Difference between revisions of "2003 AIME I Problems/Problem 14"
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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. | 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>). |
== See also == | == See also == | ||
Revision as of 23:05, 29 June 2021
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 
).
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
