1968 AHSME Problems/Problem 15
Problem
Let be the product of any three consecutive positive odd integers. The largest integer dividing all such is:
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.
See also
1968 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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