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M31-09

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If \(\frac{3^4}{2^3*5^6}\) is expressed as a terminating decimal, how many nonzero digits will the decimal have?

A. One
B. Two
C. Three
D. Four
E. Six
[Reveal] Spoiler: OA

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Official Solution:

If \(\frac{3^4}{2^3*5^6}\) is expressed as a terminating decimal, how many nonzero digits will the decimal have?

A. One
B. Two
C. Three
D. Four
E. Six


Multiply \(\frac{3^4}{2^3*5^6}\) by \(\frac{2^3}{2^3}\):
\(\frac{3^4}{2^3*5^6}*\frac{2^3}{2^3}=\frac{3^4*2^3}{2^6*5^6}=\frac{81*8}{10^6}=\frac{648}{10^6}=0.000648\).

Answer: C
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Re: M31-09 [#permalink]

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New post 11 Jan 2016, 13:39
In order to have a terminating decimal, you need a power of ten on the bottom.

I re-wrote original equation as:

3^4/(10^3)*5^3

(3^4)*(10^-3) / 5^3

.081/125

And then you don't even need to finish the long division to see that there will be three zeros.

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Re: M31-09 [#permalink]

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New post 25 Feb 2017, 12:37
DJ1986 wrote:
In order to have a terminating decimal, you need a power of ten on the bottom.

I re-wrote original equation as:

3^4/(10^3)*5^3

(3^4)*(10^-3) / 5^3

.081/125

And then you don't even need to finish the long division to see that there will be three zeros.


You don't need a power of ten on the bottom. It has to be a power of 2, 5, or both.

E.g
1/2 = .5, which terminates
1/5 = .2, which terminates
1/20 = .1, which terminates

Anything other than a power of 2, 5, or a a combination of those two powers will not terminate. E.g. 1/3, 1/6, 1/7, 1/9, etc.

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Re: M31-09 [#permalink]

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New post 26 Oct 2017, 14:42
DJ1986 wrote:
In order to have a terminating decimal, you need a power of ten on the bottom.

I re-wrote original equation as:

3^4/(10^3)*5^3

(3^4)*(10^-3) / 5^3

.081/125

And then you don't even need to finish the long division to see that there will be three zeros.


Question asks for the number of non-zeros. Brunels explanation is the easiest way to sort this and make it "visual", unless I'm missing something in your explanation?

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Re: M31-09   [#permalink] 26 Oct 2017, 14:42
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