Difference between revisions of "2001 AIME I Problems/Problem 1"
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It must be a multiple of 9, so only 99 works. | It must be a multiple of 9, so only 99 works. | ||
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== See also == | == See also == | ||
{{AIME box|year=2001|n=I|before=First Question|num-a=2}} | {{AIME box|year=2001|n=I|before=First Question|num-a=2}} | ||
Revision as of 21:43, 21 December 2007
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.
See also
| 2001 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
