I want to know my mistake in the method
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1
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Here's the question:
Let $N$ be a positive integer, not divisible by $6$. Suppose
$N$ has $6$ positive divisors, the number of positive divisors of $9N$ is:
I know how approach these questions by adding $1$ to the powers of the exponent and then multiplying them.
Here the result of the result will be $6*3$ , so the result will be 18 divisors because multiplication of N with $9$ will be $3^2$ which means extra power $2$.
I want to know where I'm wrong in my approach towards this concept.
The question has no options.
prime-numbers prime-factorization
asked Nov 19 at 19:14
Saksham
905614
add a comment |
up vote
1
down vote
favorite
Here's the question:
Let $N$ be a positive integer, not divisible by $6$. Suppose
$N$ has $6$ positive divisors, the number of positive divisors of $9N$ is:
I know how approach these questions by adding $1$ to the powers of the exponent and then multiplying them.
Here the result of the result will be $6*3$ , so the result will be 18 divisors because multiplication of N with $9$ will be $3^2$ which means extra power $2$.
I want to know where I'm wrong in my approach towards this concept.
The question has no options.
prime-numbers prime-factorization
asked Nov 19 at 19:14
Saksham
905614
For all we know from the question, we could for example have $N = 3 cdot 5^2$ or $N = 3^2 cdot 5$.
– Daniel Schepler
Nov 19 at 19:22
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
Nov 19 at 19:25
1
The problem as stated doesn't have enough information to give a unique answer. The best we could do, without more context, would be to give a small set of possible answers.
– Daniel Schepler
Nov 19 at 19:28
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
Nov 19 at 19:38
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here's the question:
Let $N$ be a positive integer, not divisible by $6$. Suppose
$N$ has $6$ positive divisors, the number of positive divisors of $9N$ is:
I know how approach these questions by adding $1$ to the powers of the exponent and then multiplying them.
Here the result of the result will be $6*3$ , so the result will be 18 divisors because multiplication of N with $9$ will be $3^2$ which means extra power $2$.
I want to know where I'm wrong in my approach towards this concept.
The question has no options.
prime-numbers prime-factorization
asked Nov 19 at 19:14
Saksham
905614
Here's the question:
Let $N$ be a positive integer, not divisible by $6$. Suppose
$N$ has $6$ positive divisors, the number of positive divisors of $9N$ is:
I know how approach these questions by adding $1$ to the powers of the exponent and then multiplying them.
Here the result of the result will be $6*3$ , so the result will be 18 divisors because multiplication of N with $9$ will be $3^2$ which means extra power $2$.
I want to know where I'm wrong in my approach towards this concept.
The question has no options.
prime-numbers prime-factorization
prime-numbers prime-factorization
asked Nov 19 at 19:14
Saksham
905614
asked Nov 19 at 19:14
Saksham
905614
edited Nov 20 at 18:25
asked Nov 19 at 19:14
Saksham
905614
asked Nov 19 at 19:14
Saksham
905614
asked Nov 19 at 19:14
Saksham
905614
905614
For all we know from the question, we could for example have $N = 3 cdot 5^2$ or $N = 3^2 cdot 5$.
– Daniel Schepler
Nov 19 at 19:22
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
Nov 19 at 19:25
1
The problem as stated doesn't have enough information to give a unique answer. The best we could do, without more context, would be to give a small set of possible answers.
– Daniel Schepler
Nov 19 at 19:28
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
Nov 19 at 19:38
add a comment |
For all we know from the question, we could for example have $N = 3 cdot 5^2$ or $N = 3^2 cdot 5$.
– Daniel Schepler
Nov 19 at 19:22
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
Nov 19 at 19:25
1
The problem as stated doesn't have enough information to give a unique answer. The best we could do, without more context, would be to give a small set of possible answers.
– Daniel Schepler
Nov 19 at 19:28
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
Nov 19 at 19:38
For all we know from the question, we could for example have $N = 3 cdot 5^2$ or $N = 3^2 cdot 5$.
– Daniel Schepler
Nov 19 at 19:22
For all we know from the question, we could for example have $N = 3 cdot 5^2$ or $N = 3^2 cdot 5$.
– Daniel Schepler
Nov 19 at 19:22
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
Nov 19 at 19:25
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
Nov 19 at 19:25
1
1
The problem as stated doesn't have enough information to give a unique answer. The best we could do, without more context, would be to give a small set of possible answers.
– Daniel Schepler
Nov 19 at 19:28
The problem as stated doesn't have enough information to give a unique answer. The best we could do, without more context, would be to give a small set of possible answers.
– Daniel Schepler
Nov 19 at 19:28
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
Nov 19 at 19:38
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
Nov 19 at 19:38
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
If $N$ has $6$ divisors it is of the form $p^2q$ or $p^5$ for $p,q$ prime. The fact that it is not divisible by $6$ means that at least one of $p,q$ is neither $2$ nor $3$, but that doesn't really help with the problem because we only care about factors of $3$. $9N$ could now be $3^4q,p^23^3,3^2p^2q, 3^7,text { or } 3^2p^5$which have $10, 12,18,8text { and } 18$ divisors respectively.
