Japan Junior Mathematical Olympiad

$$\boxed{913}$$ The given equation can be transformed to $(a-cd)(b-1)=0$, so it is enough to determine quadruples $(a,b,c,d)$ satisfying either $a-cd=0$ or $b-1=0$.

For each $a=1,2,\dots ,9$ the number of pairs $(c,d)$ satisfying $a=cd$ is the same as the number of positive factors of $a$. Therefore, the number of triplets $(a,b,c)$ for which $a=cd$ is satisfied is $1+2+2+3+2+4+2+4+3=23$.

When $b=1$, we can choose $a,b,c,d$ arbitrarily to satisfy the condition. So, there are $9\times 9\times 9=729$ ways of quadruples of the form $(a,1,c,d)$ to satisfy the condition, while if $b\ne 1$, then for each such $b$ there are 23 ways to choose $(a,c,d)$ so that $(a,b,c,d)$ becomes a quadruple satisfying the condition. Since $b\ne 1$ can be chosen 8 different ways, we see that the number of quadruples satisfying the condition is $729+8\times 23=913$.