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# M31-09

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Math Expert
Joined: 02 Sep 2009
Posts: 47157

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07 Jun 2015, 09:20
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Difficulty:

45% (medium)

Question Stats:

67% (00:59) correct 33% (01:24) wrong based on 48 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 47157

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07 Jun 2015, 09:20
2
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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 $$\frac{2^3}{2^3}$$:
$$\frac{3^4}{2^3*5^6}*\frac{2^3}{2^3}=\frac{3^4*2^3}{2^6*5^6}=\frac{81*8}{10^6}=\frac{648}{10^6}=0.000648$$.

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Manager
Joined: 05 Jul 2015
Posts: 104
GMAT 1: 600 Q33 V40
GPA: 3.3

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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.
Manager
Joined: 23 Nov 2016
Posts: 76
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51

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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.
Intern
Joined: 02 Aug 2017
Posts: 6
GMAT 1: 710 Q46 V41
GMAT 2: 600 Q39 V33

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

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