AIME 2003 I · 第 14 题

Solution

To find the smallest value of $n$, we consider when the first three digits after the decimal point are $0.251\ldots$.

Otherwise, suppose the number is in the form of $\frac{m}{n} = 0.X251 \ldots$, where $X$ is a string of $k$ digits and $n$ is small as possible. Then $10^k \cdot \frac{m}{n} - X = \frac{10^k m - nX}{n} = 0.251 \ldots$. Since $10^k m - nX$ is an integer and $\frac{10^k m - nX}{n}$ is a fraction between $0$ and $1$, we can rewrite this as $\frac{10^k m - nX}{n} = \frac{p}{q}$, where $q \le n$. Then the fraction $\frac pq = 0.251 \ldots$ suffices.

Thus we have $\frac{m'}{n} = 0.251\ldots$, or

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

As $4m' > n$, we know that the minimum value of $4m' - n$ is $1$; hence we need $250 < 2n \Longrightarrow 125 < n$. Since $4m' - n = 1$, we need $n + 1$ to be divisible by $4$, and this first occurs when $n = \boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\frac {32}{127}\approx 0.25196 \ldots$).

Solution 2

Rewrite the problem as having the smallest $n$ such that we can find an positive integer $m$ such that $0<\frac{m}{n}-\frac{251}{1000}<\frac{1}{1000}$.

We can rewrite the expression as $\frac{1000m-251n}{1000n}$, and we need $251n+x$ (where $x$ is the difference in the fraction, and ranging from $(1,2,...n-1)$ to be $0$ mod $1000$. We see that $n$ must be $3$ mod $4$ to have this happen (as this reduces the distance between the expression and $1000$.

Rewriting $n$ as $4k+3$, we get that $251(4k+3)+(4k+2)$ turns into $8k+755$, and this has to be greater than or equal to $1000$. The least $k$ that satisfies this is $31$, and we consequently get that the least value of $n$ is $127$. -dragoon -minor edits by Mathkiddie