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If d=1/(2^3*5^7) is expressed as a terminating decimal, how

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If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink] New post 20 Dec 2012, 06:11
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If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten
[Reveal] Spoiler: OA
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink] New post 20 Dec 2012, 06:12
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Walkabout wrote:
If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten


Given: d=\frac{1}{2^3*5^7}.

Multiply by \frac{2^4}{2^4} --> d=\frac{2^4}{(2^3*5^7)*2^4}=\frac{2^4}{2^7*5^7}=\frac{2^4}{10^7}=\frac{16}{10^7}=0.0000016. Hence d will have two non-zero digits, 16, when expressed as a decimal.

Answer: B.
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how   [#permalink] 20 Dec 2012, 06:12
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