Competition problem — Algebra — Maths Olympiad Prep

Library / /4 of 46

Algebra Difficulty 5.0 AIME, harder Prove it Russia

Prove that there exists a positive integer $n > 1$ such that the product of some $n$ consecutive positive integers equals the product of some $n + 100$ consecutive positive integers.

Solution

For example, let $n = (1 \cdot 2 \cdot 3 \cdots 101) - 101$ . Then the product of the first $n + 100$ natural numbers equals the product of $n$ consecutive numbers starting from $102$ and ending at $n + 101$ .

Indeed, after cancellation, the equality
$1 \cdot 2 \cdot 3 \cdots (n+100) = 102 \cdot 103 \cdots (n+101)$
reduces to
$1 \cdot 2 \cdot 3 \cdots 101 = n + 101,$
which is true.

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: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.