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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 ==
  
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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>159</math>, and the sum is <math>1+5+9=15</math>. Hence the sum of digits will be at most <math>15</math>.
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=== 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>.
  
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We only have five possibilities left for the sum of the digits: <math>3</math>, <math>6</math>, <math>9</math>, <math>12</math>, and <math>15</math>. These lead to the integers <math>18</math>, <math>36</math>, <math>54</math>, <math>72</math>, and <math>90</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>, and <math>90</math>, and we are left with just <math>\boxed{1}</math> solution: the number <math>54</math>.
+
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 ===
 +
 
 +
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

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 $3$.

We only have six possibilities left for the sum of the digits: $3$, $6$, $9$, $12$, $15$, and $18$. These lead to the integers $18$, $36$, $54$, $72$, $90$, and $108$. But for $18$ the sum of digits is $1+8=9$, which is not $3$, therefore $18$ is not a solution. Similarly we can throw away $36$, $72$, $90$, and $108$, and we are left with just $\boxed{1}$ solution: the number $54$.

Solution 2

We can write each integer between $1$ and $999$ inclusive as $\overline{abc}=100a+b+c$ 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 $54$.

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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