# Mock AIME 1 2007-2008 Problems/Problem 4

## Problem

If $x$ is an odd number, then find the largest integer that always divides the expression $$(10x+2)(10x+6)(5x+5)$$

## Solution

Rewrite the expression as $$4(5x + 1)(5x + 3)(5x+5)$$ Since $x$ is odd, let $x = 2n-1$. The expression becomes $$4(10n-4)(10n-2)(10n)=32(5n-2)(5n-1)(5n)$$ Consider just the product of the last three terms, $5n-2,5n-1,5n$, which are consecutive. At least one term must be divisible by $2$ and one term must be divisible by $3$ then. Also, since there is the $5n$ term, the expression must be divisible by $5$. Therefore, the minimum integer that always divides the expression must be $32 \cdot 2 \cdot 3 \cdot 5 = \Rightarrow{\boxed{960}}$.

To prove that the number is the largest integer to work, consider when $x=1$ and $x = 5$. These respectively evaluate to be $1920,\ 87360$; their greatest common factor is indeed $960$.

## See also

 Mock AIME 1 2007-2008 (Problems, Source) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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