Distributive Challenge

Question

Would appreciate any hints as to how to approach this:

I multiplied a 3-digit number by 1002 and got AB007C, where A, B, and C stand for digits. What was my original 3-digit number?

Answer 1

We have that $2$ times a $3$-digit number ends in $...07C$, so if the $3$-digit number is $x$, then $2x=107C$, so $x$ could be $535, 536, 537, 538,$ or $539$. (The first digit of $2x$ can't be more than $1$, or the multiplicand would be over $3$ digits.)

Only one of those $5$ options works.