2003 AIME I Problems/Problem 14
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
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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