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
Solution 4
By multiplying both sides by we get
since ,
,
If ,
, so
If ,
, so
. Since
,
and
Therefore, .
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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