Difference between revisions of "2001 AIME I Problems/Problem 1"

Revision as of 16:21, 11 June 2008

Problem

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

Solution

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 $495 + 120 + 15 = \boxed{630}$ .

Using casework, we can list out all of these numbers: $11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-We cannot have a 0 in the number if we want a real number.
+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).
-We split this up into cases:
+*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>.
-Case 1: 11-19
+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>.
-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.
-<math>11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}</math>
 == See also ==
 {{AIME box|year=2001|n=I|before=First Question|num-a=2}}
+[[Category:Introductory Number Theory Problems]]

2001 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2001 AIME I Problems/Problem 1"

Revision as of 16:21, 11 June 2008

Problem

Solution

See also