1. ## Proof

91 five-digit numbers are written on a blackboard. Prove that one can find 3 numbers on the blackboard such that the sums of their digits are equal.

I was thinking of writing the three numbers as $10000a + 1000b + 100c + 10d + e$, $10000f + 1000g + 100h + 10i + j$ and $10000k + 1000l + 100m + 10n + o$, but how do I proceed? Also, I know that there are $9*10^{4}$ different 5-digit numbers possible, but is that information going to help? This doesn't look like a permutations/combinations question to me...

Thanks!

ILoveMaths07

2. Hello, ILoveMaths07!

It's easier than you think . . . the "pigeonhole principle".

91 five-digit numbers are written on a blackboard.
Prove that one can find 3 numbers on the blackboard
such that the sums of their digits are equal.

Five-digit numbers have the range: . $[10000, 99999].$

Hence, their digit-sums have the range: . $[1,45]$

. . That is, there are only 45 possible digit-sums.

With 90 numbers of the board, it is possible they have two each
. . of the 45 different digit-sums: $\{1,1,2,2,3,3,\,\hdots\, 45,45\}$

The 91st number must duplicate one of the digit-sums.

There will be 3 numbers with the same digit-sum.