Concerning the product of all unique positive divisors
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
edited Nov 7 '18 at 8:13
Brahadeesh
6,11742361
asked Nov 7 '18 at 7:09
j.doe
62
add a comment |
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
edited Nov 7 '18 at 8:13
Brahadeesh
6,11742361
asked Nov 7 '18 at 7:09
j.doe
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
add a comment |
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
edited Nov 7 '18 at 8:13
Brahadeesh
6,11742361
asked Nov 7 '18 at 7:09
j.doe
62
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
integers
edited Nov 7 '18 at 8:13
Brahadeesh
6,11742361
asked Nov 7 '18 at 7:09
j.doe
62
edited Nov 7 '18 at 8:13
Brahadeesh
6,11742361
asked Nov 7 '18 at 7:09
j.doe
62
edited Nov 7 '18 at 8:13
Brahadeesh
6,11742361
edited Nov 7 '18 at 8:13
Brahadeesh
6,11742361
edited Nov 7 '18 at 8:13
Brahadeesh
6,11742361
6,11742361
asked Nov 7 '18 at 7:09
j.doe
62
asked Nov 7 '18 at 7:09
j.doe
62
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
add a comment |
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
add a comment |
2 Answers
2
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oldest
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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
$$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$
answered Dec 26 '18 at 23:44
Anurag A
25.6k12249
add a comment |
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 $n^2= p^2q^2$.
answered Nov 7 '18 at 8:12
Rohan
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$
answered Dec 26 '18 at 23:44
Anurag A
25.6k12249
add a comment |
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
$$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$
answered Dec 26 '18 at 23:44
Anurag A
25.6k12249
add a comment |
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
$$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$
answered Dec 26 '18 at 23:44
Anurag A
25.6k12249
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
$$(n^2)^{frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$
answered Dec 26 '18 at 23:44
Anurag A
25.6k12249
edited Dec 27 '18 at 20:24
answered Dec 26 '18 at 23:44
Anurag A
25.6k12249
answered Dec 26 '18 at 23:44
Anurag A
25.6k12249
answered Dec 26 '18 at 23:44
Anurag A
25.6k12249
25.6k12249
add a comment |
add a comment |
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 $n^2= p^2q^2$.
answered Nov 7 '18 at 8:12
Rohan
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
add a comment |
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 $n^2= p^2q^2$.
answered Nov 7 '18 at 8:12
Rohan
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
add a comment |
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 $n^2= p^2q^2$.
answered Nov 7 '18 at 8:12
Rohan
27.7k42444
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 $n^2= p^2q^2$.
answered Nov 7 '18 at 8:12
Rohan
27.7k42444
answered Nov 7 '18 at 8:12
Rohan
27.7k42444
answered Nov 7 '18 at 8:12
Rohan
27.7k42444
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
add a comment |
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
add a comment |
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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