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

(Created page with "Notice repeating decimals can be written as the following: <math>0.\overline{ab}=\frac{10a+b}{99}</math> <math>0.\overline{abc}=\frac{100a+10b+c}{999}</math> where a,b,c are t...")
 
(Solution 5)
 
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==Problem==
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The repeating decimals <math>0.abab\overline{ab}</math> and <math>0.abcabc\overline{abc}</math> satisfy
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<cmath>0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},</cmath>
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where <math>a</math>, <math>b</math>, and <math>c</math> are (not necessarily distinct) digits. Find the three digit number <math>abc</math>.
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==Solution 1==
 
Notice repeating decimals can be written as the following:
 
Notice repeating decimals can be written as the following:
  
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<math>\frac{10a+b}{99}+\frac{100a+10b+c}{999}=\frac{33}{37}</math>
 
<math>\frac{10a+b}{99}+\frac{100a+10b+c}{999}=\frac{33}{37}</math>
  
Multiply both sides by 999*99. This helps simplify the right side as well because 999=111*9=37*3*9:
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Multiply both sides by <math>999*99.</math> This helps simplify the right side as well because <math>999=111*9=37*3*9</math>:
  
 
<math>9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99</math>
 
<math>9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99</math>
  
Dividing both sides by 9 and simplifying gives:
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Dividing both sides by <math>9</math> and simplifying gives:
  
 
<math>2210a+221b+11c=99^2=9801</math>
 
<math>2210a+221b+11c=99^2=9801</math>
  
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:
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At this point, seeing the <math>221</math> factor common to both a and b is crucial to simplify. This is because taking <math>mod 221</math> to both sides results in:
  
 
<math>2210a+221b+11c  \equiv 9801 \mod 221 \iff 11c  \equiv 77 \mod 221</math>
 
<math>2210a+221b+11c  \equiv 9801 \mod 221 \iff 11c  \equiv 77 \mod 221</math>
  
Notice that we arrived to the result <math>9801 \equiv 77 \mod 221</math> 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:
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Notice that we arrived to the result <math>9801 \equiv 77 \mod 221</math> by simply dividing <math>9801</math> by <math>221</math> and seeing <math>9801=44*221+77.</math> Okay, now it's pretty clear to divide both sides by <math>11</math> in the modular equation but we have to worry about <math>221</math> being multiple of <math>11.</math> Well, <math>220</math> is a multiple of <math>11</math> so clearly, <math>221</math> couldn't be. Also, <math>221=13*17.</math> Now finally we simplify and get:
  
 
<math>c \equiv 7 \mod 221</math>
 
<math>c \equiv 7 \mod 221</math>
  
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:
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But we know <math>c</math> is between <math>0</math> and <math>9</math> because it is a digit, so <math>c</math> must be <math>7.</math> Now it is straightforward from here to find <math>a</math> and <math>b</math>:
  
 
<math>2210a+221b+11(7)=9801 \iff 221(10a+b)=9724 \iff 10a+b=44</math>
 
<math>2210a+221b+11(7)=9801 \iff 221(10a+b)=9724 \iff 10a+b=44</math>
  
and since a and b are both between 0 and 9, we have a=b=4. Finally we have the 3 digit integer <math>\boxed{447}</math>
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and since a and b are both between <math>0</math> and <math>9</math>, we have <math>a=b=4</math>. Finally we have the <math>3</math> digit integer <math>\boxed{447}</math>
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==Solution 2==
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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
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<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>
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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>
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==Solution 3==
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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>.
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Then
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<cmath>\frac{u}{99} + \frac{10u + c}{999} = \frac{33}{37}</cmath>
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<cmath>\frac{u}{11} + \frac{10u + c}{111} = \frac{9\cdot 33}{37}</cmath>
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<cmath>\frac{221u + 11c}{11\cdot 111} = \frac{9\cdot 33}{37}</cmath>
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<cmath>221u + 11c = \frac{9\cdot 33\cdot 11\cdot 111}{37}</cmath>
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<cmath>221u + 11c = 9\cdot 33^2.</cmath>
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Solving for <math>c</math> gives
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<cmath>c = 3\cdot 9\cdot 33 - \frac{221u}{11}</cmath>
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<cmath>c = 891 - \frac{221u}{11}</cmath>
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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>
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==Solution 4==
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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
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<cmath>\frac{10a + b}{99} + \frac{100a + 10b + c}{999} = \frac{33}{37} = \frac{891}{999}.</cmath>
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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
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<cmath>\frac{111a}{999} + \frac{110a+c}{999} = \frac{891}{999}.</cmath>
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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>
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== See also ==
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{{AIME box|year=2014|n=II|num-b=3|num-a=5}}
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 10:40, 16 December 2021

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 (ProblemsAnswer KeyResources)
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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