Prove that $frac21n+414n+3$ is in lowest terms for any natural $n$. [closed]

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Prove that the fraction $dfrac21n+414n+3$ is in lowest terms for any natural value of $n$.

number-theory elementary-number-theory contest-math

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edited Jul 30 at 17:15

Peter

44.9k938119

asked Jul 30 at 17:06

QuestionEverything

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closed as off-topic by Batominovski, Peter, Siong Thye Goh, dxiv, Robert Jul 30 at 17:51

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." – Batominovski, Peter, Siong Thye Goh, Robert

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

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Complex numbers are not numbers that are complicated! :-o
  – Brian Tung
  Jul 30 at 17:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Anyway, use the Euclidean algorithm to find GCD of $21n+4$ and $14n+3$. What should the GCD be, if the fraction is always irreducible?
  – Brian Tung
  Jul 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @BrianTung You just made me choke.
  – Batominovski
  Jul 30 at 17:11
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $$2(21n+4)-3(14n+3)=?$$
  – lab bhattacharjee
  Jul 30 at 17:12
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  For reference, this is IMO 1959, Problem 1---the first ever IMO problem. See artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/….
  – Batominovski
  Jul 30 at 17:13
  
  

  
  
  
  

 | 
show 5 more comments

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Prove that the fraction $dfrac21n+414n+3$ is in lowest terms for any natural value of $n$.

number-theory elementary-number-theory contest-math

share|cite|improve this question

edited Jul 30 at 17:15

Peter

44.9k938119

asked Jul 30 at 17:06

QuestionEverything

132

closed as off-topic by Batominovski, Peter, Siong Thye Goh, dxiv, Robert Jul 30 at 17:51

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." – Batominovski, Peter, Siong Thye Goh, Robert

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

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Complex numbers are not numbers that are complicated! :-o
  – Brian Tung
  Jul 30 at 17:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Anyway, use the Euclidean algorithm to find GCD of $21n+4$ and $14n+3$. What should the GCD be, if the fraction is always irreducible?
  – Brian Tung
  Jul 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @BrianTung You just made me choke.
  – Batominovski
  Jul 30 at 17:11
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $$2(21n+4)-3(14n+3)=?$$
  – lab bhattacharjee
  Jul 30 at 17:12
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  For reference, this is IMO 1959, Problem 1---the first ever IMO problem. See artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/….
  – Batominovski
  Jul 30 at 17:13
  
  

  
  
  
  

 | 
show 5 more comments

up vote
0
down vote

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1

up vote
0
down vote

favorite

1

1

Prove that the fraction $dfrac21n+414n+3$ is in lowest terms for any natural value of $n$.

number-theory elementary-number-theory contest-math

share|cite|improve this question

edited Jul 30 at 17:15

Peter

44.9k938119

asked Jul 30 at 17:06

QuestionEverything

132

Prove that the fraction $dfrac21n+414n+3$ is in lowest terms for any natural value of $n$.

number-theory elementary-number-theory contest-math

share|cite|improve this question

edited Jul 30 at 17:15

Peter

44.9k938119

asked Jul 30 at 17:06

QuestionEverything

132

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 17:15

Peter

44.9k938119

edited Jul 30 at 17:15

Peter

44.9k938119

edited Jul 30 at 17:15

Peter

44.9k938119

44.9k938119

asked Jul 30 at 17:06

QuestionEverything

132

asked Jul 30 at 17:06

QuestionEverything

132

asked Jul 30 at 17:06

QuestionEverything

132

132

closed as off-topic by Batominovski, Peter, Siong Thye Goh, dxiv, Robert Jul 30 at 17:51

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." – Batominovski, Peter, Siong Thye Goh, Robert

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

closed as off-topic by Batominovski, Peter, Siong Thye Goh, dxiv, Robert Jul 30 at 17:51

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." – Batominovski, Peter, Siong Thye Goh, Robert

