If $PQRSPQRS$ is an eight-digit number with $56$ divisors, find the number of divisors of $PQRS$.

Question

$PQRS$ is a four-digit number where $P,Q,R,S$ are the digits of the number. If $PQRSPQRS$ is an eight-digit number with $56$ divisors (including $1$ and $PQRSPQRS$), find the number of divisors of $PQRS$ (including $1$ and $PQRS$).

If a number can be written as ${a_1}^{n_1}{a_2}^{n_2}...a_m^{n_m}$, where $a_1,a_2,...a_m$ are prime numbers then the number of divisors of the given number is $(n_1+1)(n_2+1)...(n_m+1)$

Using this, $56=8\times7=(7+1)(6+1)$. So, $PQRSPQRS=a_1^7a_2^6$

Or, $56=4\times2\times7=(3+1)(1+1)(6+1)$. So, $PQRSPQRS=a_1^3a_2^1a_3^6$

Or, $56=2\times2\times2\times7=(1+1)(1+1)(1+1)(6+1)$. So, $PQRSPQRS=a_1^1a_2^1a_3^1a_4^6$

$(a_1,a_2,a_3..$ etc are different in different cases.$)$

Also, $PQRS=10^3\times P+10^2\times Q+10^1\times R+S$

And $PQRSPQRS=(10^7+10^3)\times P+(10^6+10^2)\times Q+(10^5+10)\times R+(10^4+1)S$

I don't know how to relate divisors with this. I am not even sure if I am going in the right direction. After this I thought of subtracting $PQRS$ from $PQRSPQRS$. Again, not sure if that's helping.

Answer 1

Per the comments' reflection on $(137 \times 73)$, since the prime number $(7)$ divides $(56)$, it stands to reason that the 8 digit number must be divisible by some number of the form $(p^6) ~: ~p~$ prime.

This begs the question whether the 8 digit number has form $(p_1)^1 \times (p_2)^6 \times 73 \times 137 ~: ~p_1, p_2~$ primes, or whether the 8 digit number has form $(73)^3 \times 137 \times (p_2)^6~: ~p_2 ~$ prime.

Since $(2 \times [73]^3 \times 137)$ is a 9 digit number, the latter possibility is ruled out. The former possibility implies that the 4 digit number has $(2 \times 7)$ factors.

For what it's worth, identifying $(10001 = 73 \times 137)$ is routinely done with a pocket calculator, by examining prime numbers less than $(101).$