2001 AIME I Problems/Problem 1

Revision as of 21:43, 21 December 2007 by Serialk11r (talk | contribs) (Solution)

Problem

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

Solution

We cannot have a 0 in the number if we want a real number.

We split this up into cases:

Case 1: 11-19 Then the units digit must be a factor of the number, so 11, 12, and 15 work.

Case 2: 21-29

The number must be even, so 22 and 24 work.

Case 3: 31-39

The number must be a multiple of 3, so only 33 and 36 work.

Case 4: 41-49

It must be a multiple of 4, so 44 and 48 work.

Case 5: 51-59

It must be a multiple of 5, so only 55 works.

Case 6: 61-69

It must be a multiple of 6, so only 66 works.

Case 7: 71-79

It must be a multiple of 7, so 77 is the only one that works.

Case 8: 81-89

It must be a multiple of 8, so only 88 works.

Case 9: 91-99

It must be a multiple of 9, so only 99 works.

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

See also

2001 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AIME Problems and Solutions
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