Determine the number of positive integers $a$ such that $amid 9!$ and gcd $(a, 3600)=180$.












0














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







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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


















0














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







share|cite|improve this question




















  • 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
















0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 22:53







hunnybuns

















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
















  • 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












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.






share|cite|improve this answer































    0














    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$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      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.






      share|cite|improve this answer




























        2














        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.






        share|cite|improve this answer


























          2












          2








          2






          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 23:01









          Bernard

          118k639112




          118k639112










          answered Dec 12 '18 at 22:53









          plattyplatty

          3,370320




          3,370320























              0














              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$






              share|cite|improve this answer


























                0














                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$






                share|cite|improve this answer
























                  0












                  0








                  0






                  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$






                  share|cite|improve this answer












                  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$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '18 at 0:23









                  Bill DubuqueBill Dubuque

                  209k29190631




                  209k29190631






























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