2 = 16 > 10, hence 10 ^ (1 / 2)

[@maxhagenauer24]

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?

[@adb012]

@@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.

[@maxhagenauer24]

@@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.

[@adb012]

@@maxhagenauer24 Well, what is closer to an integer? log_10 (9.8) or log_10 (1.02)?

[@ChrisWilson-gl1ff]

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 ;)

[@PoppySuzumi1223]

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. 😬

[@masterpanda064]

I also did though the same thing

[@ssarmazi]

me too.

[@manudude02]

It's a little lucky, but it is the right answer, the last 3 numbers themselves don't need to for a square number.

[@kicorse]

All perfect squares that end in 5 end in 25. That is correct. Not 225 though, e.g. 25^2 = 625.

[@Nikioko]

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 (11n - 5)² = 121n² - 110n + 25 (10n + 5)² = 10n · 11n + 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.

[@grgfrg7]

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

[@notsus8537]

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.

[@willzhao5889]

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

[@marcellmate425]

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.

[@notsus8537]

@willzhao5889 yeah, I did notice some of that afterward, but that's my first solution of solving this

[@Nikioko]

Minutes or seconds?

[@maxhagenauer24]

How does that mean they aren't perfect squares though?

[@KyriZee]

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.

[@deleted_handle]

In problem 1 I got the right answer only from vibes alone.

[@AbeIJnst]

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.

[@JaimeWarlock]

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.

[@erenerbay]

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

[@petros3207]

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.

[@rationalsceptic7634]

Beautiful Maths ❤ thx😊

[@Alfaomegabetagamma]

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).

[@samuelluc132]

They're all square numbers. I don't make the rules.

[@erikkonstas]

"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.

[@zanti4132]

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.

[@TaseMagu]

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.

[@Kastagaar]

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.

[@RoderickEtheria]

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.

[@marcogiai-coletti354]

The second problem was fun.

[@yurenchu]

5:47 Let's evaluate each of the five options: (a) 2β = 2*log₁₀(3) = log₁₀(9) = 0 + log₁₀(9) (b) 5α + β = 5*log₁₀(2) + log₁₀(3) = log₁₀(2⁵ * 3) = log₁₀(96) = 1 + log₁₀(9.6) (c) α + 2γ = log₁₀(2) + 2*log₁₀(7) = log₁₀(2 * 7^2) = log₁₀(98) = 1 + log₁₀(9.8) (d) 2α + 5β = 2*log₁₀(2) + 5*log₁₀(3) = log₁₀(2^2 * 3^5) = log₁₀(972) = 2 + log₁₀(9.72) (e) 2α + β + γ = 2*log₁₀(2) + log₁₀(3) + log₁₀(7) = log₁₀(2^2 * 3 * 7) = log₁₀(84) = 1 + log₁₀(8.4) Note that 0.5 ≈ log₁₀(3.16) < log₁₀(8.4) < log₁₀(9) < log₁₀(9.6) < log₁₀(9.72) < log₁₀(9.8) < log₁₀(10) = 1 Therefore, out of the five given options, (c) is the closest to being an integer.

[@akuunreach]

0:39 My answer is C Now to see if the root number starts with 8

[@cmuller1441]

To justify Q2 we could add than log10(x) is strictly increasing and log10(1)=0 log10(10)=1 So the answer is the x closer to 1 or to 10. Etc...

[@smdgch]

all correct answers are (c)

[@SergioRamos-iv6qw]

gonna Ace my tests with this info

[@shakedeilstein2448]

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

[@zaelgreen1670]

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.

[@OlavRH]

solved it in 30 sec by elimination. Not (a) because that's one less than a square (because of the conjugate sentence or whatever it's called in english) Not (b) because it's divisible by 3, but not 9. Saw that from taking the sum of the digits. Definitely 0 mod 3, but 6 mod 9. Not (e) because it has an odd number of 0s at the end Not (d) because it's divisible by 5 but not 25. So it has to be c

[@Thrillzrobloxbedwars]

The first question was easy to guess as 64 is a square number so option c seems the most probable.

[@WayneASchneider]

I found this pattern interesting. The square of the numbers 1-4 (forget 0 & 5 for now) are 1, 4, 9, 6. The square of the numbers 6-9 are 36, 49, 64, and 81. The last digit of the second group follows the reverse order of the first. Also, if you look at the last digit of the cube each of the numbers 0-9, you will find that all ten numbers 0-9 are used (0, 1, 8, (2)7, (6)4, (12)5, (21)6, (34)3, (51)2, and (72)9), and this cycle repeats for numbers higher than 9. Not profound or anything. Just interesting.

