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Difference between revisions of "1964 IMO Problems/Problem 1"
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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>.
 
'''(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>.
 +
 +
 +
Solution 1.1:
 +
The solution is clearer and easier to understand.
 +
(1) Since we know that <math>2^n-1</math> is congruent to 0 (mod 7), we know that <math>2^n</math> is congruent to 8 mod 7, which means <math>2^n</math> is congruent to 1 mod 7.
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Experimenting with the residue of <math>2^n</math> mod 7:
 +
n=1: 2
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n=2: 4
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n=3: 1 (this is because when <math>2^n</math> is doubled to <math>2^(n+1)</math>, the residue doubles too, but <math>4*2</math> is congruent to 1 (mod 7).
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n=4: 2
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n=5: 4
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n=6: 1
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Through induction, we easy show that this is true since the residue doubles every time you double <math>2^n</math>.
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So, the residue of <math>2^n</math> mod 7 cycles in 2, 4, 1. Therefore, <math>n</math> must be a multiple of 3. Proved.
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(2) According to part (1), the residue of <math>2^n</math> cycles in 2, 4, 1. If <math>2^n+1</math> is congruent to 0 mod 7, then <math>2^n</math> must be congruent to 6 mod 7, but this is not possible due to how <math>2^n</math> mod 7 cycles. Therefore, there is no solution. Proved.
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~hastapasta
  
 
== See Also ==  
 
== See Also ==  
 
{{IMO box|year=1964|before=First question|num-a=2}}
 
{{IMO box|year=1964|before=First question|num-a=2}}

Revision as of 18:40, 7 January 2022

Problem

(a) Find all positive integers $n$ for which $2^n-1$ is divisible by $7$.

(b) Prove that there is no positive integer $n$ for which $2^n+1$ is divisible by $7$.

Solution

We see that $2^n$ is equivalent to $2, 4,$ and $1$ $\pmod{7}$ for $n$ congruent to $1$, $2$, and $0$ $\pmod{3}$, respectively.

(a) From the statement above, only $n$ divisible by $3$ work.

(b) Again from the statement above, $2^n$ can never be congruent to $-1$ $\pmod{7}$, so there are no solutions for $n$.


Solution 1.1: The solution is clearer and easier to understand. (1) Since we know that $2^n-1$ is congruent to 0 (mod 7), we know that $2^n$ is congruent to 8 mod 7, which means $2^n$ is congruent to 1 mod 7. Experimenting with the residue of $2^n$ mod 7: n=1: 2 n=2: 4 n=3: 1 (this is because when $2^n$ is doubled to $2^(n+1)$, the residue doubles too, but $4*2$ is congruent to 1 (mod 7). n=4: 2 n=5: 4 n=6: 1 Through induction, we easy show that this is true since the residue doubles every time you double $2^n$. So, the residue of $2^n$ mod 7 cycles in 2, 4, 1. Therefore, $n$ must be a multiple of 3. Proved. (2) According to part (1), the residue of $2^n$ cycles in 2, 4, 1. If $2^n+1$ is congruent to 0 mod 7, then $2^n$ must be congruent to 6 mod 7, but this is not possible due to how $2^n$ 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
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