This is problem 22 on the grade 11 and 12 2025 math kangaroo contest. John wrote an arbitrary two-digit integer on the blackboard, then he erased the last digit of the integer. As a result, the initial integer decreased by P%. Which of the following is the closest to the largest possible value of P? We have A, 10, B, 50, C, 90, D, 95, and E, 99. You know, take your time, pause the video here, attempt the problem. Okay. So, there are a few ways to go about this, but I think what we should try is to sort of split the two values and sort of attack them independently of one another, such that we don't have to deal with, like, a bunch of different cases with, like, A and B relying on one another. So, let's start with the second, for no reason in particular. Okay, so yeah, notably, we're decreasing this number by P. So, if our number is AB, we can sort of write that as 10A plus B. And then, after our, like, sort of process, we're going to A, which means we're subtracting off 9A plus B in the process. So, this, as a percent, is actually 9A plus B over 10A plus B. So, we can write P percent is equal to this. Okay, well, then we can write this down here as P percent equals, and just rewrite this fraction as 1 minus A over 10A plus B. And notably, no matter what I put in for A, this, like, ratio of A to 10A doesn't change. So, then, if we want to minimize this second term here, and thereby maximize P percent, which, you know, is what we're trying to do. We're looking for the largest possible value of P. We want to maximize B. So, yeah, our guide is telling us to maximize B. And to maximize B, well, we can just set it to any digit we want. So, then B should be equal to 9. Okay, so we know that B is equal to 9. Now, we can try another thing. So, A9, A, pardon me, 9, goes to just A. Then, we want to make this such that, like, A compared to A concatenated with 9 is the smallest that it can possibly be. So, we can write this, again, as, like, A over 10A plus 9, or, you know, 1 minus A over 10A plus 9. And then, from there, what we can do is just sort of, like, examine which number would allow 9 to have the biggest effect on it. So, we know that A can't be 0 because it's a two-digit integer. And, you know, therefore, A has to be 1 or greater. But if we set, like, A is equal to 1, then we have 1 over 19. If we set A is equal to 2, then we get 2 over 29. A is equal to 3, 3 over 39. If we sort of, like, denote these fractions, this becomes, you know, 1 over 19. This becomes 1 over 14.5. This becomes 1 over 13. And this sequence sort of continues. So, like, the denominators get smaller and smaller and smaller. And this portion gets bigger and bigger and bigger, which we don't want. So, we have that when A is equal to 1, this is actually, the second term is minimized, which is the ideal. So, we want A to be 1. So, then, yeah, our two-digit number should be 19. So, 19 converts to 1. Therefore, we lose 18 over 19. And 18 over 19 is really close to, but admittedly less than, 19 over 20. And 19 over 20, if we recognize, is just 95%. So, this should be somewhere within the ballpark of 95%. If you want to verify this, I think it comes out to, like, 94.7% or something like that. But very close to 95%. So, that's our answer.

largest percentage decrease

two-digit number

digit erased

maximize P

option D