Difference between revisions of "2011 USAMO Problems/Problem 4"
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<cmath>\textstyle 2^{2^{25}} \equiv 2^7 \pmod {2^{25} - 1}.</cmath> | <cmath>\textstyle 2^{2^{25}} \equiv 2^7 \pmod {2^{25} - 1}.</cmath> | ||
Since <math>\textstyle 2^7</math> is not a power of 4, we are done. | Since <math>\textstyle 2^7</math> is not a power of 4, we are done. | ||
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+ | {{MAA Notice}} | ||
==See also== | ==See also== | ||
{{USAMO newbox|year=2011|num-b=3|num-a=5}} | {{USAMO newbox|year=2011|num-b=3|num-a=5}} |
Revision as of 20:56, 3 July 2013
This problem is from both the 2011 USAJMO and the 2011 USAMO, so both problems redirect here.
Problem
Consider the assertion that for each positive integer , the remainder upon dividing by is a power of 4. Either prove the assertion or find (with proof) a counterexample.
Solution
We will show that is a counter-example.
Since , we see that for any integer , . Let be the residue of . Note that since and , necessarily , and thus the remainder in question is . We want to show that is an odd power of 2 for some , and thus not a power of 4.
Let for some odd prime . Then . Since 2 is co-prime to , we have and thus
Therefore, for a counter-example, it suffices that be odd. Choosing , we have . Therefore, and thus Since is not a power of 4, we are done.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2011 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |