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: $11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}$.

See also

2001 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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