1999 AHSME Problems/Problem 25
Contents
Problem
There are unique integers such that
where for . Find .
Solution 1(Modular Functions)
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 .
Solution 2(Basic Algebra and Bashing)
We start by multiplying both sides by , and we get: After doing some guess and check, we find that the answer is .
~aopspandy
Solution 3 (The Easiest and Most Intuitive)
Let's clear up the fractions: Notice that if we divide everything by then we would have: Since and must be an integer, then we have , so .
Similarly, if we divide everything by , then we would have: Again, since and must be an integer, we have , so .
The pattern repeats itself, so in the end we have , , , , , . So
~BurpSuite, with a help from ostriches88
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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