# 2014 AIME II Problems/Problem 4

## Problem

The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy $$0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},$$

where $a$, $b$, and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$.

## Solution 1

Notice repeating decimals can be written as the following: $0.\overline{ab}=\frac{10a+b}{99}$ $0.\overline{abc}=\frac{100a+10b+c}{999}$

where a,b,c are the digits. Now we plug this back into the original fraction: $\frac{10a+b}{99}+\frac{100a+10b+c}{999}=\frac{33}{37}$

Multiply both sides by $999*99.$ This helps simplify the right side as well because $999=111*9=37*3*9$: $9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99$

Dividing both sides by $9$ and simplifying gives: $2210a+221b+11c=99^2=9801$

At this point, seeing the $221$ factor common to both a and b is crucial to simplify. This is because taking $mod 221$ to both sides results in: $2210a+221b+11c \equiv 9801 \mod 221 \iff 11c \equiv 77 \mod 221$

Notice that we arrived to the result $9801 \equiv 77 \mod 221$ by simply dividing $9801$ by $221$ and seeing $9801=44*221+77.$ Okay, now it's pretty clear to divide both sides by $11$ in the modular equation but we have to worry about $221$ being multiple of $11.$ Well, $220$ is a multiple of $11$ so clearly, $221$ couldn't be. Also, $221=13*17.$ Now finally we simplify and get: $c \equiv 7 \mod 221$

But we know $c$ is between $0$ and $9$ because it is a digit, so $c$ must be $7.$ Now it is straightforward from here to find $a$ and $b$: $2210a+221b+11(7)=9801 \iff 221(10a+b)=9724 \iff 10a+b=44$

and since a and b are both between $0$ and $9$, we have $a=b=4$. Finally we have the $3$ digit integer $\boxed{447}$

## Solution 2

Note that $\frac{33}{37}=\frac{891}{999} = 0.\overline{891}$. Also note that the period of $0.abab\overline{ab}+0.abcabc\overline{abc}$ is at most $6$. Therefore, we only need to worry about the sum $0.ababab+ 0.abcabc$. Adding the two, we get $$\begin{array}{ccccccc}&a&b&a&b&a&b\\ +&a&b&c&a&b&c\\ \hline &8&9&1&8&9&1\end{array}$$ From this, we can see that $a=4$, $b=4$, and $c=7$, so our desired answer is $\boxed{447}$

## Solution 3

Noting as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999}$, let $u = 10a + b$. Then $$\frac{u}{99} + \frac{10u + c}{999} = \frac{33}{37}$$ $$\frac{u}{11} + \frac{10u + c}{111} = \frac{9\cdot 33}{37}$$ $$\frac{221u + 11c}{11\cdot 111} = \frac{9\cdot 33}{37}$$ $$221u + 11c = \frac{9\cdot 33\cdot 11\cdot 111}{37}$$ $$221u + 11c = 9\cdot 33^2.$$

Solving for $c$ gives $$c = 3\cdot 9\cdot 33 - \frac{221u}{11}$$ $$c = 891 - \frac{221u}{11}$$

Because $c$ must be integer, it follows that $u$ must be a multiple of $11$ (because $221$ clearly is not). Inspecting the equation, one finds that only $u = 44$ yields a digit $c, 7$. Thus $abc = 10u + c = \boxed{447}.$

## Solution 4

We note as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999},$ so $$\frac{10a + b}{99} + \frac{100a + 10b + c}{999} = \frac{33}{37} = \frac{891}{999}.$$

As $\frac{10a + b}{99}$ has a factor of $11$ in the denominator while the other two fractions don't, we need that $11$ to cancel, so $11$ divides $10a + b.$ It follows that $a = b,$ so $\frac{10a + b}{99} = \frac{11a}{99} = \frac{111a}{999},$ so $$\frac{111a}{999} + \frac{110a+c}{999} = \frac{891}{999}.$$

Then $111a + 110a + c = 891,$ or $221a + c = 891.$ Thus $a = b = 4$ and $c = 7,$ so the three-digit integer $abc$ is $\boxed{447}.$

## See also

 2014 AIME II (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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