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

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Complex numbers are not numbers that are complicated! :-o
  – Brian Tung
  Jul 30 at 17:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Anyway, use the Euclidean algorithm to find GCD of $21n+4$ and $14n+3$. What should the GCD be, if the fraction is always irreducible?
  – Brian Tung
  Jul 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @BrianTung You just made me choke.
  – Batominovski
  Jul 30 at 17:11
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $$2(21n+4)-3(14n+3)=?$$
  – lab bhattacharjee
  Jul 30 at 17:12
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  For reference, this is IMO 1959, Problem 1---the first ever IMO problem. See artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/….
  – Batominovski
  Jul 30 at 17:13
  
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  Complex numbers are not numbers that are complicated! :-o
  – Brian Tung
  Jul 30 at 17:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Anyway, use the Euclidean algorithm to find GCD of $21n+4$ and $14n+3$. What should the GCD be, if the fraction is always irreducible?
  – Brian Tung
  Jul 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @BrianTung You just made me choke.
  – Batominovski
  Jul 30 at 17:11
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $$2(21n+4)-3(14n+3)=?$$
  – lab bhattacharjee
  Jul 30 at 17:12
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  For reference, this is IMO 1959, Problem 1---the first ever IMO problem. See artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/….
  – Batominovski
  Jul 30 at 17:13
  
  

  
  
  
  

5

5

Complex numbers are not numbers that are complicated! :-o
– Brian Tung
Jul 30 at 17:09

Complex numbers are not numbers that are complicated! :-o
– Brian Tung
Jul 30 at 17:09

3

3

Anyway, use the Euclidean algorithm to find GCD of $21n+4$ and $14n+3$. What should the GCD be, if the fraction is always irreducible?
– Brian Tung
Jul 30 at 17:10

Anyway, use the Euclidean algorithm to find GCD of $21n+4$ and $14n+3$. What should the GCD be, if the fraction is always irreducible?
– Brian Tung
Jul 30 at 17:10

3

3

@BrianTung You just made me choke.
– Batominovski
Jul 30 at 17:11

@BrianTung You just made me choke.
– Batominovski
Jul 30 at 17:11

5

5

$$2(21n+4)-3(14n+3)=?$$
– lab bhattacharjee
Jul 30 at 17:12

$$2(21n+4)-3(14n+3)=?$$
– lab bhattacharjee
Jul 30 at 17:12

6

6

For reference, this is IMO 1959, Problem 1---the first ever IMO problem. See artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/….
– Batominovski
Jul 30 at 17:13

For reference, this is IMO 1959, Problem 1---the first ever IMO problem. See artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/….
– Batominovski
Jul 30 at 17:13

 | 
show 5 more comments

2 Answers
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up vote
6
down vote

Writing $(n,m)$ for the GCD of $m$ and $n$, immediately one has $(n,m)=(n,m-kn)$ for any $kinmathbbZ$. Then
$$
(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1.
$$

Thus $21n+4$ and $14n+3$ are coprime, so their ratio is in lowest terms.

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answered Jul 30 at 17:37

BWW

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up vote
2
down vote

For starters such as yourself, you can begin by assuming otherwise. That is the fraction is reducible. So $exists k in mathbbN, k > 1$ such that $k mid 21n + 4, k mid 14n+3implies 21n+4 = ak, 14n+3 = bk$ for some natural numbers $a,b$. Thus: $42n+8 = 2ak, 42n+9 = 3bkimplies 3bk - 2ak = 1implies k(3b-2a) = 1implies k = 1$, contradicting the assumption that $k > 1$. This means $textgcd(21n+4,14n+3) = 1$ or $dfrac21n+414n+3$ is irreducible.

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answered Jul 30 at 17:21

DeepSea

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote

Writing $(n,m)$ for the GCD of $m$ and $n$, immediately one has $(n,m)=(n,m-kn)$ for any $kinmathbbZ$. Then
$$
(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1.
$$

Thus $21n+4$ and $14n+3$ are coprime, so their ratio is in lowest terms.

share|cite|improve this answer

answered Jul 30 at 17:37

BWW

8,70122135

add a comment | 

up vote
6
down vote

Writing $(n,m)$ for the GCD of $m$ and $n$, immediately one has $(n,m)=(n,m-kn)$ for any $kinmathbbZ$. Then
$$
(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1.
$$

Thus $21n+4$ and $14n+3$ are coprime, so their ratio is in lowest terms.

share|cite|improve this answer

answered Jul 30 at 17:37

BWW

8,70122135

add a comment | 

up vote
6
down vote

up vote
6
down vote

Writing $(n,m)$ for the GCD of $m$ and $n$, immediately one has $(n,m)=(n,m-kn)$ for any $kinmathbbZ$. Then
$$
(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1.
$$

Thus $21n+4$ and $14n+3$ are coprime, so their ratio is in lowest terms.

share|cite|improve this answer

answered Jul 30 at 17:37

BWW

8,70122135

Writing $(n,m)$ for the GCD of $m$ and $n$, immediately one has $(n,m)=(n,m-kn)$ for any $kinmathbbZ$. Then
$$
(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1.
$$

Thus $21n+4$ and $14n+3$ are coprime, so their ratio is in lowest terms.

share|cite|improve this answer

answered Jul 30 at 17:37

BWW

8,70122135

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 17:37

BWW

8,70122135

answered Jul 30 at 17:37

BWW

8,70122135

answered Jul 30 at 17:37

BWW

8,70122135

8,70122135

add a comment | 

add a comment | 

up vote
2
down vote

For starters such as yourself, you can begin by assuming otherwise. That is the fraction is reducible. So $exists k in mathbbN, k > 1$ such that $k mid 21n + 4, k mid 14n+3implies 21n+4 = ak, 14n+3 = bk$ for some natural numbers $a,b$. Thus: $42n+8 = 2ak, 42n+9 = 3bkimplies 3bk - 2ak = 1implies k(3b-2a) = 1implies k = 1$, contradicting the assumption that $k > 1$. This means $textgcd(21n+4,14n+3) = 1$ or $dfrac21n+414n+3$ is irreducible.

share|cite|improve this answer

answered Jul 30 at 17:21

DeepSea

68.8k54284

add a comment | 

up vote
2
down vote

For starters such as yourself, you can begin by assuming otherwise. That is the fraction is reducible. So $exists k in mathbbN, k > 1$ such that $k mid 21n + 4, k mid 14n+3implies 21n+4 = ak, 14n+3 = bk$ for some natural numbers $a,b$. Thus: $42n+8 = 2ak, 42n+9 = 3bkimplies 3bk - 2ak = 1implies k(3b-2a) = 1implies k = 1$, contradicting the assumption that $k > 1$. This means $textgcd(21n+4,14n+3) = 1$ or $dfrac21n+414n+3$ is irreducible.

share|cite|improve this answer

answered Jul 30 at 17:21

DeepSea

68.8k54284

add a comment | 

up vote
2
down vote

up vote
2
down vote

For starters such as yourself, you can begin by assuming otherwise. That is the fraction is reducible. So $exists k in mathbbN, k > 1$ such that $k mid 21n + 4, k mid 14n+3implies 21n+4 = ak, 14n+3 = bk$ for some natural numbers $a,b$. Thus: $42n+8 = 2ak, 42n+9 = 3bkimplies 3bk - 2ak = 1implies k(3b-2a) = 1implies k = 1$, contradicting the assumption that $k > 1$. This means $textgcd(21n+4,14n+3) = 1$ or $dfrac21n+414n+3$ is irreducible.

share|cite|improve this answer

answered Jul 30 at 17:21

DeepSea

68.8k54284

For starters such as yourself, you can begin by assuming otherwise. That is the fraction is reducible. So $exists k in mathbbN, k > 1$ such that $k mid 21n + 4, k mid 14n+3implies 21n+4 = ak, 14n+3 = bk$ for some natural numbers $a,b$. Thus: $42n+8 = 2ak, 42n+9 = 3bkimplies 3bk - 2ak = 1implies k(3b-2a) = 1implies k = 1$, contradicting the assumption that $k > 1$. This means $textgcd(21n+4,14n+3) = 1$ or $dfrac21n+414n+3$ is irreducible.

share|cite|improve this answer

answered Jul 30 at 17:21

DeepSea

68.8k54284

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 17:21

DeepSea

68.8k54284

answered Jul 30 at 17:21

DeepSea

68.8k54284

answered Jul 30 at 17:21

DeepSea

68.8k54284

68.8k54284

add a comment | 

add a comment |