Determine the number of positive integers $a$ such that $amid 9!$ and gcd $(a, 3600)=180$.
Determine the number of positive integers $a$ such that $amid 9!$ and gcd $(a, 3600)=180$.
What I know as of now is that $180mid 9!$ and that $180le ale9!$.
The prime factorization of 180 is $(2^2)(3^2)(5^1)$
and the prime factorization of 9! is $(2^7)(3^4)(5^1)(7^1)$.
So I think I figured it out, there are exactly 7 positive integers for $a$. I'm not sure if this is correct, but this is what I have concluded.
divisibility greatest-common-divisor prime-factorization
asked Dec 12 '18 at 22:44
hunnybunshunnybuns
15
add a comment |
Determine the number of positive integers $a$ such that $amid 9!$ and gcd $(a, 3600)=180$.
What I know as of now is that $180mid 9!$ and that $180le ale9!$.
The prime factorization of 180 is $(2^2)(3^2)(5^1)$
and the prime factorization of 9! is $(2^7)(3^4)(5^1)(7^1)$.
So I think I figured it out, there are exactly 7 positive integers for $a$. I'm not sure if this is correct, but this is what I have concluded.
divisibility greatest-common-divisor prime-factorization
asked Dec 12 '18 at 22:44
hunnybunshunnybuns
15
2
Well, one thing that you could do to make the life of the reader even easier is adding the prime factorization of 3600 to your post as well!
– Vincent
Dec 12 '18 at 22:48
add a comment |
Determine the number of positive integers $a$ such that $amid 9!$ and gcd $(a, 3600)=180$.
What I know as of now is that $180mid 9!$ and that $180le ale9!$.
The prime factorization of 180 is $(2^2)(3^2)(5^1)$
and the prime factorization of 9! is $(2^7)(3^4)(5^1)(7^1)$.
So I think I figured it out, there are exactly 7 positive integers for $a$. I'm not sure if this is correct, but this is what I have concluded.
divisibility greatest-common-divisor prime-factorization
asked Dec 12 '18 at 22:44
hunnybunshunnybuns
15
Determine the number of positive integers $a$ such that $amid 9!$ and gcd $(a, 3600)=180$.
What I know as of now is that $180mid 9!$ and that $180le ale9!$.
The prime factorization of 180 is $(2^2)(3^2)(5^1)$
and the prime factorization of 9! is $(2^7)(3^4)(5^1)(7^1)$.
So I think I figured it out, there are exactly 7 positive integers for $a$. I'm not sure if this is correct, but this is what I have concluded.
divisibility greatest-common-divisor prime-factorization
divisibility greatest-common-divisor prime-factorization
asked Dec 12 '18 at 22:44
hunnybunshunnybuns
15
asked Dec 12 '18 at 22:44
hunnybunshunnybuns
15
edited Dec 12 '18 at 22:53
hunnybuns
asked Dec 12 '18 at 22:44
hunnybunshunnybuns
15
asked Dec 12 '18 at 22:44
hunnybunshunnybuns
15
asked Dec 12 '18 at 22:44
hunnybunshunnybuns
15
15
2
Well, one thing that you could do to make the life of the reader even easier is adding the prime factorization of 3600 to your post as well!
– Vincent
Dec 12 '18 at 22:48
add a comment |
2
Well, one thing that you could do to make the life of the reader even easier is adding the prime factorization of 3600 to your post as well!
– Vincent
Dec 12 '18 at 22:48
2
2
Well, one thing that you could do to make the life of the reader even easier is adding the prime factorization of 3600 to your post as well!
– Vincent
Dec 12 '18 at 22:48
Well, one thing that you could do to make the life of the reader even easier is adding the prime factorization of 3600 to your post as well!
– Vincent
Dec 12 '18 at 22:48
add a comment |
2 Answers
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What we know:
- $180 mid a$
$a$ and $3600$ share no common factors greater than $180$. In particular, any prime divisor of $3600$ cannot occur more times in $a$ than it does in $180$.- $a mid 9!$
So one strategy for counting the number of possible $a$'s can be counting the number of ways to build one, using these facts. First we know that $180 mid a$, so $a$'s prime factorization must contain $(2^2)(3^2)(5^1)$. Additionally, it cannot contain any more $2$'s or $5$'s, as this would give a larger gcd with $3600$ (however, it can contain more $3$'s).
