# 2001 AIME I Problems/Problem 1

## Problem

Find the sum of all positive two-digit integers that are divisible by each of their digits.

## Solution 1

Let our number be $10a + b$, $a,b \neq 0$. Then we have two conditions: $10a + b \equiv 10a \equiv 0 \pmod{b}$ and $10a + b \equiv b \pmod{a}$, or $a$ divides into $b$ and $b$ divides into $10a$. Thus $b = a, 2a,$ or $5a$ (note that if $b = 10a$, then $b$ would not be a digit).

• For $b = a$, we have $n = 11a$ for nine possibilities, giving us a sum of $11 \cdot \frac {9(10)}{2} = 495$.
• For $b = 2a$, we have $n = 12a$ for four possibilities (the higher ones give $b > 9$), giving us a sum of $12 \cdot \frac {4(5)}{2} = 120$.
• For $b = 5a$, we have $n = 15a$ for one possibility (again, higher ones give $b > 9$), giving us a sum of $15$.

If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = \boxed{630}$.

## Solution 2

Using casework, we can list out all of these numbers: $$11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.$$

## Solution 3

To further expand on solution 2, it would be tedious to test all $90$ two-digit numbers. We can reduce the amount to look at by focusing on the tens digit. First, we cannot have any number that is a multiple of $10$. We also note that any number with the same digits is a number that satisfies this problem. This gives $$11, 22, 33, ... 99.$$ We start from each of these numbers and constantly add the digit of the tens number of the respective number. For example, we look at numbers $11, 12, 13, ... 19$ and numbers $22, 24, 26, 28$. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of $5$ or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.