prove 3 separate subsets of 90 numbers with similar sums

Question

"Given a set of 90 numbers , each with 3 digits , prove that there exist 3 subsets which are each separate , that have the same sum (sum of the numbers)." I know that I should use the pigeonhole principle and somehow divide the subsets into different parts, and a hint says that I should consider 2-membered subsets, but I am unable to correctly divide these subsets to use the pigeonhole principle. Thanks!

Answer 1

There are $\dbinom{90}{2}=4005$ possible $2$-element subsets of $90$ element set. Alternatively,

Notice we can make $\{a_1,a_2\},\dots,\{a_1,a_{90}\}$ then $\{a_2,a_3\},\dots,\{a_2,a_{90}\}$ then $\dots$

Which is $(90-1)+(90-2)+\dots+(90-90)=\dfrac12 (90-1) 90=4005$ as claimed.

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Because all numbers in the set have $3$ digits, we see that:

The smallest and largest possible sums are $100+101=201$ and $999+998=1997$.

This means every sum of those $2$-subsets is in $[201,1997]$ which is $1797$ possibilities.

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We see now by pigeonhole that there exist $\left\lceil\dfrac{4005}{1797}\right\rceil=3$ many $2$-subsets with same sum.