# Thread: 16 digit from titu andreescu book

1. ## 16 digit from titu andreescu book

Q Prove that among any 16 distinct digit +ve integer not exceeding 100 there are four distinct ones a,b,c,d such that a+b=c+d

SIR some body has given this hint to me but i am not able to explain the thing that how to prove that all the four integers are distinct
hint given
Rearrange a+b=c+d, it is equivalent to finding two sets of differences |a'-b'|=|c'-d'| for distinct a',b',c',d' in our 16. Now how big can the set be, when ?

There are 120 differences $\displaystyle |x-y|$, so some of them must coincide. But a pair that coincide need not use four distinct integers. It could happen that $\displaystyle a<c<b$, with $\displaystyle c-a = b-c$. But suppose that an integer $\displaystyle c$ is the middle integer of more than one such triple, say $\displaystyle a<c<b$ and $\displaystyle a'<c<b'$. Then the four distinct numbers $\displaystyle a,\ a',\ b,\ b'$ satisfy |a-a'| = |b-b'|.
So we may assume that there are at most 14 such triples, one for each of the 16 available numbers except for the smallest and the largest (which cannot be the middle element of a triple). That still leaves more than 100 differences $\displaystyle |x-y|$. So there must be at least one pair of equal differences using four distinct integers.