Difference between revisions of "2013 AMC 12A Problems/Problem 10"

Latest revision as of 18:38, 30 April 2021

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 $11+33+99=143$

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

Solution 4

As in previous solutions, we have $n|99$ and $\overline{ab} = 99/n$ . If we had $a=b$ , the decimal would be $0.\overline{a}$ , which is characterized by $n|9$ and $a = 9/n$ . So we seek the sum of the factors of 99 that are not also factors of 9.

Since $99 = 3^2 \cdot 11$ , the sum is $(1 + 3 + 9)(1 + 11) - (1 + 3 + 9) = 13(12 - 1) = \textbf{(D)} 143$ .

Video Solution

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

~sugar_rush

@@ Line 38: / Line 38: @@
 <math>\boxed{\textbf{(D)} 143}</math>
+==Solution 4==
+As in previous solutions, we have <math>n|99</math> and <math>\overline{ab} = 99/n</math>. If we had <math>a=b</math>, the decimal would be <math>0.\overline{a}</math>, which is characterized by <math>n|9</math> and <math>a = 9/n</math>. So we seek the sum of the factors of 99 that are not also factors of 9.
+Since <math>99 = 3^2 \cdot 11</math>, the sum is <math>(1 + 3 + 9)(1 + 11) - (1 + 3 + 9) = 13(12 - 1) = \textbf{(D)} 143</math>.
+==Video Solution==
+https://www.youtube.com/watch?v=XQpQaomC2tA
+~sugar_rush
 == See also ==
 {{AMC12 box|year=2013|ab=A|num-b=9|num-a=11}}
 {{MAA Notice}}

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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