Problem: Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number? Solution: I tried solving this using $a,b,c$ and $x,y$ to represent digits multiplied by powers of $10$. I later stumbled upon a 1000-IQ solution. Simply let $x$ be the two digit number and $y$ be the three digit number. We have that $1000x+y=9xy$. This looks simple. Now, divide by the insane choice of $xy$ to get $\frac{1000}{y}=9-\frac{1}{x}.$ This means that $\frac{1000}{y}$ is just a tiny bit smaller than $9$. This leads us to find $y = 112$ with $x = 14$ so we get a sum of $126$.