Difference between revisions of "2013 AMC 12A Problems/Problem 10"
(→See also) |
(→Solution) |
||
| Line 5: | Line 5: | ||
<math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad </math> | <math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad </math> | ||
| − | |||
==Solution 1== | ==Solution 1== | ||
Note that <math>\frac{1}{11} = 0.\overline{09}</math>. | Note that <math>\frac{1}{11} = 0.\overline{09}</math>. | ||
Revision as of 23:16, 9 March 2018
Contents
Problem
Let 
be the set of positive integers 
for which 
has the repeating decimal representation 
with 
and 
different digits. What is the sum of the elements of ?
Solution 1
Note that 
.
Dividing by 3 gives 
, and dividing by 9 gives 
.
The answer must be at least 
, but cannot be 
since no 
other than 
satisfies the conditions, so the answer is .
Solution 2
Let us begin by working with the condition . Let 
. So, 
. In order for this fraction 
to be in the form 
, 
must be a multiple of 
. Hence the possibilities of are 
. Checking each of these, 
and 
. So the only values of that have distinct and are 
and . So, 
See also
| 2013 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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. 
