Problem
ALG-B1-M05-P025 Six Consecutive Integers
#25
★★★★★ Level 5 of 5
Six consecutive positive integers are given. Prove that they can be labelled \(a,b,c,d,e,f\) so that \(\frac a{b+c}+\frac d{e+f}=1\).
Inspired by regional olympiad method · 2026 · Grade 10 · Problem 1
Write the numbers as \(n,n+1,\ldots,n+5\) and make equal denominators.
Take \(a=n\), \(d=n+5\), \(b=n+1\), \(c=n+4\), \(e=n+2\), \(f=n+3\).
Then \(b+c=e+f=2n+5\), and \(a+d=2n+5\). Hence \(\frac a{b+c}+\frac d{e+f}=\frac{a+d}{2n+5}=1\).
The idea is to split six consecutive numbers symmetrically.