1951 AHSME Problems/Problem 15
Contents
Problem
The largest number by which the expression is divisible for all possible integral values of , is:
Solution 1
Factoring the polynomial gives According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3. Therefore must divide the given expression. Plugging in yields . So the largest possibility is .
Clearly the answer is
Solution 2
In general, | were and are integers. So here | - always for any integer .Hence,the correct answer is .
~geometry wizard.
See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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