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Difference between revisions of "2014 AIME II Problems/Problem 4"

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==Solution 2==
 
==Solution 2==
 
Note that <math>\frac{33}{37}=\frac{891}{999} = 0.\overline{891}</math>.  Also note that the period of <math>0.abab\overline{ab}+0.abcabc\overline{abc}</math> is at most <math>6</math>.  Therefore, we only need to worry about the sum <math>0.ababab+ 0.abcabc</math>.  Adding the two, we get
 
Note that <math>\frac{33}{37}=\frac{891}{999} = 0.\overline{891}</math>.  Also note that the period of <math>0.abab\overline{ab}+0.abcabc\overline{abc}</math> is at most <math>6</math>.  Therefore, we only need to worry about the sum <math>0.ababab+ 0.abcabc</math>.  Adding the two, we get
−
<cmath> \begin{tabular}{ccccccc}&A&B&A&B&A&B\\ +&A&B&C&A&B&C\\ \hline &8&9&1&8&9&1\end{tabular} </cmath>
+
<cmath> \begin{array}{ccccccc}&a&b&a&b&a&b\\ +&a&b&c&a&b&c\\ \hline &8&9&1&8&9&1\end{array} </cmath>
−
From this, we can see that <math>A=4</math>, <math>B=4</math>, and <math>C=7</math>, so our desired answer is <math>\boxed{447}</math>
+
From this, we can see that <math>a=4</math>, <math>b=4</math>, and <math>c=7</math>, so our desired answer is <math>\boxed{447}</math>
 +
 
 +
==Solution 3==
 +
Noting as above that <math>0.\overline{ab} = \frac{10a + b}{99}</math> and <math>0.\overline{abc} = \frac{100a + 10b + c}{999}</math>, let <math>u = 10a + b</math>.
 +
Then
 +
<cmath>\frac{u}{99} + \frac{10u + c}{999} = \frac{33}{37}</cmath>
 +
 
 +
<cmath>\frac{u}{11} + \frac{10u + c}{111} = \frac{9\cdot 33}{37}</cmath>
 +
 
 +
<cmath>\frac{221u + 11c}{11\cdot 111} = \frac{9\cdot 33}{37}</cmath>
 +
 
 +
<cmath>221u + 11c = \frac{9\cdot 33\cdot 11\cdot 111}{37}</cmath>
 +
 
 +
<cmath>221u + 11c = 9\cdot 33^2.</cmath>
 +
 
 +
Solving for <math>c</math> gives
 +
 
 +
<cmath>c = 3\cdot 9\cdot 33 - \frac{221u}{11}</cmath>
 +
 
 +
<cmath>c = 891 - \frac{221u}{11}</cmath>
 +
 
 +
Because <math>c</math> must be integer, it follows that <math>u</math> must be a multiple of <math>11</math> (because <math>221</math> clearly is not). Inspecting the equation, one finds that only <math>u = 44</math> yields a digit <math>c, 7</math>. Thus <math>abc = 10u + c = \boxed{447}.</math>
 +
 
 +
==Solution 4==
 +
We note as above that <math>0.\overline{ab} = \frac{10a + b}{99}</math> and <math>0.\overline{abc} = \frac{100a + 10b + c}{999},</math> so
 +
 
 +
<cmath>\frac{10a + b}{99} + \frac{100a + 10b + c}{999} = \frac{33}{37} = \frac{891}{999}.</cmath>
 +
 
 +
As <math>\frac{10a + b}{99}</math> has a factor of <math>11</math> in the denominator while the other two fractions don't, we need that <math>11</math> to cancel, so <math>11</math> divides <math>10a + b.</math> It follows that <math>a = b,</math> so <math>\frac{10a + b}{99} = \frac{11a}{99} = \frac{111a}{999},</math> so
 +
 
 +
<cmath>\frac{111a}{999} + \frac{110a+c}{999} = \frac{891}{999}.</cmath>
 +
 
 +
Then <math>111a + 110a + c = 891,</math> or <math>221a + c = 891.</math> Thus <math>a = b = 4</math> and <math>c = 7,</math> so the three-digit integer <math>abc</math> is <math>\boxed{447}.</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/7g5dztxGUrk
 +
 
 +
~savannahsolver
 +
 
 +
== See also ==
 +
{{AIME box|year=2014|n=II|num-b=3|num-a=5}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 07:05, 11 February 2023

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 $\boxed{447}.$

Video Solution

https://youtu.be/7g5dztxGUrk

~savannahsolver

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 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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