How do you prove that $5(7^n)+3(11^n)-8$ is divisible by $3$ for all natural numbers $n$? [closed]

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How do you prove that $5(7^n)+3(11^n)-8$ is divisible by $3$ for all natural numbers $n$? I have been trying to figure this out and kept reaching dead ends. I would be grateful for some help

number-theory

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edited Jul 27 at 11:20

N. F. Taussig

38.1k93053

asked Jul 27 at 11:16

Tom Allen

195

closed as off-topic by José Carlos Santos, Xander Henderson, Mostafa Ayaz, amWhy, Adrian Keister Jul 28 at 0:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Xander Henderson, Mostafa Ayaz, amWhy, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You only need to check $n=0,1,2$ (since the powers of $7,11$ are periodic $pmod 3$).
  – lulu
  Jul 27 at 11:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
  – N. F. Taussig
  Jul 27 at 11:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By induction you can prove that $5cdot 7^n-8$ is divisible by $3$. The $3cdot 11^n$ is multiple of $3$ explicitly: For $n=0$, $5cdot 7^0-8=-3$, which is a multiple of $3$. If you assume that $5cdot 7^n-8$ is multiple of $3$, then $5cdot 7^n+1-8=3cdot 2cdot 5cdot 7^n+ (5cdot 7^n-8)$, which is a sum of two multiples of $3$.
  – user578878
  Jul 27 at 11:26
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

2

How do you prove that $5(7^n)+3(11^n)-8$ is divisible by $3$ for all natural numbers $n$? I have been trying to figure this out and kept reaching dead ends. I would be grateful for some help

number-theory

share|cite|improve this question

edited Jul 27 at 11:20

N. F. Taussig

38.1k93053

asked Jul 27 at 11:16

Tom Allen

195

closed as off-topic by José Carlos Santos, Xander Henderson, Mostafa Ayaz, amWhy, Adrian Keister Jul 28 at 0:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Xander Henderson, Mostafa Ayaz, amWhy, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You only need to check $n=0,1,2$ (since the powers of $7,11$ are periodic $pmod 3$).
  – lulu
  Jul 27 at 11:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
  – N. F. Taussig
  Jul 27 at 11:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By induction you can prove that $5cdot 7^n-8$ is divisible by $3$. The $3cdot 11^n$ is multiple of $3$ explicitly: For $n=0$, $5cdot 7^0-8=-3$, which is a multiple of $3$. If you assume that $5cdot 7^n-8$ is multiple of $3$, then $5cdot 7^n+1-8=3cdot 2cdot 5cdot 7^n+ (5cdot 7^n-8)$, which is a sum of two multiples of $3$.
  – user578878
  Jul 27 at 11:26
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

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2

up vote
-1
down vote

favorite

2

2

How do you prove that $5(7^n)+3(11^n)-8$ is divisible by $3$ for all natural numbers $n$? I have been trying to figure this out and kept reaching dead ends. I would be grateful for some help

number-theory

share|cite|improve this question

edited Jul 27 at 11:20

N. F. Taussig

38.1k93053

asked Jul 27 at 11:16

Tom Allen

195

How do you prove that $5(7^n)+3(11^n)-8$ is divisible by $3$ for all natural numbers $n$? I have been trying to figure this out and kept reaching dead ends. I would be grateful for some help