Added: $N$ could have $0,1,2, text {or } 5$ factors of $3$. Adding two more multiplies the number of factors by $3,2,frac 53, text { or } frac 43$ respectively, giving $18,12,10,8$ factors for $9N$
answered Nov 19 at 20:21
Ross Millikan
288k23195365
If $N$ already has $6$ factors that means the product of the exponets (after adding $1$) of the prime factors of $N$ is $6$ and then $9N$ would bring $3^2$ which means the earlier product of exponents $ + 1 * 3$ (since $3^2$) and $2 + 1$ is $3$. Also $3^4$ can be broken down as $3^2 . 3^2 . q$ and hence the number of divisors will be $3 * 3 * 2$ $i.e.$ $18$ . I hope it's readable now.
– Saksham
Nov 20 at 3:57
Yes, it is readable now, but I don't understand the point. I added another bit that might help. $9N$ could have as many as $18$ factors, which would be the case if $N$ has no factors of $3$. If $N$ does have some factors of $3$ the number of factors of $9N$ will be less.
– Ross Millikan
Nov 20 at 4:16
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If $N$ has $6$ divisors it is of the form $p^2q$ or $p^5$ for $p,q$ prime. The fact that it is not divisible by $6$ means that at least one of $p,q$ is neither $2$ nor $3$, but that doesn't really help with the problem because we only care about factors of $3$. $9N$ could now be $3^4q,p^23^3,3^2p^2q, 3^7,text { or } 3^2p^5$which have $10, 12,18,8text { and } 18$ divisors respectively.
Added: $N$ could have $0,1,2, text {or } 5$ factors of $3$. Adding two more multiplies the number of factors by $3,2,frac 53, text { or } frac 43$ respectively, giving $18,12,10,8$ factors for $9N$
answered Nov 19 at 20:21
Ross Millikan
288k23195365
If $N$ already has $6$ factors that means the product of the exponets (after adding $1$) of the prime factors of $N$ is $6$ and then $9N$ would bring $3^2$ which means the earlier product of exponents $ + 1 * 3$ (since $3^2$) and $2 + 1$ is $3$. Also $3^4$ can be broken down as $3^2 . 3^2 . q$ and hence the number of divisors will be $3 * 3 * 2$ $i.e.$ $18$ . I hope it's readable now.
– Saksham
Nov 20 at 3:57
Yes, it is readable now, but I don't understand the point. I added another bit that might help. $9N$ could have as many as $18$ factors, which would be the case if $N$ has no factors of $3$. If $N$ does have some factors of $3$ the number of factors of $9N$ will be less.
– Ross Millikan
Nov 20 at 4:16
add a comment |
up vote
1
down vote
If $N$ has $6$ divisors it is of the form $p^2q$ or $p^5$ for $p,q$ prime. The fact that it is not divisible by $6$ means that at least one of $p,q$ is neither $2$ nor $3$, but that doesn't really help with the problem because we only care about factors of $3$. $9N$ could now be $3^4q,p^23^3,3^2p^2q, 3^7,text { or } 3^2p^5$which have $10, 12,18,8text { and } 18$ divisors respectively.
Added: $N$ could have $0,1,2, text {or } 5$ factors of $3$. Adding two more multiplies the number of factors by $3,2,frac 53, text { or } frac 43$ respectively, giving $18,12,10,8$ factors for $9N$
answered Nov 19 at 20:21
Ross Millikan
288k23195365
If $N$ already has $6$ factors that means the product of the exponets (after adding $1$) of the prime factors of $N$ is $6$ and then $9N$ would bring $3^2$ which means the earlier product of exponents $ + 1 * 3$ (since $3^2$) and $2 + 1$ is $3$. Also $3^4$ can be broken down as $3^2 . 3^2 . q$ and hence the number of divisors will be $3 * 3 * 2$ $i.e.$ $18$ . I hope it's readable now.
– Saksham
Nov 20 at 3:57
Yes, it is readable now, but I don't understand the point. I added another bit that might help. $9N$ could have as many as $18$ factors, which would be the case if $N$ has no factors of $3$. If $N$ does have some factors of $3$ the number of factors of $9N$ will be less.
– Ross Millikan
Nov 20 at 4:16
add a comment |
up vote
1
down vote
up vote
1
down vote
If $N$ has $6$ divisors it is of the form $p^2q$ or $p^5$ for $p,q$ prime. The fact that it is not divisible by $6$ means that at least one of $p,q$ is neither $2$ nor $3$, but that doesn't really help with the problem because we only care about factors of $3$. $9N$ could now be $3^4q,p^23^3,3^2p^2q, 3^7,text { or } 3^2p^5$which have $10, 12,18,8text { and } 18$ divisors respectively.
