I don't know what the pidgeon-hole principle is, but I imagine the solution will go something like this:
The largest five-digit number is 99999, the sum of whose digits is 45.
The smallest five-digit number is 10000, the sum of whose digits is 1.
You can find any sum between those values by some choice of 45 five-digit numbers, so if you choose 91 such numbers, then whatever they are three of them must have the same sum, because there are two lots of 45 numbers plus 1 in 91.
Using the Pigeon-hole principle...
91 five-digit numbers are written on a blackboard. Prove that one can find three numbers on the blackboard such that the sums of their digits are equal.
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