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

2001 AIME I (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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2001 AIME I Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

See also