Mixing
I want to know my mistake in the method
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 2 days ago
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 2 days ago
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
2 days ago
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
2 days ago
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
2 days ago
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
2 days ago
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 2 days ago
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 2 days ago
Saksham
905614
asked 2 days ago
Saksham
905614
edited yesterday
asked 2 days ago
Saksham
905614
asked 2 days ago
Saksham
905614
asked 2 days ago
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
2 days ago
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
2 days ago
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
2 days ago
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
2 days ago
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
2 days ago
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
2 days ago
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
2 days ago
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
2 days ago
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
2 days ago
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
2 days ago
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
2 days ago
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
2 days ago
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
2 days ago
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
2 days ago
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
2 days ago
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
2 days ago
add a comment |
1 Answer
1
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 2 days ago
Ross Millikan
287k23195364
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
yesterday
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
yesterday
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 2 days ago
Ross Millikan
287k23195364
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
yesterday
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
yesterday
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 2 days ago
Ross Millikan
287k23195364
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
yesterday
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
yesterday
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 2 days ago
Ross Millikan
287k23195364
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 2 days ago
Ross Millikan
287k23195364
edited yesterday
answered 2 days ago
Ross Millikan
287k23195364
answered 2 days ago
Ross Millikan
287k23195364
answered 2 days ago
Ross Millikan
287k23195364
287k23195364
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005379%2fi-want-to-know-my-mistake-in-the-method%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
2 days ago
@DanielSchepler not necessarily everytime this will be the case, how can you generalise this?
– Saksham
2 days ago
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
2 days ago
@DanielSchepler this is the only data given in the question. This is an olympiad question.
– Saksham
2 days ago