2009 AMC 12A Problems/Problem 12

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 $54$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2009 AMC 12A Problems/Problem 12

Contents

Problem

Solution

Solution 1

Solution 2

See Also