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

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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 $\overline{abc}=100a+10b+c$ 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 $\boxed{1}$ .

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 === Solution 3 ===
-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>. Next 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>.
+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 ==

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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