A TMUA Paper 2 Style Question (by an Oxford Graduate)

Ajouté : 2026-05-12

[00:00:00]Today I'm going to be solving a problem from the MAT, the Oxford University's former mathematics admissions test. And this problem is kind of a logic-based problem, so really useful if you're preparing for the TMUA, as with other MAT problems. Let's have a look. Let n be bigger than one be an integer, and let pi of n denote the number of distinct prime factors of n, and let xn denote the final digit of n. For example, pi of 8 equals 1 and pi of 6 equals 2. Which of the following statements must be false? We've got five statements here. If pi of n is 1, then there are some values of xn which mean that n cannot be prime. If pi of n equals 1, there are some values of xn that mean n must be prime. If pi of n equals 1, there are some values of xn which which are impossible. If pi of n plus x of n equals 2, we cannot tell if n is prime. And finally, if pi of n equals 2, all values of xn are possible. Do pause the video now and give this problem a go for yourself. I'm going to dive right into a solution here.

[00:00:56]So, as with most of these kind of problems where you're given a bunch of statements and you want to know whether each of those statements individually are true or false. In this case, exactly one of them will be false and four will be true. We just kind of want to assess these one by one. So, I'm going to start with a statement A here. If pi of n equals 1, there are some values of xn that mean n cannot be prime. So, let's just break down what this means. So, if pi of n equals 1, that's the number of distinct prime factors of n. That just means that n therefore must have exactly one prime factor. So, that means that n is p to the power of k. So, a prime number to the power of some positive integer.

[00:01:32]And we want to know, is it the case that some values of xn mean that n cannot be prime?

[00:01:38]Now, there's kind of a lot to digest here. We've got n, we've got p, we've got k, we've got pi of n, we've got x of n. There's loads of things floating about and it can be very easy uh it can become very easy to be intimidated by this problem. So, if I was doing this in an exam, I'd just break it down into cases. So, we're looking at some values of xn that mean n cannot be prime. So, let's just look at some values of XN.

[00:01:58]Let's start with one.

[00:02:00]So, if XN was one, and I know that N must be in the form P to the K.

[00:02:05]Are there any values of N that mean N cannot be prime?

[00:02:11]Are there Sorry, is it possible for N Sorry, let me start that again. If I just presume XN equals one, does that necessarily mean N cannot be prime? Or does that mean N must be prime? Or whatever.

[00:02:22]And the answer is, well, no, because we could have N is 11. That means N is prime.

[00:02:28]But equally, we could have N is 121, which is prime. And so, here XN being one doesn't mean N cannot be prime.

[00:02:37]Um because N could just be 11, for example.

[00:02:40]Um so, okay.

[00:02:43]Cool, but that's only one value of XN.

[00:02:45]What about if XN was two?

[00:02:47]Does that mean N cannot be prime? Well, no, we could have N is two. So, I don't even in fact even need to look at ones that aren't prime. I just need to find examples that are prime.

[00:02:55]So, if XN was two, well, N could just be the prime number two. Obviously, ends in a two and is prime.

[00:03:01]Okay, what about three?

[00:03:03]So, if XN So, if N ends in a three, well, N could just be three. And so, that's a prime number. But, if XN is four, well, the only way that N can If If N ends in a four, that means N certainly must be even. And that means if N is in the form P to the K, the P must be two.

[00:03:22]So, we must have something something something something four equals two to the K. And the only way this is possible if K is bigger than one.

[00:03:30]And so, therefore, N wouldn't be prime.

[00:03:33]So, statement A is true, so we can eliminate it because there are some values of XN, namely XN equaling four, that mean N cannot be prime. If I told you pi of N was one and XN equals four, then you know that N cannot be prime. In fact, you don't even need this bit of information. If I told you XN was four, you know n ends in a four and therefore cannot be prime.

[00:03:56]Cool. So, statement A is true and thus we can eliminate statement A.

[00:04:00]Okay, I'm going to skip B for the time being and I'm going to jump to C. If pi of n equals one, there are some values of xn which are impossible.

[00:04:09]And the trick to thinking about this is actually by thinking about the contrapositive.

[00:04:13]So, what if pi of n wasn't one?

[00:04:17]So, is it possible for some values of xn to be high that pi of n can't be one?

[00:04:22]And we can just think about this. So, if I told you what the last digit of a number n is, from that information alone, could you deduce that n has more than one prime factor? You can start thinking about possible last digits and if you think about the last digit zero, you'll realize, ah, if a number ends in a zero, it must be a multiple of 10. And thus must contain the prime factors two and five at the bare minimum. Maybe contain other prime factors, but it must contain two and five as prime factors.

[00:04:49]And so, if pi of n equals one, it's impossible for xn to be zero. So, this statement is true and thus we can eliminate C.

[00:04:57]Okay, let's jump to D. If pi of n plus x of n equals two, we cannot tell if n is prime. Okay, well, pi of n we know is the number of distinct prime factors of n and so must be a positive integer. So, and obviously xn is a non-negative integer. So, the only way this is possible is if we have one plus one or two plus zero.

[00:05:17]And we want to know, you know, we is it true that we cannot tell n is prime?

[00:05:22]Uh and the answer is yes. Again, we can just find um a counterexample here. So, we basically want to find a value of n um that means pi of n plus x of n is two and n is prime and equally a value of n with pi of n plus x of n equaling two and n is not prime.

[00:05:40]And so, if we just focus on the one plus one case, we need pi of n to be one and the last digit of n to be one. Well, what's a prime a number that's in the form p to the k and ends in a one? Well, we could have 11. That's a prime number.

