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 $32 \cdot 2 \cdot 3 \cdot 5 = \boxed{960}$ .

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 $960$ .

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

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Mock AIME 1 2007-2008 Problems/Problem 4

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