Added: $N$ could have $0,1,2, text {or } 5$ factors of $3$. Adding two more multiplies the number of factors by $3,2,frac 53, text { or } frac 43$ respectively, giving $18,12,10,8$ factors for $9N$
answered Nov 19 at 20:21
Ross Millikan
288k23195365
If $N$ has $6$ divisors it is of the form $p^2q$ or $p^5$ for $p,q$ prime. The fact that it is not divisible by $6$ means that at least one of $p,q$ is neither $2$ nor $3$, but that doesn't really help with the problem because we only care about factors of $3$. $9N$ could now be $3^4q,p^23^3,3^2p^2q, 3^7,text { or } 3^2p^5$which have $10, 12,18,8text { and } 18$ divisors respectively.
Added: $N$ could have $0,1,2, text {or } 5$ factors of $3$. Adding two more multiplies the number of factors by $3,2,frac 53, text { or } frac 43$ respectively, giving $18,12,10,8$ factors for $9N$
answered Nov 19 at 20:21
Ross Millikan
288k23195365
edited Nov 20 at 3:09
answered Nov 19 at 20:21
Ross Millikan
288k23195365
answered Nov 19 at 20:21
Ross Millikan
288k23195365
answered Nov 19 at 20:21
Ross Millikan
288k23195365
288k23195365
If $N$ already has $6$ factors that means the product of the exponets (after adding $1$) of the prime factors of $N$ is $6$ and then $9N$ would bring $3^2$ which means the earlier product of exponents $ + 1 * 3$ (since $3^2$) and $2 + 1$ is $3$. Also $3^4$ can be broken down as $3^2 . 3^2 . q$ and hence the number of divisors will be $3 * 3 * 2$ $i.e.$ $18$ . I hope it's readable now.
– Saksham
Nov 20 at 3:57
Yes, it is readable now, but I don't understand the point. I added another bit that might help. $9N$ could have as many as $18$ factors, which would be the case if $N$ has no factors of $3$. If $N$ does have some factors of $3$ the number of factors of $9N$ will be less.
– Ross Millikan
Nov 20 at 4:16
add a comment |
If $N$ already has $6$ factors that means the product of the exponets (after adding $1$) of the prime factors of $N$ is $6$ and then $9N$ would bring $3^2$ which means the earlier product of exponents $ + 1 * 3$ (since $3^2$) and $2 + 1$ is $3$. Also $3^4$ can be broken down as $3^2 . 3^2 . q$ and hence the number of divisors will be $3 * 3 * 2$ $i.e.$ $18$ . I hope it's readable now.
– Saksham
Nov 20 at 3:57
Yes, it is readable now, but I don't understand the point. I added another bit that might help. $9N$ could have as many as $18$ factors, which would be the case if $N$ has no factors of $3$. If $N$ does have some factors of $3$ the number of factors of $9N$ will be less.
– Ross Millikan
Nov 20 at 4:16
If $N$ already has $6$ factors that means the product of the exponets (after adding $1$) of the prime factors of $N$ is $6$ and then $9N$ would bring $3^2$ which means the earlier product of exponents $ + 1 * 3$ (since $3^2$) and $2 + 1$ is $3$. Also $3^4$ can be broken down as $3^2 . 3^2 . q$ and hence the number of divisors will be $3 * 3 * 2$ $i.e.$ $18$ . I hope it's readable now.
– Saksham
Nov 20 at 3:57
If $N$ already has $6$ factors that means the product of the exponets (after adding $1$) of the prime factors of $N$ is $6$ and then $9N$ would bring $3^2$ which means the earlier product of exponents $ + 1 * 3$ (since $3^2$) and $2 + 1$ is $3$. Also $3^4$ can be broken down as $3^2 . 3^2 . q$ and hence the number of divisors will be $3 * 3 * 2$ $i.e.$ $18$ . I hope it's readable now.
– Saksham
Nov 20 at 3:57
Yes, it is readable now, but I don't understand the point. I added another bit that might help. $9N$ could have as many as $18$ factors, which would be the case if $N$ has no factors of $3$. If $N$ does have some factors of $3$ the number of factors of $9N$ will be less.
– Ross Millikan
Nov 20 at 4:16
Yes, it is readable now, but I don't understand the point. I added another bit that might help. $9N$ could have as many as $18$ factors, which would be the case if $N$ has no factors of $3$. If $N$ does have some factors of $3$ the number of factors of $9N$ will be less.
– Ross Millikan
Nov 20 at 4:16
add a comment |
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For all we know from the question, we could for example have $N = 3 cdot 5^2$ or $N = 3^2 cdot 5$.
– Daniel Schepler
Nov 19 at 19:22
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
Nov 19 at 19:25
1
The problem as stated doesn't have enough information to give a unique answer. The best we could do, without more context, would be to give a small set of possible answers.
– Daniel Schepler
Nov 19 at 19:28
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
Nov 19 at 19:38