2013 AMC 12A Problems/Problem 10

Problem

Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits. What is the sum of the elements of $S$?

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad$

Solution 1

Note that $\frac{1}{11} = 0.\overline{09}$.

Dividing by 3 gives $\frac{1}{33} = 0.\overline{03}$, and dividing by 9 gives $\frac{1}{99} = 0.\overline{01}$.

$S = \{11, 33, 99\}$

$11 + 33 + 99 = 143$

The answer must be at least $143$, but cannot be $155$ since no $n \le 12$ other than $11$ satisfies the conditions, so the answer is $143$.

Solution 2

Let us begin by working with the condition $0.\overline{ab} = 0.ababab\cdots,$. Let $x = 0.ababab\cdots$. So, $100x-x = ab \Rightarrow x = \frac{ab}{99}$. In order for this fraction $x$ to be in the form $\frac{1}{n}$, $99$ must be a multiple of $ab$. Hence the possibilities of $ab$ are $1,3,9,11,33,99$. Checking each of these, $\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},$ and $\frac{99}{99} = 1$. So the only values of $n$ that have distinct $a$ and $b$ are $11,33,$ and $99$. So, $11+33+99= \boxed{\textbf{(D)} 143}$


Solution 3

Notice that we have $\frac{100}{n}= ab.\overline{ab}$

We can subtract $\frac{1}{n}=00.\overline{ab}$ to get \[\frac{99}{n}=ab\]

From this we determine $n$ must be a positive factor of $99$


The factors of $99$ are $1,3,9,11,33,$ and $99$.

For $n=1,3,$ and $9$ however, they yield $ab=99,33$ and $11$ which doesn't satisfy $a$ and $b$ being distinct.

For $n=11,33$ and $99$ we have $ab=09,03$ and $01$. (Notice that $a$ or $b$ can be zero)

The sum of these $n$ are $11+33+99=143$

$\boxed{\textbf{(D)} 143}$

Video Solution

https://www.youtube.com/watch?v=XQpQaomC2tA

~sugar_rush

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS