Find the smallest digit $d$ so that $437{,}d03$ is divisible by 9.
I know that a number is divisible by 9 if and only if the sum of its digits is divisible by 9. So I need to find $d$ such that $4 + 3 + 7 + d + 0 + 3$ is a multiple of 9. Adding up the given digits, I get $4 + 3 + 7 + 0 + 3 = 17$. So I need to add $d$ to 17 and get a multiple of 9. The smallest multiple of 9 that is greater than 17 is 18. So I need $d$ to be $18 - 17 = 1$. This is the smallest possible value of $d$ that makes the number divisible by 9. 