Difference between revisions of "2003 AIME I Problems/Problem 14"
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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}{m}-\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>. | ||
== See also == | == See also == |
Revision as of 15:26, 19 December 2023
Contents
[hide]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 .
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.