2003 AIME I Problems/Problem 14

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Problem

The decimal representation of $m/n,$ where $m$ and $n$ are relatively prime positive integers and $m < n,$ contains the digits $2, 5$ , and $1$ consecutively, and in that order. Find the smallest value of $n$ for which this is possible.

Solution

To find the smallest value of $n$ , we consider when the first three digits after the decimal point are $0.251\ldots$ .

Otherwise, suppose the number is in the form of $\frac{m}{n} = 0.X251 \ldots$ , were $X$ is a string of $k$ digits and $n$ is small as possible. Then $10^k \cdot \frac{m}{n} - X = \frac{10^k m - nX}{n} = 0.251 \ldots$ . Since $10^k m - nX$ is an integer and $\frac{10^k m - nX}{n}$ is a fraction between $0$ and $1$ , we can rewrite this as $\frac{10^k m - nX}{n} = \frac{p}{q}$ , where $q \le n$ . Then the fraction $\frac pq = 0.251 \ldots$ suffices.

Thus we have $\frac{m}{n} = 0.251\ldots$ , or

$\frac{251}{1000} \le \frac{m}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m < 252n \Longleftrightarrow n \le...$

As $4m > n$ , we know that the minimum value of $4m - n$ is $1$ ; hence we need $250 < 2n \Longrightarrow 125 < n$ . Since $4m - n = 1$ , we need $n + 1$ to be divisible by $4$ , and this first occurs when $n = \boxed{ 127 }$ (note that if $4m-n > 1$ , then $n > 250$ ). Indeed, this gives $m = 32$ and the fraction $\frac {32}{127}\approx 0.25196 \ldots$ ).

2003 AIME I (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2003 AIME I Problems/Problem 14

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Problem

Solution

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