number-theory

share|cite|improve this question

edited Jul 27 at 11:20

N. F. Taussig

38.1k93053

asked Jul 27 at 11:16

Tom Allen

195

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 11:20

N. F. Taussig

38.1k93053

edited Jul 27 at 11:20

N. F. Taussig

38.1k93053

edited Jul 27 at 11:20

N. F. Taussig

38.1k93053

38.1k93053

asked Jul 27 at 11:16

Tom Allen

195

asked Jul 27 at 11:16

Tom Allen

195

asked Jul 27 at 11:16

Tom Allen

195

195

closed as off-topic by José Carlos Santos, Xander Henderson, Mostafa Ayaz, amWhy, Adrian Keister Jul 28 at 0:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Xander Henderson, Mostafa Ayaz, amWhy, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Xander Henderson, Mostafa Ayaz, amWhy, Adrian Keister Jul 28 at 0:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Xander Henderson, Mostafa Ayaz, amWhy, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You only need to check $n=0,1,2$ (since the powers of $7,11$ are periodic $pmod 3$).
  – lulu
  Jul 27 at 11:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
  – N. F. Taussig
  Jul 27 at 11:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By induction you can prove that $5cdot 7^n-8$ is divisible by $3$. The $3cdot 11^n$ is multiple of $3$ explicitly: For $n=0$, $5cdot 7^0-8=-3$, which is a multiple of $3$. If you assume that $5cdot 7^n-8$ is multiple of $3$, then $5cdot 7^n+1-8=3cdot 2cdot 5cdot 7^n+ (5cdot 7^n-8)$, which is a sum of two multiples of $3$.
  – user578878
  Jul 27 at 11:26
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You only need to check $n=0,1,2$ (since the powers of $7,11$ are periodic $pmod 3$).
  – lulu
  Jul 27 at 11:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
  – N. F. Taussig
  Jul 27 at 11:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By induction you can prove that $5cdot 7^n-8$ is divisible by $3$. The $3cdot 11^n$ is multiple of $3$ explicitly: For $n=0$, $5cdot 7^0-8=-3$, which is a multiple of $3$. If you assume that $5cdot 7^n-8$ is multiple of $3$, then $5cdot 7^n+1-8=3cdot 2cdot 5cdot 7^n+ (5cdot 7^n-8)$, which is a sum of two multiples of $3$.
  – user578878
  Jul 27 at 11:26
  
  

  
  
  
  

3

3

You only need to check $n=0,1,2$ (since the powers of $7,11$ are periodic $pmod 3$).
– lulu
Jul 27 at 11:19

You only need to check $n=0,1,2$ (since the powers of $7,11$ are periodic $pmod 3$).
– lulu
Jul 27 at 11:19

1

1

Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Jul 27 at 11:21

Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Jul 27 at 11:21

1

1

By induction you can prove that $5cdot 7^n-8$ is divisible by $3$. The $3cdot 11^n$ is multiple of $3$ explicitly: For $n=0$, $5cdot 7^0-8=-3$, which is a multiple of $3$. If you assume that $5cdot 7^n-8$ is multiple of $3$, then $5cdot 7^n+1-8=3cdot 2cdot 5cdot 7^n+ (5cdot 7^n-8)$, which is a sum of two multiples of $3$.
– user578878
Jul 27 at 11:26

By induction you can prove that $5cdot 7^n-8$ is divisible by $3$. The $3cdot 11^n$ is multiple of $3$ explicitly: For $n=0$, $5cdot 7^0-8=-3$, which is a multiple of $3$. If you assume that $5cdot 7^n-8$ is multiple of $3$, then $5cdot 7^n+1-8=3cdot 2cdot 5cdot 7^n+ (5cdot 7^n-8)$, which is a sum of two multiples of $3$.
– user578878
Jul 27 at 11:26

add a comment | 

2 Answers
2

active

oldest

votes

up vote
4
down vote



accepted

If you are unfamiliar with modulo arithmetic:

• Since $7^n=(6+1)^n = 1^n + *cdot 1^n-16^1 + cdots + *cdot 1cdot 6^n-1 + * cdot 6^n$, (where it's not important which numbers are in the positions marked by an asterisk), you can see that $7^n = 3k_1 + 1$ for some $k_1inmathbb N$


• Similarly, you can show that $11^k = (12-1)^n = 3k_2 + (-1)^n$ for some $k_2inmathbb N$, but in this case, that doesn't even matter, because $3cdot 11^n$ is always divisible by $3$.

