Let d be a positive integer. Show that among any group of d +1 (not necessary consecutive) integers, there are two with exactly the same remainder when divided by d.
for the sake of illustration, let's say d = 1, that is a positive integer. so d + 1 would be 2. there is no guarantee, by the pigeonhole principle or otherwise, that any arbitrary group of two integers would have two elements that are congruent mod 4
TPH used that principle. We have n + 1 integers being divided by n, so the possible remainders are 0, 1, 2, 3, ..., (n - 1), one less than the number we divided by. if you count starting from 0 up to (n - 1) you get n remainders. look at my illustrations above, you will find if you divide by 2 you have 2 possible remainders, if you divide by 5, you have 5 possible remainders. so here, we have n possible remainders. but there are (n + 1) integers, meaning when you get to the nth integer, you would have exhausted the n remainders, so the next (n + 1)th integer must reuse one of the remainders already used, so there will be two numbers with the same remainder when divided by n
see here for more on the Pigeonhole Principle
if you pick an integer and it has a remainder of 0 when divided by n, then you put it in the hole for 0, if it has a remainder of 1 when divided by n, you put it in the hole for 1, and so on. eventually, after n consecutive integers, you would have filled all the holes, and the last integer must be put into a hole that is already used. (if the integers are not necessarily consecutive, and it is the case that you haven't filled the holes by the nth integer, it simply means you already reused one of the holes at least once, so there are already at least two integers with exactly the same remainder)