How to find all the positive integers n of 4 digits such that all its digits are perfect squares and n is a multiple of 2, 3, 5 and 7? I was trying to use the divisibility rules of 2,3,5 and 7 but I becomes very tedious and couldn's solve the problem. I think there could be a faster way to solve it or to apply those rules. Please help me and thank you very much.

Hint: The last digit has to be $0$ to be divisible by $2$ and $5$ simultaneously. Thus, The number you are seeking is $ABC0$ where $A,B,C\in\{0,1,4,9\}$ but $A \ne 0$.
Also, $A+B+C=3k$ for divisibility via $3$-Equation $(1)$
And $A+B+C =7 \lambda$-Equation$(2)$
Although I'd recommend checking divisibility rules for $7$ before taking equation $(2)$ into account.
or in short, it should be divisible by $21$.
I can see only one number here i.e. $4410$
Disclaimer: But there can be more. I haven't check the whole sample space yet.
