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2003 AIME I Problems/Problem 14: Difference between revisions

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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, and in that order. Find the smallest value of <math> n </math> for which this is possible.
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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==Solution 2==
==Solution 2==


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}{m}-\frac{251}{1000}<\frac{1}{1000}</math>.
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>.


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>.
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>.


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
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

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 $\frac{m}{n} = 0.X251 \ldots$, where $X$ 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

$\frac{251}{1000} \le \frac{m'}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m' < 252n \Longleftrightarrow n \le 250(4m'-n) < 2n.$

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 $\frac {32}{127}\approx 0.25196 \ldots$).

Solution 2

Rewrite the problem as having the smallest $n$ such that we can find an positive integer $m$ such that $0<\frac{m}{n}-\frac{251}{1000}<\frac{1}{1000}$.

We can rewrite the expression as $\frac{1000m-251n}{1000n}$, and we need $251n+x$ (where $x$ is the difference in the fraction, and ranging from (1,2,...n-1) to be $0$ mod $1000$. We see that $n$ must be $3$ mod $4$ to have this happen (as this reduces the distance between the expression and $1000$.

Rewriting $n$ as $4k+3$, we get that $251(4k+3)+(4k+2)$ turns into $8k+755$, and this has to be greater than or equal to $1000$. The least $k$ that satisfies this is $31$, and we consequently get that the least value of $n$ is $127$. -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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