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Difference between revisions of "2009 AMC 12A Problems/Problem 12"
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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>.
 
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>.
  
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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>.
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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>. Thus, the number is divisible by <math>18</math>.
  
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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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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>, but since the number is divisible by <math>18</math>, the digits can only add to <math>9</math> or <math>18</math>. This leads to the integers <math>18</math>, <math>36</math>, <math>54</math>, <math>72</math>, <math>90</math>, and <math>108</math> being possibilities. We can check to see that <math>\boxed{1}</math> solution: the number <math>54</math> is the only solution that satisfies the conditions in the problem.
  
 
=== Solution 2 ===
 
=== 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>.
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We can write each integer between <math>1</math> and <math>999</math> inclusive as <math>\overline{abc}=100a+10b+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>.
 
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>.
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=== Solution 3 ===
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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>. Since the number is <math>6</math> times the sum of its digits, it must be divisible by <math>6</math>, therefore also by <math>3</math>, therefore the sum of its digits must be divisible by <math>3</math>. With this in mind we can conclude that the number must be divisible by <math>18</math>, not just by <math>6</math>. Since the number is divisible by <math>18</math>, it is also divisible by <math>9</math>, therefore the sum of its digits is divisible by <math>9</math>, therefore the number is divisible by <math>54</math>, which leaves us with <math>54</math>, <math>108</math> and <math>162</math>. Only <math>54</math> is <math>6</math> times its digits, hence the answer is <math>\boxed{1}</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}}
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{{MAA Notice}}

Latest revision as of 19:06, 17 December 2017

Problem

How many positive integers less than $1000$ are $6$ times the sum of their digits?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 12$

Solution

Solution 1

The sum of the digits is at most $9+9+9=27$. Therefore the number is at most $6\cdot 27 = 162$. Out of the numbers $1$ to $162$ the one with the largest sum of digits is $99$, and the sum is $9+9=18$. Hence the sum of digits will be at most $18$.

Also, each number with this property is divisible by $6$, therefore it is divisible by $3$, and thus also its sum of digits is divisible by $3$. Thus, the number is divisible by $18$.

We only have six possibilities left for the sum of the digits: $3$, $6$, $9$, $12$, $15$, and $18$, but since the number is divisible by $18$, the digits can only add to $9$ or $18$. This leads to the integers $18$, $36$, $54$, $72$, $90$, and $108$ being possibilities. We can check to see that $\boxed{1}$ solution: the number $54$ is the only solution that satisfies the conditions in the problem.

Solution 2

We can write each integer between $1$ and $999$ inclusive as $\overline{abc}=100a+10b+c$ where $a,b,c\in\{0,1,\dots,9\}$ and $a+b+c>0$. The sum of digits of this number is $a+b+c$, hence we get the equation $100a+10b+c = 6(a+b+c)$. This simplifies to $94a + 4b - 5c = 0$. Clearly for $a>0$ there are no solutions, hence $a=0$ and we get the equation $4b=5c$. This obviously has only one valid solution $(b,c)=(5,4)$, hence the only solution is the number $54$.

Solution 3

The sum of the digits is at most $9+9+9=27$. Therefore the number is at most $6\cdot 27 = 162$. Since the number is $6$ times the sum of its digits, it must be divisible by $6$, therefore also by $3$, therefore the sum of its digits must be divisible by $3$. With this in mind we can conclude that the number must be divisible by $18$, not just by $6$. Since the number is divisible by $18$, it is also divisible by $9$, therefore the sum of its digits is divisible by $9$, therefore the number is divisible by $54$, which leaves us with $54$, $108$ and $162$. Only $54$ is $6$ times its digits, hence the answer is $\boxed{1}$.

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

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