[@handanyldzhan9232]

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.

[@johnosullivan675]

E is divisible by 10^2.

[@handanyldzhan9232]

​@@johnosullivan675 E/10^2 is divisible by 10 but not 100, so the point stands.

[@brollout7406]

At the time I saw the video I was impressed that youtube showed that you had 3.14M subscribers :D

[@2pairOfAces]

For the first one, I couldn't say with certainty that C was a square, but I could easily say the other 4 weren't just based on the last 2 nonzero digits of each. I get that you like to explain different ways of things, but the quick explanation is "there's no way to square any digit and end up with 99, 33, 35, or 54".

[@wholesomeparas1]

Rule 1 - If last digit of a perfect square is odd, then the second last digit must be even. Option A is eliminated Rule 2 - Perfect squares don't end with 2, 3, 7, 8. Option B is eliminated Rule 3 - if last digit of a number is 5, then second last digit must be 2 for the number to be a perfect square. Option D eliminated Rule 4 - perfect sqaures ending with 0 always end up with even number of consecutive zeros, like 00, 0000, 000000, and similar. Option E eliminated

[@hellgazer8370]

SPOILER BLOCK!!

[@mananbachani5189]

for 2nd problem: a)log9 b)log96~1+log9.6 c)log98~1+log9.8 d)2 + log9.72 e)log84~1+log8.4 all options are now comparable, c is the correct answer

[@wholesomeparas1]

There can be only 22 possibilities of last two digits for it to be a perfect square. 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96 If last two digits of a number are not among them, then that number can't be a perfect square

[@jc5c515]

What happened with the music at the end? It's missing in the last videos

[@AlexTikhonovich]

Remember similar problem on your channel. Thus didn't have much problem solving it

[@mananbachani5189]

its pretty easy question, a) a number just above this is square of 10,000, therefore its not possible for 99,999,999 to be a square too b) there is no square which ends with 3 d)last digit is 5, we can only get 5 in a square if the number has 5 at the end too, but that square always has 25 at the end, here it is 35 therefore it cant be square e)there are odd numbers of 0 at the end in this number, and there is no such number squared like that, it always is even, therefore one of its factors doesnt get square rooted with integer perfectly, which means c is the correct option, this question would become very interesting if none of these was an option, which will force us to calculate

[@omm-o6i]

Did the same thing bro

[@omm-o6i]

You could type that much in 2 min 🤔

[@mananbachani5189]

also there's a trick to squaring number with 5 at the end for eg 25square=2×3 and 25 joint together which is 625 knowing this we can take n(n+1)=6494852, and find its roots n will come as 2548, therefore its square is 25485, ofcourse this should be done to a bit smaller number, then it would be worth it, otherwise just finding the square manually is our best way

[@daniellerosalie2155]

I did the same thing

[@yurenchu]

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 .

[@valentinadenisevich1630]

Какие лёгкие задания!

[@daniellerosalie2155]

The first problem is C.

[@lazyvector]

The hint: 2^3×3^5 = 9.72×10² Me: ah, so the answer is d then

[@9adam4]

No perfect square has a digit sum of 3 or 6 (in base 10).

[@Haseena-yv9fl]

Ooh god c was the answer i give in just,3 sec

[@Broke_Stacker]

My mind is weird. I knew 100% that it was C based in the number configurations, but could have never explained to you why.

[@LearningSiteGuy]

0:51 it's not FORTY FIVE THOUSAND, it is FIFTY FOUR THOUSAND

[@verkuilb]

Right after that, he says, “This problem is not testing your trivia knowledge of perfect squares; it’s not testing whether you can extract a square root.” That’s correct-it’s actually testing your ability to read a nine-digit number aloud. 😅

[@gutwrenchingdeaths]

c?

[@NikunjGoel-su3bf]

Not first

[@o4saken1]

nsquared -1 would work if n was 1 :)

[@o4saken1]

0, 1, 4, 9, 16, etc. Well as you see 0 and 1 are two squares right next to each other, :)

[@chrisglosser7318]

Even at Oxford, the answer is always (c)…

[@christopherscarpino8994]

the square root of 225 is 15, so this is the answer.

[@Square875]

Second

[@peterpinel5251]

E

[@spicemasterii6775]

All of those are square numbers.