Considering the factorization of $9! = (2^7)(3^4)(5^1)(7^1)$, and removing the $180$ which is already fixed leaves the factorization $(2^5)(3^2)(7^1)$ to work with. But we can't multiply by $2$, as noted above. So we can really multiply by any factor of $(3^2)(7^1)$. There are $6$ such numbers -- we have $3$ ways to choose a power of $3$ and $2$ ways to choose a power of $7$ to use.
edited Dec 12 '18 at 23:01
Bernard
118k639112
answered Dec 12 '18 at 22:53
plattyplatty
3,370320
add a comment |
Hint $ $ Let $,d={large frac{a}{180}},$ and cancel $180$ to get: $,(d,20)=1 $ & $ d:{largemid, frac{9!}{180}}! = 2^5cdot 3^2cdot 7iff dmid 3^2cdot 7$
answered Dec 26 '18 at 0:23
Bill DubuqueBill Dubuque
209k29190631
add a comment |
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2 Answers
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2 Answers
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What we know:
- $180 mid a$
$a$ and $3600$ share no common factors greater than $180$. In particular, any prime divisor of $3600$ cannot occur more times in $a$ than it does in $180$.- $a mid 9!$
So one strategy for counting the number of possible $a$'s can be counting the number of ways to build one, using these facts. First we know that $180 mid a$, so $a$'s prime factorization must contain $(2^2)(3^2)(5^1)$. Additionally, it cannot contain any more $2$'s or $5$'s, as this would give a larger gcd with $3600$ (however, it can contain more $3$'s).
Considering the factorization of $9! = (2^7)(3^4)(5^1)(7^1)$, and removing the $180$ which is already fixed leaves the factorization $(2^5)(3^2)(7^1)$ to work with. But we can't multiply by $2$, as noted above. So we can really multiply by any factor of $(3^2)(7^1)$. There are $6$ such numbers -- we have $3$ ways to choose a power of $3$ and $2$ ways to choose a power of $7$ to use.
edited Dec 12 '18 at 23:01
Bernard
118k639112
answered Dec 12 '18 at 22:53
plattyplatty
3,370320
add a comment |
What we know:
- $180 mid a$
$a$ and $3600$ share no common factors greater than $180$. In particular, any prime divisor of $3600$ cannot occur more times in $a$ than it does in $180$.- $a mid 9!$
So one strategy for counting the number of possible $a$'s can be counting the number of ways to build one, using these facts. First we know that $180 mid a$, so $a$'s prime factorization must contain $(2^2)(3^2)(5^1)$. Additionally, it cannot contain any more $2$'s or $5$'s, as this would give a larger gcd with $3600$ (however, it can contain more $3$'s).
Considering the factorization of $9! = (2^7)(3^4)(5^1)(7^1)$, and removing the $180$ which is already fixed leaves the factorization $(2^5)(3^2)(7^1)$ to work with. But we can't multiply by $2$, as noted above. So we can really multiply by any factor of $(3^2)(7^1)$. There are $6$ such numbers -- we have $3$ ways to choose a power of $3$ and $2$ ways to choose a power of $7$ to use.
edited Dec 12 '18 at 23:01
Bernard
118k639112
answered Dec 12 '18 at 22:53
plattyplatty
3,370320
add a comment |
What we know:
- $180 mid a$
$a$ and $3600$ share no common factors greater than $180$. In particular, any prime divisor of $3600$ cannot occur more times in $a$ than it does in $180$.- $a mid 9!$
So one strategy for counting the number of possible $a$'s can be counting the number of ways to build one, using these facts. First we know that $180 mid a$, so $a$'s prime factorization must contain $(2^2)(3^2)(5^1)$. Additionally, it cannot contain any more $2$'s or $5$'s, as this would give a larger gcd with $3600$ (however, it can contain more $3$'s).
