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

(Solution 2)
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=== Solution 2 ===
 
=== 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>.
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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>.
  

Revision as of 21:47, 24 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 $\overline{abc}=100a+10b+c$ 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 $54$.

See Also

2009 AMC 12A (ProblemsAnswer KeyResources)
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