Find all natural n such that gcd(n, 1260)=70 and the number of positive divisors of n is 30

Question

I'm very unsure about my solution for this problem, this is what I did, and I would appreciate a lot if someone can check it:

$gcd(n,1260)$ divides $n$, so write $n=70*m$.

$70=2*5*7$, then $\#Div_+(70) = 2^3=8$. So $\#Div_+(n) = 8*\#Div_+(m)$.

But we know that $\#Div_+(n) = 30$, and $\nexists k$ (natural) such that $8*k=30$.

Then, we can conclude that there is no such natural $n$.

Thanks for reading!

Answer 1

A sketch of a method:

1. You can compute the gcd of two numbers by writing their prime factorizations and then taking the smaller exponent of each prime factor. Let's write $n = 2^{e_2} \times 3^{e_3} \times 5^{e_5} \times 7^{e_7} \times \cdots$ (where all but a finite number of the exponents $e_p$ are zero). Now, $70$ factors into $2 \times 5 \times 7$ and $1260$ factors into $2^2 \times 3^2 \times 5 \times 7$. This means that $\min(e_2, 2) = 1$, $\min(e_3, 2) = 0$, $\min(e_5, 1) = 1$, and $\min(e_7, 1) = 1$.

2. The number of divisors of $n$ is $(e_2 + 1)(e_3+1)(e_5+1)(e_7+1)(e_{11} + 1)(e_{13} + 1)\cdots$. If this product is $30$, then $(e_2+1)(e_3+1)(e_5+1)(e_7+1)$ has to divide $30 = 2 \times 3 \times 5$. It's not too hard to make a (very short) list of all the possible values of $e_2, e_3, e_5, e_7$ that fit the equations in part (1) and such that $(e_2+1)(e_3+1)(e_5+1)(e_7+1)$ divides $30$. It will turn out that this product in fact equals $30$ for all the possible values of $e_2, e_3, e_5, e_7$, which means that $n$ can't have any prime factors above 7.

Answer 2

Thank to everyone for the feedback! This is my answer based on your help:

$1260=2^2*3^2*5*7$ and $70 = 2*5*7$, so $n = 2*5^{1+x_1}*7^{1+x^2}*p_1^{a_1}*...*p_k^{a_k}$.

$\#Div_+(n)=(1+1)(1+x_1+1)(1+x_2+1)(a_1+1)...(a_k+1)=30=2*3*5$

So we have that $(1+x_1+1)(1+x_2+1) = 3*5$, and either $x_1=1$ and $x_2=3$ or $x_1=3$ and $x_2=1$.

And then $n=2*5^2*7⁴$ or $n=2*5^4*7^2$.

Answer 3

By dividing $n$ by $70$, the problem is equivalent to finding those $m$ coprime to $1260/70=18$ for which $70m$ has exactly $30$ divisors.

Since $m$ is coprime to $18$, it is already not going to be divisible by $2$, so we only need to treat the cases where it is divisible by $5$ or $7$ separately.

If $m$ is coprime to $70$, then the number of divisors of $70m$ is $8$ times the number of divisors of $m$. Clearly, since $30$ is not divisible by $8$, $70m$ could not have exactly $30$ divisors in this case.

Suppose that either $m=5^k \cdot x$ or $m=7^k \cdot x$ where $k>0$ and neither $5$ nor $7$ divides $x$. Then, the number of divisors of $70m$ is $4$ times $k+2$ times the number of divisors of $x$. Clearly, since $30$ is not divisible by $4$, $70m$ could not have exactly $30$ divisors in this case either.

Suppose that $m$ is divisible by $35$, so $m=5^j \cdot 7^k \cdot x$ where $j$ and $k$ are positive and neither $5$ nor $7$ divides $x$. Then, the number of divisors of $70m$ is $2$ times $j+2$ times $k+2$ times the number of divisors of $x$. For this to equal $30$, $j+2$ times $k+2$ times the number of divisors of $x$ would have to be $15$.

Examining the factorizations of $15$ into $3$ positive factors (treating rearrangements as different factorizations), here are all the possibilities:

• $1 \cdot 1 \cdot 15$
• $1 \cdot 3 \cdot 5$
• $1 \cdot 5 \cdot 3$
• $1 \cdot 15 \cdot 1$
• $3 \cdot 1 \cdot 5$
• $3 \cdot 5 \cdot 1$
• $5 \cdot 1 \cdot 3$
• $5 \cdot 3 \cdot 1$
• $15 \cdot 1 \cdot 1$

Both $j+2$ and $k+2$ must be at least $3$, so one can eliminate the factorizations where one or both of the first $2$ factors are $1$. This leaves only two cases:

• Either $j+2=3$ (so $j=1$) and $k+2=5$ (so $k=3$), or
• $j+2=5$ (so $j=3$) and $k+2=3$ (so $k=1$)

In both cases, the last factor is $1$, so $x$ has exactly $1$ divisor, so $x$ must equal $1$.

So, the $2$ values of $n$ are $70 \cdot 5 \cdot 7^3=120,050$ and $70 \cdot 5^3 \cdot 7=61,250$.