Considering the factorization of $9! = (2^7)(3^4)(5^1)(7^1)$, and removing the $180$ which is already fixed leaves the factorization $(2^5)(3^2)(7^1)$ to work with. But we can't multiply by $2$, as noted above. So we can really multiply by any factor of $(3^2)(7^1)$. There are $6$ such numbers -- we have $3$ ways to choose a power of $3$ and $2$ ways to choose a power of $7$ to use.
edited Dec 12 '18 at 23:01
Bernard
118k639112
answered Dec 12 '18 at 22:53
plattyplatty
3,370320
What we know:
- $180 mid a$
$a$ and $3600$ share no common factors greater than $180$. In particular, any prime divisor of $3600$ cannot occur more times in $a$ than it does in $180$.- $a mid 9!$
So one strategy for counting the number of possible $a$'s can be counting the number of ways to build one, using these facts. First we know that $180 mid a$, so $a$'s prime factorization must contain $(2^2)(3^2)(5^1)$. Additionally, it cannot contain any more $2$'s or $5$'s, as this would give a larger gcd with $3600$ (however, it can contain more $3$'s).
Considering the factorization of $9! = (2^7)(3^4)(5^1)(7^1)$, and removing the $180$ which is already fixed leaves the factorization $(2^5)(3^2)(7^1)$ to work with. But we can't multiply by $2$, as noted above. So we can really multiply by any factor of $(3^2)(7^1)$. There are $6$ such numbers -- we have $3$ ways to choose a power of $3$ and $2$ ways to choose a power of $7$ to use.
edited Dec 12 '18 at 23:01
Bernard
118k639112
answered Dec 12 '18 at 22:53
plattyplatty
3,370320
edited Dec 12 '18 at 23:01
Bernard
118k639112
edited Dec 12 '18 at 23:01
Bernard
118k639112
edited Dec 12 '18 at 23:01
Bernard
118k639112
118k639112
answered Dec 12 '18 at 22:53
plattyplatty
3,370320
answered Dec 12 '18 at 22:53
plattyplatty
3,370320
answered Dec 12 '18 at 22:53
plattyplatty
3,370320
3,370320
add a comment |
add a comment |
Hint $ $ Let $,d={large frac{a}{180}},$ and cancel $180$ to get: $,(d,20)=1 $ & $ d:{largemid, frac{9!}{180}}! = 2^5cdot 3^2cdot 7iff dmid 3^2cdot 7$
answered Dec 26 '18 at 0:23
Bill DubuqueBill Dubuque
209k29190631
add a comment |
Hint $ $ Let $,d={large frac{a}{180}},$ and cancel $180$ to get: $,(d,20)=1 $ & $ d:{largemid, frac{9!}{180}}! = 2^5cdot 3^2cdot 7iff dmid 3^2cdot 7$
answered Dec 26 '18 at 0:23
Bill DubuqueBill Dubuque
209k29190631
add a comment |
Hint $ $ Let $,d={large frac{a}{180}},$ and cancel $180$ to get: $,(d,20)=1 $ & $ d:{largemid, frac{9!}{180}}! = 2^5cdot 3^2cdot 7iff dmid 3^2cdot 7$
answered Dec 26 '18 at 0:23
Bill DubuqueBill Dubuque
209k29190631
Hint $ $ Let $,d={large frac{a}{180}},$ and cancel $180$ to get: $,(d,20)=1 $ & $ d:{largemid, frac{9!}{180}}! = 2^5cdot 3^2cdot 7iff dmid 3^2cdot 7$
answered Dec 26 '18 at 0:23
Bill DubuqueBill Dubuque
209k29190631
answered Dec 26 '18 at 0:23
Bill DubuqueBill Dubuque
209k29190631
answered Dec 26 '18 at 0:23
Bill DubuqueBill Dubuque
209k29190631
answered Dec 26 '18 at 0:23
Bill DubuqueBill Dubuque
209k29190631
209k29190631
add a comment |
add a comment |
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2
Well, one thing that you could do to make the life of the reader even easier is adding the prime factorization of 3600 to your post as well!
– Vincent
Dec 12 '18 at 22:48