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

 
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== Problem ==
 
== Problem ==
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Find the sum of all positive two-digit integers that are divisible by each of their digits.
  
== Solution ==
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== Solution 1 ==
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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). 
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*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>. 
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*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>.
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*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>.
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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>.
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== 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>.
  
 
== See also ==
 
== See also ==
* [[2001 AIME I Problems]]
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{{AIME box|year=2001|n=I|before=First Question|num-a=2}}
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Revision as of 14:30, 30 May 2017

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

See also

2001 AIME I (ProblemsAnswer KeyResources)
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

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