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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 12 Sep 2004, 00:40
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-d-1-2-3-5-7-is-expressed-as-a-terminating-decimal-how-144440.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jun 2013, 21:47, edited 1 time in total.
Edited the question.
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 [#permalink] New post 12 Sep 2004, 10:13
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I can only think of the following method...may not be most efficient

Step 1: Looking at the denominator = no. of two's and 5's will get as many 10's (we have three 2's and three 5's(amonf seven 5's) , so I will rewrite it as

(1000)*5^4

=> d= 0.001*1/(5^4)

step 2: I will rewrite 1/(5^4) as = (2/10)*(2/10)*(2/10)*(2/10) => 16*(1/10000)

step 3: We now have d =0.0000016.
So, 2 non zero digits.
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 [#permalink] New post 12 Sep 2004, 14:49
d = 1 / (2^3 x 5^7)
= 1 / (1000 x 5^4)
= 10^-3 x (1/625)
= 10^-7 x (10000/625)
= 10^-7 x 16

So 2 digits
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Re: If d = 1/(2^3*5^7) is expressed as a terminating decimal, [#permalink] New post 17 Jun 2013, 21:09
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Another way to do it is :

We know x^a*y^a=(X*Y)^a

given = 1/(2^3*5^7)
= Multiply and divide by 2^4
=2^4/(2^3*2^4*5^7)
=2^4/10^7
=> non zero digits are 16 => Ans B
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink] New post 17 Jun 2013, 21:48
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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

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-d-1-2-3-5-7-is-expressed-as-a-terminating-decimal-how-144440.html
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how   [#permalink] 17 Jun 2013, 21:48
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