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Digits 251 in sequence (Posted on 2009-01-05) |
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When written as a decimal, the fraction m/n (with m < n, both positive integers) contains the consecutive digits 2, 5, 1 (in this order). Find the smallest possible n.
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Submitted by pcbouhid
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Solution:
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If n * 0. ...251... = integer, then n * 0.251... = integer (where we simply delete the digits before 251). So it is sufficient to consider 0.251...
If n * 0.251 = integer, then n must be a multiple of 1000.
So assume n ≤ 1000. We require that [n * 0.251] = m, [n * 0.252] = m+1, for some m.
Let {x} denote the fractional part of x. Then since n * 0.252 - n * 0.251 = n/1000, we must have {n * 0.251} ≥ 1 - n/1000.
If n is a multiple of 4, then {n * 0.251} = {n/1000}, so we need n > 500.
If n = 1 mod 4, then {n * 0.251} = {0.25 + n/1000}, so n > 375.
Similarly, if n = 2 mod 4, then n > 250, and if n = 3 mod 4, then n > 125.
Thus the smallest candidate is 127, and 32/127 = 0.251... |
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