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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==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}{ | + | 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>. | + | 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 14:23, 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
![$\frac{251}{1000} \le \frac{m'}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m' < 252n \Longleftrightarrow n \le 250(4m'-n) < 2n.$](http://latex.artofproblemsolving.com/9/2/5/9257b7a9db1cb1693beff21d7784e977268df528.png)
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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