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Difference between revisions of "1964 IMO Problems/Problem 1"

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(Ph.D degree, IMO coach，https://www.youtube.com/@math000)
 
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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 ==
  
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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.
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We claim <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.
  
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'''(a)''' From the statement above, only <math>n</math> divisible by <math>3</math> work.
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'''(a)''' From the statement above, only <math>n</math> divisible by <math>3</math> will work.
  
 
'''(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>.
  
  
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=Solution 1.1=:
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=== Solution 1.1 ===
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The solution is clearer and easier to understand.
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This 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.
 
(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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<math>n</math>=6: 1
 
<math>n</math>=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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Through induction, we easily show that this is true since the residue doubles every time you double <math>2^n</math>.
  
 
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.
 
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.  
 
(2) According to part (1), the residue of <math>2^n</math> cycles in 2, 4, 1.  
  
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If <math>2*2^n</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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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.
  
 
~hastapasta
 
~hastapasta

Latest revision as of 18:20, 22 August 2024

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 $7$.

Video Solution

https://youtu.be/rap6C3ks29s

Solution 1

We claim $2^n$ 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 $3$ will 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

This 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 $2*2^n$, the residue doubles too, but $4*2=8$ is congruent to 1 (mod 7).

$n$=4: 2

$n$=5: 4

$n$=6: 1

Through induction, we easily 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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