For smallest value n=127, the decimal representation is possible.
When the first three digits after the decimal point are 0.251,
we consider the smallest value of n....
Assume the number is otherwise in the form of
[tex]\frac{m}{n} = 0.X251 \ldots,[/tex]
where X is a k-digit string and n is as small as possible.
Then
[tex]10^k \cdot \frac{m}{n} - X = \frac{10^k m - nX}{n} = 0.251 \ldots[/tex]
. Since [tex]10^k m - nX[/tex] is an integer and [tex]\frac{10^k m - nX}{n}[/tex] is a fraction between 0 and 1, we can rewrite this as [tex]\frac{10^k m - nX}{n} = \frac{p}{q}[/tex], where [tex]q \le n[/tex]. Then the fraction [tex]$\frac pq = 0.251 \ldots$[/tex] suffices.
Thus we have[tex]\frac{m'}{n} = 0.251\ldots[/tex], or
[tex]\frac{251}{1000} \le \frac{m'}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m' < 252n \Longleftrightarrow n \le 250(4m'-n) < 2n.[/tex]
As [tex]$4m' > n$[/tex], we understand that the minimum value of [tex]$4m' - n$[/tex] is 1; hence we need [tex]$250 < 2n \Longrightarrow 125 < n$[/tex].
Since [tex]$4m' - n = 1$[/tex], we need [tex]$n + 1$[/tex] to be divisible by four, and this occurs for the first time when n = [tex]\boxed{ 127 }[/tex]
(note that if[tex]4m'-n > 1,[/tex] then [tex]$n > 250$[/tex]).
this gives [tex]$m' = 32$[/tex] and the fraction [tex]$\frac {32}{127}\approx 0.25196 \ldots$[/tex]).
Thus, the smallest value for n is 127.
To learn more about relatively prime numbers refer here
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