Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AIME II Problems/Problem 4"
(→‎Solution 1)
Line 18: Line 18:
 
<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:
+
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:
+
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:
+
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:
+
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:
+
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>
+
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>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 20:32, 20 May 2014

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 \[\begin{tabular}{ccccccc}&A&B&A&B&A&B\\ +&A&B&C&A&B&C\\ \hline &8&9&1&8&9&1\end{tabular}\] From this, we can see that $A=4$, $B=4$, and $C=7$, so our desired answer is $\boxed{447}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_4&oldid=62059"