Difference between revisions of "1964 IMO Problems/Problem 1"
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'''(b)''' Prove that there is no positive integer <math>n</math> for which <math>2^n+1</math> is divisible by <math>7</math>. | '''(b)''' Prove that there is no positive integer <math>n</math> for which <math>2^n+1</math> is divisible by <math>7</math>. | ||
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+ | == Video Solution == | ||
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+ | https://youtu.be/rap6C3ks29s | ||
== Solution 1 == | == Solution 1 == |
Latest revision as of 18:20, 22 August 2024
Problem
(a) Find all positive integers for which is divisible by .
(b) Prove that there is no positive integer for which is divisible by .
Video Solution
Solution 1
We claim is equivalent to and for congruent to , , and , respectively.
(a) From the statement above, only divisible by will work.
(b) Again from the statement above, can never be congruent to , so there are no solutions for .
Solution 1.1
This solution is clearer and easier to understand.
(1) Since we know that is congruent to 0 (mod 7), we know that is congruent to 8 mod 7, which means is congruent to 1 mod 7.
Experimenting with the residue of mod 7:
=1: 2
=2: 4
=3: 1 (this is because when is doubled to , the residue doubles too, but is congruent to 1 (mod 7).
=4: 2
=5: 4
=6: 1
Through induction, we easily show that this is true since the residue doubles every time you double .
So, the residue of mod 7 cycles in 2, 4, 1. Therefore, must be a multiple of 3. Proved.
(2) According to part (1), the residue of cycles in 2, 4, 1.
If is congruent to 0 mod 7, then must be congruent to 6 mod 7, but this is not possible due to how mod 7 cycles. Therefore, there is no solution. Proved.
~hastapasta
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 |