Difference between revisions of "1964 IMO Problems/Problem 1"
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== Solution == | == Solution == | ||
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We see that <math>2^n</math> is equivalent to <math>2, 4,</math> and <math>1</math> <math>\pmod{7}</math> for <math>n</math> congruent to <math>1</math>, <math>2</math>, and <math>0</math> <math>\pmod{3}</math>, respectively. | We see that <math>2^n</math> is equivalent to <math>2, 4,</math> and <math>1</math> <math>\pmod{7}</math> for <math>n</math> congruent to <math>1</math>, <math>2</math>, and <math>0</math> <math>\pmod{3}</math>, respectively. | ||
− | (a) From the statement above, only <math>n</math> divisible by <math>3</math> work. | + | '''(a)''' From the statement above, only <math>n</math> divisible by <math>3</math> work. |
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+ | '''(b)''' Again from the statement above, <math>2^n</math> can never be congruent to <math>-1</math> <math>\pmod{7}</math>, so there are no solutions for <math>n</math>. | ||
− | + | == See Also == | |
+ | {{IMO box|year=1964|before=First question|num-a=2}} |
Revision as of 12:45, 29 January 2021
Problem
(a) Find all positive integers for which is divisible by .
(b) Prove that there is no positive integer for which is divisible by .
Solution
We see that is equivalent to and for congruent to , , and , respectively.
(a) From the statement above, only divisible by work.
(b) Again from the statement above, can never be congruent to , so there are no solutions for .
See Also
1964 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |