Mixing
prime factorization of rare and distinct rational numbers [closed]
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The prime factorization of a natural number n can be written as n=pr ^2 where p and rare distinct and prime numbers. How many factors does n have, including 1 and itself?
prime-numbers
asked Feb 2 at 19:55
student9696student9696
6
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closed as off-topic by max_zorn, YiFan, Gibbs, Leucippus, Cesareo Feb 3 at 1:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
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The prime factorization of a natural number n can be written as n=pr ^2 where p and rare distinct and prime numbers. How many factors does n have, including 1 and itself?
prime-numbers
asked Feb 2 at 19:55
student9696student9696
6
$endgroup$
closed as off-topic by max_zorn, YiFan, Gibbs, Leucippus, Cesareo Feb 3 at 1:10
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – max_zorn, YiFan, Gibbs, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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I guess meant is "p and r are". In this case, it is a very easy exercise.
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– Peter
Feb 3 at 9:50
add a comment |
$begingroup$
The prime factorization of a natural number n can be written as n=pr ^2 where p and rare distinct and prime numbers. How many factors does n have, including 1 and itself?
prime-numbers
asked Feb 2 at 19:55
student9696student9696
6
$endgroup$
The prime factorization of a natural number n can be written as n=pr ^2 where p and rare distinct and prime numbers. How many factors does n have, including 1 and itself?
prime-numbers
prime-numbers
asked Feb 2 at 19:55
student9696student9696
6
asked Feb 2 at 19:55
student9696student9696
6
asked Feb 2 at 19:55
student9696student9696
6
asked Feb 2 at 19:55
student9696student9696
6
asked Feb 2 at 19:55
student9696student9696
6
6
closed as off-topic by max_zorn, YiFan, Gibbs, Leucippus, Cesareo Feb 3 at 1:10
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – max_zorn, YiFan, Gibbs, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by max_zorn, YiFan, Gibbs, Leucippus, Cesareo Feb 3 at 1:10
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – max_zorn, YiFan, Gibbs, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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I guess meant is "p and r are". In this case, it is a very easy exercise.
$endgroup$
– Peter
Feb 3 at 9:50
add a comment |
$begingroup$
I guess meant is "p and r are". In this case, it is a very easy exercise.
$endgroup$
– Peter
Feb 3 at 9:50
$begingroup$
I guess meant is "p and r are". In this case, it is a very easy exercise.
$endgroup$
– Peter
Feb 3 at 9:50
$begingroup$
I guess meant is "p and r are". In this case, it is a very easy exercise.
$endgroup$
– Peter
Feb 3 at 9:50
add a comment |
1 Answer
1
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The number of divisors of any number $n=p_1^{alpha_1}p_2^{alpha_2}cdots p_k^{alpha_k}$ (often called $tau(n)$) is equal to $(alpha_1+1)(alpha_2+1)cdots (alpha_k+1)$
In your case, $alpha_1=1, alpha_2=2$ so the number of divisors is $2cdot 3=6$
As suggested in the comment, such a simple set can be readily listed: $1,p,r,r^2,pr,pr^2$
answered Feb 2 at 21:25
Keith BackmanKeith Backman
1,5511812
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I was just re stating the question from a test word for word
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– student9696
Feb 3 at 2:13
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number of divisors of any number $n=p_1^{alpha_1}p_2^{alpha_2}cdots p_k^{alpha_k}$ (often called $tau(n)$) is equal to $(alpha_1+1)(alpha_2+1)cdots (alpha_k+1)$
In your case, $alpha_1=1, alpha_2=2$ so the number of divisors is $2cdot 3=6$
As suggested in the comment, such a simple set can be readily listed: $1,p,r,r^2,pr,pr^2$
answered Feb 2 at 21:25
Keith BackmanKeith Backman
1,5511812
$endgroup$
$begingroup$
I was just re stating the question from a test word for word
$endgroup$
– student9696
Feb 3 at 2:13
add a comment |
$begingroup$
The number of divisors of any number $n=p_1^{alpha_1}p_2^{alpha_2}cdots p_k^{alpha_k}$ (often called $tau(n)$) is equal to $(alpha_1+1)(alpha_2+1)cdots (alpha_k+1)$
In your case, $alpha_1=1, alpha_2=2$ so the number of divisors is $2cdot 3=6$
As suggested in the comment, such a simple set can be readily listed: $1,p,r,r^2,pr,pr^2$
answered Feb 2 at 21:25
Keith BackmanKeith Backman
1,5511812
$endgroup$
$begingroup$
I was just re stating the question from a test word for word
$endgroup$
– student9696
Feb 3 at 2:13
add a comment |
$begingroup$
The number of divisors of any number $n=p_1^{alpha_1}p_2^{alpha_2}cdots p_k^{alpha_k}$ (often called $tau(n)$) is equal to $(alpha_1+1)(alpha_2+1)cdots (alpha_k+1)$
In your case, $alpha_1=1, alpha_2=2$ so the number of divisors is $2cdot 3=6$
As suggested in the comment, such a simple set can be readily listed: $1,p,r,r^2,pr,pr^2$
answered Feb 2 at 21:25
Keith BackmanKeith Backman
1,5511812
$endgroup$
The number of divisors of any number $n=p_1^{alpha_1}p_2^{alpha_2}cdots p_k^{alpha_k}$ (often called $tau(n)$) is equal to $(alpha_1+1)(alpha_2+1)cdots (alpha_k+1)$
In your case, $alpha_1=1, alpha_2=2$ so the number of divisors is $2cdot 3=6$
As suggested in the comment, such a simple set can be readily listed: $1,p,r,r^2,pr,pr^2$
answered Feb 2 at 21:25
Keith BackmanKeith Backman
1,5511812
answered Feb 2 at 21:25
Keith BackmanKeith Backman
1,5511812
answered Feb 2 at 21:25
Keith BackmanKeith Backman
1,5511812
answered Feb 2 at 21:25
Keith BackmanKeith Backman
1,5511812
1,5511812
$begingroup$
I was just re stating the question from a test word for word
$endgroup$
– student9696
Feb 3 at 2:13
add a comment |
$begingroup$
I was just re stating the question from a test word for word
$endgroup$
– student9696
Feb 3 at 2:13
$begingroup$
I was just re stating the question from a test word for word
$endgroup$
– student9696
Feb 3 at 2:13
$begingroup$
I was just re stating the question from a test word for word
$endgroup$
– student9696
Feb 3 at 2:13
add a comment |
$begingroup$
I guess meant is "p and r are". In this case, it is a very easy exercise.
$endgroup$
– Peter
Feb 3 at 9:50