You also have $8 = 3cdot 2 + 2$, so you have

$$beginalign5cdot 7^n + 3cdot 11^n - 8 &= 5cdot (3k_1 + 1) + 3cdot 11^n - 3cdot 2 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 5 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 3 \&= 3cdot (5k_1 + 11^n - 2 + 1)endalign$$ which is clearly a multiple of $3$.

share|cite|improve this answer

edited Jul 27 at 11:36

answered Jul 27 at 11:32

5xum

81.8k382146

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you've got a superfluous equals-sign in your first equation.
  – Jam
  Jul 27 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jam Yeah it was supposed to be a $+$ sign
  – 5xum
  Jul 27 at 11:36
  

  
  
  
  

add a comment | 

up vote
7
down vote

Using modular arithmetic: modulo $3$,

beginalign*
5cdot 7^n + 3cdot 11^n - 8 equiv 2cdot 1^n+0 -2 = 0
endalign*

for all $n$.

share|cite|improve this answer

answered Jul 27 at 11:20

Teddan the Terran

66617

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

If you are unfamiliar with modulo arithmetic:

• Since $7^n=(6+1)^n = 1^n + *cdot 1^n-16^1 + cdots + *cdot 1cdot 6^n-1 + * cdot 6^n$, (where it's not important which numbers are in the positions marked by an asterisk), you can see that $7^n = 3k_1 + 1$ for some $k_1inmathbb N$


• Similarly, you can show that $11^k = (12-1)^n = 3k_2 + (-1)^n$ for some $k_2inmathbb N$, but in this case, that doesn't even matter, because $3cdot 11^n$ is always divisible by $3$.

You also have $8 = 3cdot 2 + 2$, so you have

$$beginalign5cdot 7^n + 3cdot 11^n - 8 &= 5cdot (3k_1 + 1) + 3cdot 11^n - 3cdot 2 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 5 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 3 \&= 3cdot (5k_1 + 11^n - 2 + 1)endalign$$ which is clearly a multiple of $3$.

share|cite|improve this answer

edited Jul 27 at 11:36

answered Jul 27 at 11:32

5xum

81.8k382146

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you've got a superfluous equals-sign in your first equation.
  – Jam
  Jul 27 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jam Yeah it was supposed to be a $+$ sign
  – 5xum
  Jul 27 at 11:36
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

If you are unfamiliar with modulo arithmetic:

• Since $7^n=(6+1)^n = 1^n + *cdot 1^n-16^1 + cdots + *cdot 1cdot 6^n-1 + * cdot 6^n$, (where it's not important which numbers are in the positions marked by an asterisk), you can see that $7^n = 3k_1 + 1$ for some $k_1inmathbb N$


• Similarly, you can show that $11^k = (12-1)^n = 3k_2 + (-1)^n$ for some $k_2inmathbb N$, but in this case, that doesn't even matter, because $3cdot 11^n$ is always divisible by $3$.

You also have $8 = 3cdot 2 + 2$, so you have

$$beginalign5cdot 7^n + 3cdot 11^n - 8 &= 5cdot (3k_1 + 1) + 3cdot 11^n - 3cdot 2 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 5 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 3 \&= 3cdot (5k_1 + 11^n - 2 + 1)endalign$$ which is clearly a multiple of $3$.

share|cite|improve this answer

edited Jul 27 at 11:36

answered Jul 27 at 11:32

5xum

81.8k382146

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you've got a superfluous equals-sign in your first equation.
  – Jam
  Jul 27 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jam Yeah it was supposed to be a $+$ sign
  – 5xum
  Jul 27 at 11:36
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

If you are unfamiliar with modulo arithmetic:

• Since $7^n=(6+1)^n = 1^n + *cdot 1^n-16^1 + cdots + *cdot 1cdot 6^n-1 + * cdot 6^n$, (where it's not important which numbers are in the positions marked by an asterisk), you can see that $7^n = 3k_1 + 1$ for some $k_1inmathbb N$


• Similarly, you can show that $11^k = (12-1)^n = 3k_2 + (-1)^n$ for some $k_2inmathbb N$, but in this case, that doesn't even matter, because $3cdot 11^n$ is always divisible by $3$.

