Can you solve these Oxford admissions questions?

Which number is a perfect square? Which is closest to an integer?

0:00 intro
0:16 problem 1
5:36 problem 2

Oxford 2023 MAT problems
Oxford 2023 MAT solutions
bprp math basics

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[skydiverclassc]

My most interesting math factoid is that if you square the number 111, 111, 111, the result is 12, 345, 678, 987, 654, 321.

[Brain_]

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

[TC-cqoc]

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.

[ChrisWilson-glff]

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 ;)

[Duck_Bidiyani]

Congratulations on 3.14 million subscribers

[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.

[PoppySuzumi]

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. 😬

[petros]

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.

[deleted_handle]

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

[romywilliamson]

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).

[MrNostril]

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.

[blakemcalevey-scurr]

Fun! I did it pretty similarly, except b:
It's divisible by 3, which gives which can't be divisible by 3 again since it's digits don't add to 0 modulo 3.

[dmdeemer]

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).

[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.

[GlobalWarmingSkeptic]

On answer 1: This one was easy but I didn't fully understand the "why" behind a lot of it, but I knew based on working with programming how the products work out, such as no squares ending in 35.

On answer 2: I did it the hard way (and in my head too, calculating the 4th root of 1000 lol). I have long forgotten how to work with logarithms and that's useful to know how to deal with multiplication in logarithms

Glad I got both right I'm not a complete dumbass.

Before I look: I am guessing (c) because of the following:
Answer A can't be right because 100 million is the square of 10, 000, thus having 1 less than that isn't possible as a perfect square as a result.
Answer B can't be right because no squares end in 3
Answer C can be right because any square of any number ending in 5 will always be 25, no exceptions.
Answer D is wrong for the same reason C is right. There are no squares that end in 35.
Answer E cannot be right becasue it has an odd number of 0s. All squares ending in 0 must have an even number of 0s.

Thus it can only be C.

2nd question:
I don't know much about converting logs but I can do some basic estimation of roots in my head. My very very rough calculations put it pretty close to an integer, as I think log of 7 is like 0.8 or something and log of 2 is a lil less than 0.25, which would put a+2y at less than 0.1 off of an integer. That's my guess but it's more up in the air than actually knowing number theory.

[elinhulldin]

10:13 Actually you first have to check if any
N < √10 ≈ 3.16 meaning
lg N < 0.5

You already made sure that
1 ≤ N ≤ 10 meaning
0 ≤ lg N ≤ 1

For N ≤ √10
lg N is the smallest distance from an integer (0)

For N ≥ √10
1 - lg N
= lg(10/N) is the smallest distance from an integer (1)

Simplified the answer the one where either N or 10/N is lower (or higher) than all other N and 10/N.


(lg means log_10 by the way)

[ShawnRavenfire]

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!)

[mikefochtman]

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.

[prufrock]

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

[HiFiGuy]

Nice and clean solutions. Just one tiny thing: at 0:48 you misspoke, transposing the 45 and 54.