Difference between revisions of "2009 AMC 12A Problems/Problem 12"

Latest revision as of 20:06, 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 $\boxed{1}$ .

@@ Line 10: / Line 10: @@
 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>. Thus, the number is divisible by <math>18</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>.
+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 ===

2009 AMC 12A (Problems • Answer Key • Resources)
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