Difference between revisions of "2003 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
− | The [[decimal]] representation of <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] positive integers and <math> m < n, </math> contains the digits <math>2, 5</math>, and <math>1</math> consecutively | + | The [[decimal]] representation of <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] positive integers and <math> m < n, </math> contains the digits <math>2, 5</math>, and <math>1</math> consecutively and in that order. Find the smallest value of <math> n </math> for which this is possible. |
== Solution == | == Solution == | ||
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− | Otherwise, suppose the number is in the form of <math>\frac{m}{n} = 0.X251 \ldots</math>, | + | Otherwise, suppose the number is in the form of <math>\frac{m}{n} = 0.X251 \ldots</math>, where <math>X</math> is a string of <math>k</math> digits and <math>n</math> is small as possible. Then <math>10^k \cdot \frac{m}{n} - X = \frac{10^k m - nX}{n} = 0.251 \ldots</math>. Since <math>10^k m - nX</math> is an integer and <math>\frac{10^k m - nX}{n}</math> is a fraction between <math>0</math> and <math>1</math>, we can rewrite this as <math>\frac{10^k m - nX}{n} = \frac{p}{q}</math>, where <math>q \le n</math>. Then the fraction <math>\frac pq = 0.251 \ldots</math> suffices. |
− | Thus we have <math>\frac{m}{n} = 0.251\ldots</math>, or | + | Thus we have <math>\frac{m'}{n} = 0.251\ldots</math>, or |
− | <center><math>\frac{251}{1000} \le \frac{m}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m < 252n \Longleftrightarrow n \le 250(4m-n) < 2n.</math></center> | + | <center><math>\frac{251}{1000} \le \frac{m'}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m' < 252n \Longleftrightarrow n \le 250(4m'-n) < 2n.</math></center> |
− | As <math>4m > n</math>, we know that the minimum value of <math>4m - n</math> is <math>1</math>; hence we need <math>250 < 2n \Longrightarrow 125 < n</math>. Since <math>4m - n = 1</math>, we need <math>n + 1</math> to be divisible by <math>4</math>, and this first occurs when <math>n = \boxed{ 127 }</math> (note that if <math>4m-n > 1</math>, then <math>n > 250</math>). Indeed, this gives <math>m = 32</math> and the fraction <math>\frac {32}{127}\approx 0.25196 \ldots</math>). | + | As <math>4m' > n</math>, we know that the minimum value of <math>4m' - n</math> is <math>1</math>; hence we need <math>250 < 2n \Longrightarrow 125 < n</math>. Since <math>4m' - n = 1</math>, we need <math>n + 1</math> to be divisible by <math>4</math>, and this first occurs when <math>n = \boxed{ 127 }</math> (note that if <math>4m'-n > 1</math>, then <math>n > 250</math>). Indeed, this gives <math>m' = 32</math> and the fraction <math>\frac {32}{127}\approx 0.25196 \ldots</math>). |
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+ | ==Solution 2== | ||
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+ | Rewrite the problem as having the smallest <math>n</math> such that we can find an positive integer <math>m</math> such that <math>0<\frac{m}{n}-\frac{251}{1000}<\frac{1}{1000}</math>. | ||
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+ | We can rewrite the expression as <math>\frac{1000m-251n}{1000n}</math>, and we need <math>251n+x</math> (where <math>x</math> is the difference in the fraction, and ranging from (1,2,...n-1) to be <math>0</math> mod <math>1000</math>. We see that <math>n</math> must be <math>3</math> mod <math>4</math> to have this happen (as this reduces the distance between the expression and <math>1000</math>. | ||
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+ | Rewriting <math>n</math> as <math>4k+3</math>, we get that <math>251(4k+3)+(4k+2)</math> turns into <math>8k+755</math>, and this has to be greater than or equal to <math>1000</math>. The least <math>k</math> that satisfies this is <math>31</math>, and we consequently get that the least value of <math>n</math> is <math>127</math>. -dragoon -minor edits by Mathkiddie | ||
== See also == | == See also == |
Latest revision as of 13:23, 4 January 2024
Contents
Problem
The decimal representation of where and are relatively prime positive integers and contains the digits , and consecutively and in that order. Find the smallest value of for which this is possible.
Solution
To find the smallest value of , we consider when the first three digits after the decimal point are .
Otherwise, suppose the number is in the form of , where is a string of digits and is small as possible. Then . Since is an integer and is a fraction between and , we can rewrite this as , where . Then the fraction suffices.
Thus we have , or
As , we know that the minimum value of is ; hence we need . Since , we need to be divisible by , and this first occurs when (note that if , then ). Indeed, this gives and the fraction ).
Solution 2
Rewrite the problem as having the smallest such that we can find an positive integer such that .
We can rewrite the expression as , and we need (where is the difference in the fraction, and ranging from (1,2,...n-1) to be mod . We see that must be mod to have this happen (as this reduces the distance between the expression and .
Rewriting as , we get that turns into , and this has to be greater than or equal to . The least that satisfies this is , and we consequently get that the least value of is . -dragoon -minor edits by Mathkiddie
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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