Difference between revisions of "2011 USAMO Problems/Problem 4"
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+ | {{duplicate|[[2011 USAJMO Problems|2011 USAJMO #6]] and [[2011 USAMO Problems|2011 USAMO #4]]}} | ||
''This problem is from both the 2011 USAJMO and the 2011 USAMO, so both problems redirect here.'' | ''This problem is from both the 2011 USAJMO and the 2011 USAMO, so both problems redirect here.'' | ||
==Problem== | ==Problem== | ||
− | Consider the assertion that for each positive integer <math>n \ge 2</math>, the remainder upon dividing <math>2^{2^n}</math> by <math>2^n-1</math> is a power of 4. Either prove the assertion or find (with proof) a | + | Consider the assertion that for each positive integer <math>n \ge 2</math>, the remainder upon dividing <math>2^{2^n}</math> by <math>2^n-1</math> is a power of 4. Either prove the assertion or find (with proof) a counter-example. |
==Solution== | ==Solution== | ||
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==Solution 2== | ==Solution 2== | ||
− | Lemma (useful for all situations): If x and y are positive integers such that <math>2^x - 1</math> divides <math>2^y - 1</math>, then x divides y. | + | Lemma (useful for all situations): If <math>x</math> and <math>y</math> are positive integers such that <math>2^x - 1</math> divides <math>2^y - 1</math>, then <math>x</math> divides <math>y</math>. |
− | Proof: <math>2^y \equiv 1 \pmod{2^x - 1}</math>. Replacing the 1 with a <math>2^x</math> and dividing out the powers of two should create an easy induction proof which will be left to the reader as an Exercise. | + | Proof: <math>2^y \equiv 1 \pmod{2^x - 1}</math>. Replacing the <math>1</math> with a <math>2^x</math> and dividing out the powers of two should create an easy induction proof which will be left to the reader as an Exercise. |
− | Consider n = 25. 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 | + | 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>. Euler's theorem gives <math>2^{24} \equiv 2^4 \equiv 16 \pmod{25}</math> and so <math>k \ge 16</math>. However, <math>2^{2k} \geq 2^{32} > 2^{25} - 1</math>, a contradiction. Thus, <math>n = 25</math> is a valid counterexample. |
{{MAA Notice}} | {{MAA Notice}} | ||
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{{USAMO newbox|year=2011|num-b=3|num-a=5}} | {{USAMO newbox|year=2011|num-b=3|num-a=5}} | ||
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+ | [[Category: Olympiad Number Theory Problems]] |
Latest revision as of 14:41, 25 April 2021
- The following problem is from both the 2011 USAJMO #6 and 2011 USAMO #4, so both problems redirect to this page.
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 . Euler's theorem 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 |