Difference between revisions of "2001 AIME I Problems/Problem 1"
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Find the sum of all positive two-digit integers that are divisible by each of their digits. | Find the sum of all positive two-digit integers that are divisible by each of their digits. | ||
| − | == Solution == | + | == Solution 1 == |
| − | + | Let our number be <math>10a + b</math>, <math>a,b \neq 0</math>. Then we have two conditions: <math>10a + b \equiv 10a \equiv 0 \pmod{b}</math> and <math>10a + b \equiv b \pmod{a}</math>, or <math>a</math> divides into <math>b</math> and <math>b</math> divides into <math>10a</math>. Thus <math>b = a, 2a,</math> or <math>5a</math> (note that if <math>b = 10a</math>, then <math>b</math> would not be a digit). | |
| − | + | *For <math>b = a</math>, we have <math>n = 11a</math> for nine possibilities, giving us a sum of <math>11 \cdot \frac {9(10)}{2} = 495</math>. | |
| + | *For <math>b = 2a</math>, we have <math>n = 12a</math> for four possibilities (the higher ones give <math>b > 9</math>), giving us a sum of <math>12 \cdot \frac {4(5)}{2} = 120</math>. | ||
| + | *For <math>b = 5a</math>, we have <math>n = 15a</math> for one possibility (again, higher ones give <math>b > 9</math>), giving us a sum of <math>15</math>. | ||
| + | If we ignore the case <math>b = 0</math> as we have been doing so far, then the sum is <math>495 + 120 + 15 = \boxed{630}</math>. | ||
| − | + | == Solution 2 == | |
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| − | + | Using casework, we can list out all of these numbers: <math>11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}</math>. | |
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| − | <math>11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}</math> | ||
== 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}} | ||
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| + | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 14:30, 30 May 2017
Contents
Problem
Find the sum of all positive two-digit integers that are divisible by each of their digits.
Solution 1
Let our number be 
, 
. Then we have two conditions: 
and 
, or 
divides into 
and divides into 
. Thus 
or 
(note that if 
, then would not be a digit).
- For
, we have 
for nine possibilities, giving us a sum of 
. - For
, we have 
for four possibilities (the higher ones give 
), giving us a sum of 
. - For
, we have 
for one possibility (again, higher ones give ), giving us a sum of 
.
, we have 
for nine possibilities, giving us a sum of 
.
, we have 
for four possibilities (the higher ones give 
), giving us a sum of 
.
, we have 
for one possibility (again, higher ones give 
.If we ignore the case 
as we have been doing so far, then the sum is 
.
Solution 2
Using casework, we can list out all of these numbers: 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
