Difference between revisions of "2009 AMC 12A Problems/Problem 12"
(New page: == Problem == How many positive integers less than <math>1000</math> are <math>6</math> times the sum of their digits? <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 ...) |
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== Solution == | == Solution == | ||
− | The sum of the digits is at most <math>9+9+9=27</math>. Therefore the number is at most <math>6\cdot 27 = 162</math>. Out of the numbers <math>1</math> to <math>162</math> the one with the largest sum of digits is <math> | + | === Solution 1 === |
+ | |||
+ | The sum of the digits is at most <math>9+9+9=27</math>. Therefore the number is at most <math>6\cdot 27 = 162</math>. Out of the numbers <math>1</math> to <math>162</math> the one with the largest sum of digits is <math>99</math>, and the sum is <math>9+9=18</math>. Hence the sum of digits will be at most <math>18</math>. | ||
Also, each number with this property is divisible by <math>6</math>, therefore it is divisible by <math>3</math>, and thus also its sum of digits is divisible by <math>3</math>. | Also, each number with this property is divisible by <math>6</math>, therefore it is divisible by <math>3</math>, and thus also its sum of digits is divisible by <math>3</math>. | ||
− | We only have | + | We only have six possibilities left for the sum of the digits: <math>3</math>, <math>6</math>, <math>9</math>, <math>12</math>, <math>15</math>, and <math>18</math>. These lead to the integers <math>18</math>, <math>36</math>, <math>54</math>, <math>72</math>, <math>90</math>, and <math>108</math>. But for <math>18</math> the sum of digits is <math>1+8=9</math>, which is not <math>3</math>, therefore <math>18</math> is not a solution. Similarly we can throw away <math>36</math>, <math>72</math>, <math>90</math>, and <math>108</math>, and we are left with just <math>\boxed{1}</math> solution: the number <math>54</math>. |
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+ | === Solution 2 === | ||
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+ | We can write each integer between <math>1</math> and <math>999</math> inclusive as <math>\overline{abc}=100a+b+c</math> where <math>a,b,c\in\{0,1,\dots,9\}</math> and <math>a+b+c>0</math>. | ||
+ | The sum of digits of this number is <math>a+b+c</math>, hence we get the equation <math>100a+10b+c = 6(a+b+c)</math>. This simplifies to <math>94a + 4b - 5c = 0</math>. Clearly for <math>a>0</math> there are no solutions, hence <math>a=0</math> and we get the equation <math>4b=5c</math>. This obviously has only one valid solution <math>(b,c)=(5,4)</math>, hence the only solution is the number <math>54</math>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2009|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2009|ab=A|num-b=11|num-a=13}} |
Revision as of 14:04, 15 February 2009
Contents
[hide]Problem
How many positive integers less than are times the sum of their digits?
Solution
Solution 1
The sum of the digits is at most . Therefore the number is at most . Out of the numbers to the one with the largest sum of digits is , and the sum is . Hence the sum of digits will be at most .
Also, each number with this property is divisible by , therefore it is divisible by , and thus also its sum of digits is divisible by .
We only have six possibilities left for the sum of the digits: , , , , , and . These lead to the integers , , , , , and . But for the sum of digits is , which is not , therefore is not a solution. Similarly we can throw away , , , and , and we are left with just solution: the number .
Solution 2
We can write each integer between and inclusive as where and . The sum of digits of this number is , hence we get the equation . This simplifies to . Clearly for there are no solutions, hence and we get the equation . This obviously has only one valid solution , hence the only solution is the number .
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |