Pigeonhole principle and sums

Question

Given 69 integers not exceeding 100, prove that you can find a,b,c,d such that a+b+c=d. Is this true for 68 numbers?

Answer 1

Consider $~a,~b,~c,~d~$ be such that $~a\lt b<c~$ and $~a+b+c=d~$.

Also let the numbers be $~1 ≤ a_1 < a_2 < \cdots < a_{69} ≤ 100~$.

Clearly $~a_1 ≤ 32~$.

Consider the sequence $$a_3 + a_1 < a_4 + a_1 < \cdots < a_{69} + a_1$$ $$a_3 − a_2 < a_4 − a_2 < \cdots < a_{69} − a_2~.$$ Each of their terms is a positive integer not exceeding $~100 + 32 = 132~$. Since the two sequences have jointly $~67+67 = 134~$ terms, there must exist indices $~i, j \in \{3, 4, \cdots , 69\}~$ such that $~a_i + a_1 = a_j − a_2~$. We have $~a_1 < a_2 < a_i~$ , and since $~a_1 + a_2 + a_i = a_j~$ , the first part is done.

A counterexample for the second part is given by the set $~\{33, 34, 35, \cdots , 100\}~$.

Ref.: https://www.awesomemath.org/wp-pdf-files/amy-online/amy_sample_wm.pdf