[@terryliu9015]

Why do they write 9.72x10^2 instead of just 972 lol

[@CheckmateSurvivor]

If we see the Moon going westwards, but our maths calculation says otherwise, which one do we trust?

[@maxhagenauer24]

How does the fact the first 10 square numbers don't end in a 3 mean all numbers that end in a 3 are not aquare numbers?

[@danmerget]

Because any number with more than one digit can be expressed as 10x+y, where x is an integer and y is a one-digit number. If we square 10x+y, we get 100x^2 +20xy + y^2. Note that the first two terms (100x^2 and 20xy) are multiples of 10, so they don't affect the final digit, and therefore the square of 10x+y has the same final digit as the square of y.

[@danmerget]

BTW, while I spelled out the process algebraically in my previous comment, the fact that you can determine the last digit of any product by looking at the last digit of the two factors is pretty intuitively obvious if you've done enough multiplications by hand.

[@maxhagenauer24]

@danmerget But if y was a number greater than 3, then it's square is a 2 digit number which would make it different. How does it specifically the same as the last digit if y instead of being y itself?

[@maxhagenauer24]

@danmerget The last digit of what 2 factors? That's already saying something different than what you said algebraicly. And wheather it is obviously to people or not doesn't answer the question.

[@danmerget]

@@maxhagenauer24 "If y was a number greater than 3, then its square is a 2 digit number which would make it different" - No, because we're only talking about the LAST digit. For example, let's say that y is 4. The square is 16, and the fact that 16 is a two-digit number means that we add "1" to the ten's digit, which might carry over into the hundreds digit, the thousands digit, and so ad infinitum. But none of that affects the LAST digit, which is "6", and none of the stuff that affects the digits to the left of the "6" have any effect on the digit "6". "The last digit of what 2 factors? That's already saying something different" - It's simply a more general case of what I showed in the first comment. The question was about squares, so I focused on multiplying a number (or, more precisely, a positive integer) by itself. But the same rule applies to multiplying ANY two numbers together: you can always tell the last digit of the product by just looking at the product of the last digit of each factor. Algebraically, let's say the two numbers are 10x+y and 10z+w, where x and z are nonnegative integers and y & w are one-digit numbers. (10x+y)(10z+w) = 100xz + 10xw + 10yz + yw, and since the first three terms (100xz, 10xw, and 10yz) are all multiples of 10, the last digit of the full product will be the same as the last digit of yw. The "perfect square" rule is just a special case in which x=z and y=w.

[@SousouCell]

100 .000.000 is a perfect square therefor that number minus one is not ..... bye a) A square of number that starts with 5 always ends with 25 ..... bye d) Theres not square number that ends with 3 ...... only 0 4 5 6 9 .... bye b) Square number that end with 0 .... must end with an odd numbers of zeros ......bye e) Therefor .... theres only c) left .......

[@shohamsen8986]

So the hint for problem 2 is a red herring? That's just sneaky.

[@kicorse]

It saves you a little bit of arithmetic. Also, if the solver were in any doubt that they should rearrange the expressions so that everything is inside a log_10, the hint would reassure them that this is the correct approach.

[@shohamsen8986]

​@kicorse it's a bad hint then. Does more harm than good

[@willzhao5889]

No it gives a shortcut for solving

[@erikkonstas]

But it's literally used...? I mean sure it's not strictly necessary, but then again "no calculators allowed".

[@TaseMagu]

It actually points you into the right direction as to what format all numbers should have.

[@rupamjha5009]

Can only be (c) since 99,999,999 just seemed too wierd to be a square number.

[@4493005]

it took me 5 seconds to solve it

[@Nikioko]

a) 99.999.999 = 100.000.000 - 1 = 10.000² - 1 ⇒ not a square b) 123.333.333: ending on 3 ⇒ not a square c) 649.485.225: ending on 25 ⇒ possibly a square d) 713.291.035: ending on 35 ⇒ not a square e) 987.654.000: ending on odd number of zeroes (multiple of 1000) ⇒ not a square Since there is only one possible square, this must be the correct answer by elimination.

[@nuhozubaedsalam]

I solved it by assuming that, as the 649485225 contains 225 = 15², 649848225 must be a perfect square as well

[@cyruschang1904]

99,999,999 = 99 x 1010101 => not a square because 1010101 is not divisible by 11 123,333,333 => no square number can have 3 as its single digit. 649,485,225 ✅ by elimination 713,291,035 => not a square because it is not divisible by 25 987,654,000 = 100 x 9876540 => not a square number because 9876540 is not divisible by 100