Competition problem — Number theory

Library / /192 of 520

Number theory Difficulty 5.9 AIME, harder Prove it

25. Let $n>1$ . Prove: $n$ can be expressed as the sum of two or more consecutive positive integers if and only if $n \neq 2^{k}$ .

Solution

25. Let $m \geqslant 1, r \geqslant 1, m+(m+1)+\cdots+(m+r)=(r+1)(2 m+r) / 2=n . r+1$ and $2 m+r$ have opposite parity, which proves the necessity. When $n=2^{k} \cdot n^{\prime}, 2 \nmid n^{\prime}>1$ , if $2^{k+1}>n^{\prime}$ , take $r=n^{\prime}-1,2 m=2^{k+1}-r$ ; if $2^{k+1}<n^{\prime}$ , take $r=2^{k+1}-1,2 m=n^{\prime}-r$ . This proves the sufficiency.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.