A five-digit number N.... A five-digit number $N$ is equal to $45$ times the product of its $5$ digits. Find $N$.
Please help. I am not sure how to solve this. I have a feeling it is simple

Let the digits of $n$ from left to right be $a,b,c,d$, and $e$, so that we’re told that
$$10^4a+10^3b+10^2c+10d+e=45abcde\;.\tag{1}$$
Clearly $n$ is a multiple of $5$, so $e$ is either $0$ or $5$. But obviously none of the digits is $0$, so $e=5$, and $(1)$ becomes
$$10^4a+10^3b+10^2c+10d+5=225abcd$$
or, after dividing through by $5$,
$$2\left(10^3a+10^2b+10c+d\right)+1=45abcd\;.\tag{2}$$
The lefthand side of $(2)$ is odd, so all of $a,b,c$, and $d$ must be odd as well. The one’s digit of $$2\left(10^3a+10^2b+10c+d\right)+1$$ is either $2d+1$ or $2d+1-10=2d-9$, depending on whether $d<5$ or not. However, the number in $(2)$ is an odd multiple of $5$, so its one’s digit is $5$, and either $2d+1=5$ or $2d-9=5$. Since $d$ is known to be odd, the former is impossible, $d=7$, and $(2)$ becomes
$$20\left(10^2a+10b+c\right)+15=315abc$$
or, after another division by $5$,
$$4\left(10^2a+10b+c\right)+3=63abc\;.\tag{3}$$
Reduce $(3)$ modulo $9$ to see that $$4\left(10^2a+10b+c\right)\equiv 6\pmod 9$$ and hence $10^2a+10b+c\equiv6\pmod9$, which further implies that $a+b+c\equiv6\pmod9$. We know that $a,b$, and $c$ are odd, so $a+b+c$ is odd, and therefore $a+b+c$ can only be $15$; the next larger possibility is $33$, which is too large, since $a+b+c\le3\cdot9=27$. The largest of $a,b$, and $c$ must therefore be at least $5$.
Now note that $100,000>n=45abcde=1575abc$, so $$abc\le\left\lfloor\frac{100,000}{1575}\right\rfloor=63\;.$$ This reduces the possibilities for $a,b$, and $c$ to $7,7,1$ (in some order) and $9,5,1$ (in some order). In the first case we have $45abcde=77175$, so $77175$ is a solution. In the second case we have $45abcde=70875$ and get no solution. The unique solution is therefore $n=77175$.
