2017 AIME I Problems/Problem 14
Contents
Problem 14
Let and satisfy and . Find the remainder when is divided by .
Solution 1
The first condition implies
So .
Putting each side to the power of :
so . Specifically,
so we have that
We only wish to find . To do this, we note that and now, by the Chinese Remainder Theorem, wish only to find . By Euler's Totient Theorem:
so
so we only need to find the inverse of . It is easy to realize that , so
Using Chinese Remainder Theorem, we get that , finishing the solution.
Solution 2 (Another way to find a)
Obviously letting will simplify a lot and to make the term simpler, let . Then,
Obviously, is times a power of . (It just makes sense.) Testing, we see satisfy the equation so . Therefore, ~Ddk001
Alternate solution
If you've found but you don't know that much number theory.
Note , so what we can do is take and keep squaring it (mod 1000).
Video Solution by mop 2024
~r00tsOfUnity
See also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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