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In any set of $181$ square integers, prove that one can always find a subset of $19$ numbers, sum of whose elements is divisible by $19$.

Someone on AOPS:

Direct checking shows that any square is $0,1,4,9,16,6,17,11,7,5 \mod [19].$ Thus there are exactly $10$ distinct possibilities. From the pigeon-hole principle, since you used $181$ numbers at least one of the class contains at least $19$ of your squares. Adding these $19$ squares from the same class leads to the result.

Can anyone explain me this solution, or some other?

Thanks!

PS: This question is from Indian National Mathematical Olympiad 1994, Problem 3

RobPratt
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Akshat Vats
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1 Answers1

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If there were at most $18$ of each of those $10$ square congruences being used in the set, then the maximum number of elements in the set would be $18 \times 10 = 180$. However, since there are $181$ elements, this means that, by the pigeon-hole principle, there must be at least one congruence, say $r$, used at least $19$ times. As the solution states, adding these $19$ elements will give a sum with a remainder of $19r$, i.e., a value divisible by $19$.

Note that, in general, you actually only need just a set of $2 \times 19 - 1 = 37$ integers, without any particular restrictions on their congruences, for there to be a subset of $19$ of them that sum to a multiple of $19$. For more information about these types of problems, please see Proving that among any $2n - 1$ integers, there's always a subset of $n$ which sum to a multiple of $n$.

John Omielan
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