1999 AHSME Problems/Problem 25
Problem
There are unique integers such that
where for . Find .
Solution
Multiply out the to get
By Wilson's Theorem (or by straightforward division), , so . Then we move to the left and divide through by to obtain
We then repeat this procedure , from which it follows that , and so forth. Continuing, we find the unique solution to be (uniqueness is assured by the Division Theorem). The answer is .
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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