1968 AHSME Problems/Problem 15

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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 this product can be written as n-2,n,n+2. Because it is 3 consecutive odd integers we know that it 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. IE: 7,9,11. 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. $\fbox{D}$

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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