Difference between revisions of "2011 USAMO Problems/Problem 4"
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Consider <math>n = 25</math>. We will prove that this case is a counterexample via contradiction. | Consider <math>n = 25</math>. We will prove that this case is a counterexample via contradiction. | ||
− | Because <math>4 = 2^2</math>, we will assume there exists a positive integer k such that <math>2^{2^n} - 2^{2k}</math> divides <math>2^n - 1</math> and <math>2^{2k} < 2^n - 1</math>. Dividing the powers of 2 from LHS gives <math>2^{2^n - 2k} - 1</math> divides <math>2^n - 1</math>. Hence, <math>2^n - 2k</math> divides n. Because n = 25 is odd, <math>2^{24} - k</math> divides 25. Modular arithmetic (in particular, Euler's totient function for <math>25</math> equals <math>20</math>) gives <math>2^{24} \equiv 2^4 \equiv 16 \pmod{25}</math> and so <math>k \ge 16</math>. However, <math>2^{2k} >= 2^{32} > 2^{25} - 1</math>, a contradiction. Thus, <math>n = 25</math> is a valid counterexample. | + | Because <math>4 = 2^2</math>, we will assume there exists a positive integer <math>k</math> such that <math>2^{2^n} - 2^{2k}</math> divides <math>2^n - 1</math> and <math>2^{2k} < 2^n - 1</math>. Dividing the powers of <math>2</math> from LHS gives <math>2^{2^n - 2k} - 1</math> divides <math>2^n - 1</math>. Hence, <math>2^n - 2k</math> divides <math>n</math>. Because <math>n = 25</math> is odd, <math>2^{24} - k</math> divides <math>25</math>. Modular arithmetic (in particular, Euler's totient function for <math>25</math> equals <math>20</math>) gives <math>2^{24} \equiv 2^4 \equiv 16 \pmod{25}</math> and so <math>k \ge 16</math>. However, <math>2^{2k} >= 2^{32} > 2^{25} - 1</math>, a contradiction. Thus, <math>n = 25</math> is a valid counterexample. |
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Revision as of 15:23, 10 March 2017
This problem is from both the 2011 USAJMO and the 2011 USAMO, so both problems redirect here.
Contents
[hide]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 counter-example.
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.
Solution 2
Lemma (useful for all situations): If and are positive integers such that divides , then divides . Proof: . Replacing the with a and dividing out the powers of two should create an easy induction proof which will be left to the reader as an Exercise.
Consider . We will prove that this case is a counterexample via contradiction.
Because , we will assume there exists a positive integer such that divides and . Dividing the powers of from LHS gives divides . Hence, divides . Because is odd, divides . Modular arithmetic (in particular, Euler's totient function for equals ) gives and so . However, , a contradiction. Thus, is a valid counterexample.
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 |