1968 AHSME Problems/Problem 15

Problem

Let $P$ be the product of any three consecutive positive odd integers. The largest integer dividing all such $P$ is:

$\text{(A) } 15\quad \text{(B) } 6\quad \text{(C) } 5\quad \text{(D) } 3\quad \text{(E) } 1$

Solution

Notice that the product $P$ can be written as $2n-1$ , $2n+1$ , $2n+3$ . Because $P$ is defined as a \textit{ 3 consecutive odd integer } product impies that $P$ must be divisible by at least 3. A is ruled out because factors of 5 only arise every 5 terms, if we were to take the 3 terms in the middle of the factors of 5 we wouldn't have a factor of 5. Obviously B is impossible because we are multiplying odd numbers and 2 would never become one of our prime factors. C is ruled out with the same logic as A. Lastly E is ruled out because we have already proved that 3 is possible, and the question asks for greatest possible. $\fbox{D}$

In first step we could also apply modular arthmetic and get the same answer.

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1968 AHSME Problems/Problem 15

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