Difference between revisions of "2014 AIME II Problems/Problem 4"
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+ | ==Problem== | ||
+ | |||
+ | The repeating decimals <math>0.abab\overline{ab}</math> and <math>0.abcabc\overline{abc}</math> satisfy | ||
+ | |||
+ | <cmath>0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},</cmath> | ||
+ | |||
+ | where <math>a</math>, <math>b</math>, and <math>c</math> are (not necessarily distinct) digits. Find the three digit number <math>abc</math>. | ||
+ | |||
+ | ==Solution 1== | ||
Notice repeating decimals can be written as the following: | Notice repeating decimals can be written as the following: | ||
Line 9: | 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== | ||
+ | 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{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> | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | == 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 10:40, 16 December 2021
Problem
The repeating decimals and satisfy
where , , and are (not necessarily distinct) digits. Find the three digit number .
Solution 1
Notice repeating decimals can be written as the following:
where a,b,c are the digits. Now we plug this back into the original fraction:
Multiply both sides by This helps simplify the right side as well because :
Dividing both sides by and simplifying gives:
At this point, seeing the factor common to both a and b is crucial to simplify. This is because taking to both sides results in:
Notice that we arrived to the result by simply dividing by and seeing Okay, now it's pretty clear to divide both sides by in the modular equation but we have to worry about being multiple of Well, is a multiple of so clearly, couldn't be. Also, Now finally we simplify and get:
But we know is between and because it is a digit, so must be Now it is straightforward from here to find and :
and since a and b are both between and , we have . Finally we have the digit integer
Solution 2
Note that . Also note that the period of is at most . Therefore, we only need to worry about the sum . Adding the two, we get From this, we can see that , , and , so our desired answer is
Solution 3
Noting as above that and , let . Then
Solving for gives
Because must be integer, it follows that must be a multiple of (because clearly is not). Inspecting the equation, one finds that only yields a digit . Thus
Solution 4
We note as above that and so
As has a factor of in the denominator while the other two fractions don't, we need that to cancel, so divides It follows that so so
Then or Thus and so the three-digit integer is
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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