Grades 11-12 Video Solutions 2025-2025_11-12

Video Transcription

This is problem 12 on the grades 11 and 12, 2025, Math Kangaroo.  Among 10 different positive integers, there are exactly 5 that are divisible by 5, and exactly 7 that are divisible by 7. Let m be the largest of these numbers.  What is the smallest possible value of m?  So there are a few different ways to go about this one.  But first, let's sort of examine this weird sort of inequality we've got. So we know that exactly 5 are divisible by 5. So 5 of them are divisible by 5.  5 divides 5 of them, and 7 divides 7 of them.  But if we notice, 5 plus 7 is greater than 10.  So by this neat little thing called the pigeonhole principle,  and this is going to be sort of the first extension,  if you aren't aware of what the pigeonhole principle,  it effectively states that if you have k greater than n balls and n boxes, at least one box will have two balls in it if you were to, like, distribute them.  That feels like kind of an obvious statement, but can be used to prove some really miraculous things. So maybe after this problem or after these videos, look this thing up.  It's very, very interesting and very, very useful.  Anyways, by the pigeonhole principle, we know that there must be  two numbers, or at least two numbers, divisible by 5 and 7. So, like, 5 divides x and 7 divides, sorry, 7 divides x.  Which, you know, is just code for 35 divides x.  So if there are two numbers like that, we know that 35 and 70 should be in our sets, because we're looking to, like, minimize the values of m.  So then if we know that 35 and 70 are in our set,  now we can just sort of, like, evenly distribute the other stuff. So now that there are three numbers remaining and five numbers remaining  that are divisible by 7 and divisible by 5,  and we can just generate really whichever ones we'd like. So for these fives, let's just generate 5, 10, 15. And for these sevens, we can just generate 7, 14, 21, 28.  And we've already used 35, so let's use 42 now.  And if you sort of notice, these all work. Like, this set of numbers totally satisfies all these conditions.  And yet 70, because 70 is our largest value, is not an answer choice. So the answer should be e, another value. And that other value is 70.  Anyways, yeah, check out the pigeonhole principle.

Video Summary

The problem involves finding the smallest possible maximum among 10 different positive integers, with exactly 5 divisible by 5 and 7 divisible by 7. Using the pigeonhole principle, at least two numbers must be divisible by both 5 and 7, meaning they are divisible by 35. Including numbers like 35 and 70 in our set and filling in the rest with smaller multiples of 5 and 7 creates a valid set of numbers. By minimizing the largest number, the solution satisfies the conditions and identifies 70 as the smallest largest number, which is the edge case given in the problem.

Keywords

smallest maximum

positive integers

pigeonhole principle

divisible by 5

divisible by 7