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