Any partition of $\{1,2,\ldots,100\}$ into seven subsets yields a subset with numbers $a,b,c,d$ such that $a+b=c+d$. [closed]
Solution 1:
By pigeonhole principle some subset $S$ must have at least $15$ members, say $$S=\{a_1,a_2,....a_k\}$$ where $k\geq 15.$
Let $$A:=\{(x,y),\;x,y\in S, x<y\}$$
Now observe a function $$f:A\longrightarrow \{1,2,...,99\}$$ which is (well) defined with $$f(x,y) = y-x$$
Clearly, since $|A|= {k\choose 2} \geq {15\choose 2} = 105$ this function is not injective. So there are $a,b,c,d$ such that $$f(a,b)= f(c,d)\implies b-a=d-c\implies b+c=a+d$$