You also have $8 = 3cdot 2 + 2$, so you have

$$beginalign5cdot 7^n + 3cdot 11^n - 8 &= 5cdot (3k_1 + 1) + 3cdot 11^n - 3cdot 2 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 5 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 3 \&= 3cdot (5k_1 + 11^n - 2 + 1)endalign$$ which is clearly a multiple of $3$.

share|cite|improve this answer

edited Jul 27 at 11:36

answered Jul 27 at 11:32

5xum

81.8k382146

If you are unfamiliar with modulo arithmetic:

• Since $7^n=(6+1)^n = 1^n + *cdot 1^n-16^1 + cdots + *cdot 1cdot 6^n-1 + * cdot 6^n$, (where it's not important which numbers are in the positions marked by an asterisk), you can see that $7^n = 3k_1 + 1$ for some $k_1inmathbb N$


• Similarly, you can show that $11^k = (12-1)^n = 3k_2 + (-1)^n$ for some $k_2inmathbb N$, but in this case, that doesn't even matter, because $3cdot 11^n$ is always divisible by $3$.

You also have $8 = 3cdot 2 + 2$, so you have

$$beginalign5cdot 7^n + 3cdot 11^n - 8 &= 5cdot (3k_1 + 1) + 3cdot 11^n - 3cdot 2 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 5 - 2 \&= 3cdot(5k_1 + 11^n - 2) + 3 \&= 3cdot (5k_1 + 11^n - 2 + 1)endalign$$ which is clearly a multiple of $3$.

share|cite|improve this answer

edited Jul 27 at 11:36

answered Jul 27 at 11:32

5xum

81.8k382146

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 11:36

edited Jul 27 at 11:36

edited Jul 27 at 11:36

answered Jul 27 at 11:32

5xum

81.8k382146

answered Jul 27 at 11:32

5xum

81.8k382146

answered Jul 27 at 11:32

5xum

81.8k382146

81.8k382146

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you've got a superfluous equals-sign in your first equation.
  – Jam
  Jul 27 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jam Yeah it was supposed to be a $+$ sign
  – 5xum
  Jul 27 at 11:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you've got a superfluous equals-sign in your first equation.
  – Jam
  Jul 27 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jam Yeah it was supposed to be a $+$ sign
  – 5xum
  Jul 27 at 11:36
  

  
  
  
  

I think you've got a superfluous equals-sign in your first equation.
– Jam
Jul 27 at 11:35

I think you've got a superfluous equals-sign in your first equation.
– Jam
Jul 27 at 11:35

@Jam Yeah it was supposed to be a $+$ sign
– 5xum
Jul 27 at 11:36

@Jam Yeah it was supposed to be a $+$ sign
– 5xum
Jul 27 at 11:36

add a comment | 

up vote
7
down vote

Using modular arithmetic: modulo $3$,

beginalign*
5cdot 7^n + 3cdot 11^n - 8 equiv 2cdot 1^n+0 -2 = 0
endalign*

for all $n$.

share|cite|improve this answer

answered Jul 27 at 11:20

Teddan the Terran

66617

add a comment | 

up vote
7
down vote

Using modular arithmetic: modulo $3$,

beginalign*
5cdot 7^n + 3cdot 11^n - 8 equiv 2cdot 1^n+0 -2 = 0
endalign*

for all $n$.

share|cite|improve this answer

answered Jul 27 at 11:20

Teddan the Terran

66617

add a comment | 

up vote
7
down vote

up vote
7
down vote

Using modular arithmetic: modulo $3$,

beginalign*
5cdot 7^n + 3cdot 11^n - 8 equiv 2cdot 1^n+0 -2 = 0
endalign*

for all $n$.

share|cite|improve this answer

answered Jul 27 at 11:20

Teddan the Terran

66617

Using modular arithmetic: modulo $3$,

beginalign*
5cdot 7^n + 3cdot 11^n - 8 equiv 2cdot 1^n+0 -2 = 0
endalign*

for all $n$.

share|cite|improve this answer

answered Jul 27 at 11:20

Teddan the Terran

66617

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 11:20

Teddan the Terran

66617

answered Jul 27 at 11:20

Teddan the Terran

66617

answered Jul 27 at 11:20

Teddan the Terran

66617

66617

add a comment | 

add a comment |