Number theory and combinatorial problem

Question

Consider seven distinct positive integers not exceeding $1706$. Prove that there are three of them, say $a,b,c$ such that $a < b+c < 4a$.

Answer 1

Hint: Let the integers in increasing order be $a_1, a_2, \ldots, a_7$. Realize that in base $4$, $1706_{10} = 122222_4$. Thus, divide $[1, 1706]$ into

$$\underbrace{[1, 2)}_{A_1} \cup \underbrace{[2, 10)}_{A_2} \cup \underbrace{[10, 42)}_{A_3} \cup \underbrace{[42, 170)}_{A_4} \cup \underbrace{[170, 682)}_{A_5} \cup \underbrace{[682, 1706]}_{A_6}$$

Can you use the Pigeonhole Principle now? There are two cases: one where there are at least three integers in a single $A$-set, and when there are two(then there is a sub-case here for $A_6$).

Answer 2

Assume that the numbers are $1\le a_1<a_2<\ldots<a_7$. Assume contrariwise that no triple satisfies the inequalities $a<b+c<4a$. This implies that for all $i=2,3,4,5,6,$ we must have $$a_{i+1}\ge4a_i-a_1,\qquad(*)$$ for otherwise $a=a_i, b=a_{i+1},c=a_1$ is a solution. Applying $(*)$ five times gives $$ \begin{aligned} a_3&\ge4a_2-a_1,\\ a_4&\ge16a_2-5a_1,\\ \cdots&\ge\cdots,\\ a_7&\ge1024a_2-341a_1. \end{aligned} $$ Because $a_2\ge a_1+1, a_1\ge1,$ this tells us that $$a_7\ge (1024-341)a_1+1024\ge683a_1+1024\ge1707.$$ A contradiction.