[00:05:54]Or we could have 121, which is 11 squared, which is also in this form, ends in a one, and is not prime. And so therefore, if, you know, someone in the street came up to you and said, "Hey, pi of n plus x of n is two. Can you tell if n is prime?" The answer's no, because uh for all you know, n could be 11, or n could be 121, or n could be a bunch of other numbers, but the fact that we have have these two and these are, you know, either side of the the prime barrier, where this is prime, this is not prime, we we don't have enough information to tell if n is prime or not. So, D is true, we cannot tell if n is prime, and thus we can eliminate D.

[00:06:30]And let's look at E. If pi of n equals two, all values of xn are possible. Now, this statement turns out to be true, and we can just do this kind of by exhaustion. So, let's just create a little table here of xn and then an example of n here. So, we want n to be uh pi of n to be two, so for n to have exactly two distinct prime factors, and we want um to just list out the values of xn. So, we'll go 0 1 2 3 4 5 6 7 8 9.

[00:06:58]So, if xn is zero, what could n be?

[00:07:00]Well, n could be 10. That ends in a zero and has two prime factors.

[00:07:05]What about one? If xn is one, can n uh what what what value of n could be We could use 21, for example, 3 * 7.

[00:07:12]Two. And here's where we can use a nice little trick. We can use 22, for example. That ends in a two and has two prime factors, two and 11. We can do the same with three and the same with four and the same with five. Um those obviously end in two, three, four, five, and they all have two prime factors. We can't do the same with six, cuz 66 is 2 * 3 * 11.

[00:07:30]Uh so, we just need to think of a different prime number, a different number that ends in a six and has two prime factors. So, 26 does the job. I'll come back to the seven in a second. For eight, we can just use uh What can we use for eight? So, prime number with I could use 88 in fact. And same with nine, use 99 for that. For seven, we can actually use 77. I don't know why I left that one. But we we can use 77 for that one. Very nice. So, all values of x n are possible if you're told only that pi of n equals two. So, statement E is true and thus statement B must be by process of elimination the false option. Of course, here I kind of cheekily skipped over B cuz I knew the answer was B in advance of making this video.

[00:08:13]But you wouldn't necessarily know to do that in an exam. So, let's let's examine why statement B is actually false.

[00:08:19]So, it says if pi of n equals one, then there are some values of x n which mean n must be prime. So, we're just going to again going to show this by exhaustion or come up with a counterexample by exhaustion. And I guess we can use the fact from when we were investigating statement C. We know that if pi of n equals one, x n can't be zero. So, we don't even need to look at that case. We just need to look at x n being one to nine.

[00:08:44]So, x n being one to nine.

[00:08:50]And then obviously coming up with a counterexample n. So, we want to show statement B is false so that for each of these values of x n, we can find an example where n isn't prime because if statement B was true, that would mean that there's at least one of these values of x n where n would have to be prime. So, to show that that statement is false, we need to show that for each of these values of n, we can come up with at least one counterexample where n isn't prime. Okay, so let's look with x n is one. A counterexample could be 121 or 11 squared.

[00:09:20]So, just to go over that, pi of n definitely equals one there.

[00:09:23]x n definitely equals one and n isn't prime.

[00:09:27]Okay, statement sorry, if x n is two, well, we could use two to the power of five, which is 32, which ends in a two and isn't prime.

[00:09:37]Three, we could use three to the power of five. So, notice how I'm just thinking of powers of like that that the the prime number. So, three three squared or three is obviously prime, so we can't use that. Three squared is nine, three cubed is 27, three to the four is 81, and then three to the five is 243.

[00:09:55]Okay, four, well, we could just use two squared for that. Uh and same for six, we could use like a power of two, two to the four. And for eight, two cubed.

[00:10:04]Uh for nine, we could use three squared.

[00:10:06]And for five, we could use five squared, which is 25, not prime, ends in a five.

[00:10:09]And for seven, well, again, we can use a similar trick. Well, I know seven's obviously a prime. Seven squared is 49, which ends in a nine. So, that tells me that 49 squared, whatever 49 squared is, ends in a one. And so, seven to the power of four is something something something one. And so, seven to the power of five will be something something something seven. And so, therefore, I can use seven to the five as my counterexample there. It's a prime number.

[00:10:34]It is Sorry, not a prime number, ends in a seven. I could have also used three cubed, but I quite like this explanation for seven to the five. Anyway, that showcases that statement B must be false because there are no values of xn that force n to be prime because for each value of xn, we found an example of n where n isn't prime.

[00:10:52]Anyway, we'll finish the video there for today. Pretty nice problem, if you ask me. And one which I think would probably take most students even you know, very able students slightly longer than the average MAT or TMUA problem. But, that's kind of the point of saving time on the earlier problems in the paper. So, even though yes, you have on average around three minutes and a bit for each question, you definitely want to be aiming to solve most questions in under two minutes. So, for questions like this where even for myself, even with explaining it, it would even if I wasn't explaining all the details to you guys, it would still probably take me at least three to four minutes to solve this one. You want to give yourself that kind of buffer from the earlier questions. Anyway, uh if you do want to see more of these types of problems, make sure you're following, subscribing, liking, comment, all that sort of stuff. I'm actually soon going to be releasing my own problems that I share with my students. So, I had taken students every single year who are looking to get into Oxford and Cambridge, and over 80% of them get into Oxford / Cambridge, and so there's definitely some good material in the resources there. So, I'll be sharing some of those very, very soon on this channel, so don't miss out on those.