Concerning the product of all unique positive divisors












1















If the product of all the unique positive divisors of $n$, a positive integer which is not a perfect cube, is $n^2$, then the product of all the unique positive divisors of $n^2$ is:
(A) $n^3$
(B) $n^4$
(C) $n^6$
(D) $n^8$
(E) $n^9$




I tried using the formula: $frac{n^{d(n)}}{2}$, where $d(n)$ is the number of factors and I got the answer as $n^7/2$.










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  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 7 '18 at 7:11
















1















If the product of all the unique positive divisors of $n$, a positive integer which is not a perfect cube, is $n^2$, then the product of all the unique positive divisors of $n^2$ is:
(A) $n^3$
(B) $n^4$
(C) $n^6$
(D) $n^8$
(E) $n^9$




I tried using the formula: $frac{n^{d(n)}}{2}$, where $d(n)$ is the number of factors and I got the answer as $n^7/2$.










share|cite|improve this question
























  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 7 '18 at 7:11














1












1








1








If the product of all the unique positive divisors of $n$, a positive integer which is not a perfect cube, is $n^2$, then the product of all the unique positive divisors of $n^2$ is:
(A) $n^3$
(B) $n^4$
(C) $n^6$
(D) $n^8$
(E) $n^9$




I tried using the formula: $frac{n^{d(n)}}{2}$, where $d(n)$ is the number of factors and I got the answer as $n^7/2$.










share|cite|improve this question
















If the product of all the unique positive divisors of $n$, a positive integer which is not a perfect cube, is $n^2$, then the product of all the unique positive divisors of $n^2$ is:
(A) $n^3$
(B) $n^4$
(C) $n^6$
(D) $n^8$
(E) $n^9$




I tried using the formula: $frac{n^{d(n)}}{2}$, where $d(n)$ is the number of factors and I got the answer as $n^7/2$.







integers






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edited Nov 7 '18 at 8:13









Brahadeesh

6,11742361




6,11742361










asked Nov 7 '18 at 7:09









j.doe

62




62












  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 7 '18 at 7:11


















  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 7 '18 at 7:11
















Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 7 '18 at 7:11




Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 7 '18 at 7:11










2 Answers
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Since the product of positive divisors of $n$ is $n^{frac{d(n)}{2}}$. We want
$$n^{frac{d(n)}{2}}=n^2.$$
This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divisors of $n^2$ is
$$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$






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    0














    Hint: First, note that for the above condition to hold, $n$ cannot be prime and hence can be considered as the product of two primes, $p$ and $q$.



    Then, try to list out the factors of $n^2= p^2q^2$.






    share|cite|improve this answer





















    • Hello, @j.doe, you can accept the answer by ticking the question mark on the side if it helped you.
      – Rohan
      Nov 7 '18 at 8:56











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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    active

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    1














    Since the product of positive divisors of $n$ is $n^{frac{d(n)}{2}}$. We want
    $$n^{frac{d(n)}{2}}=n^2.$$
    This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divisors of $n^2$ is
    $$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$






    share|cite|improve this answer




























      1














      Since the product of positive divisors of $n$ is $n^{frac{d(n)}{2}}$. We want
      $$n^{frac{d(n)}{2}}=n^2.$$
      This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divisors of $n^2$ is
      $$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$






      share|cite|improve this answer


























        1












        1








        1






        Since the product of positive divisors of $n$ is $n^{frac{d(n)}{2}}$. We want
        $$n^{frac{d(n)}{2}}=n^2.$$
        This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divisors of $n^2$ is
        $$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$






        share|cite|improve this answer














        Since the product of positive divisors of $n$ is $n^{frac{d(n)}{2}}$. We want
        $$n^{frac{d(n)}{2}}=n^2.$$
        This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divisors of $n^2$ is
        $$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 27 '18 at 20:24

























        answered Dec 26 '18 at 23:44









        Anurag A

        25.6k12249




        25.6k12249























            0














            Hint: First, note that for the above condition to hold, $n$ cannot be prime and hence can be considered as the product of two primes, $p$ and $q$.



            Then, try to list out the factors of $n^2= p^2q^2$.






            share|cite|improve this answer





















            • Hello, @j.doe, you can accept the answer by ticking the question mark on the side if it helped you.
              – Rohan
              Nov 7 '18 at 8:56
















            0














            Hint: First, note that for the above condition to hold, $n$ cannot be prime and hence can be considered as the product of two primes, $p$ and $q$.



            Then, try to list out the factors of $n^2= p^2q^2$.






            share|cite|improve this answer





















            • Hello, @j.doe, you can accept the answer by ticking the question mark on the side if it helped you.
              – Rohan
              Nov 7 '18 at 8:56














            0












            0








            0






            Hint: First, note that for the above condition to hold, $n$ cannot be prime and hence can be considered as the product of two primes, $p$ and $q$.



            Then, try to list out the factors of $n^2= p^2q^2$.






            share|cite|improve this answer












            Hint: First, note that for the above condition to hold, $n$ cannot be prime and hence can be considered as the product of two primes, $p$ and $q$.



            Then, try to list out the factors of $n^2= p^2q^2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 7 '18 at 8:12









            Rohan

            27.7k42444




            27.7k42444












            • Hello, @j.doe, you can accept the answer by ticking the question mark on the side if it helped you.
              – Rohan
              Nov 7 '18 at 8:56


















            • Hello, @j.doe, you can accept the answer by ticking the question mark on the side if it helped you.
              – Rohan
              Nov 7 '18 at 8:56
















            Hello, @j.doe, you can accept the answer by ticking the question mark on the side if it helped you.
            – Rohan
            Nov 7 '18 at 8:56




            Hello, @j.doe, you can accept the answer by ticking the question mark on the side if it helped you.
            – Rohan
            Nov 7 '18 at 8:56


















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