1957 AHSME Problems/Problem 32
Problem
The largest of the following integers which divides each of the numbers of the sequence is:
Solution
To begin, we look for a maximum value that this common divisor could have by looking at the smallest terms. , but this gives us no information because all positive integers divide zero. However, , so our answer cannot be greater than (because divided by an integer larger than is never an integer). This fact eliminates options (B) and (D).
Now, we will factorize to discover some properties: \begin{align*} n^5-n &= n(n^4-1) \\ &= n(n^2-1)(n^2+1) \\ &= n(n-1)(n+1)(n^2+1) \end{align*} One of and must be even, so divides for all . Likewise, one of , , and must be divisible by , so must divide as well. In the case of , there are, in total, three cases:
Case 1: divides either , , or . Then, clearly must divide .
Case 2: divides . In this case, , so . Then, , so (and, consequently, ) is divisible by .
Case 3: divides . In this case, , so . Thus, , so (and, consequently, ) is divisible by .
Thus, 5 divides for all .
Because , , and all divide and are relatively prime, also divides . Thus, because is the largest of our remaining choices, we choose answer .
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
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