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M31-09

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Bunuel
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M31-09 [#permalink] New post   07 Jun 2015, 09:20
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70% (01:32) correct 30% (01:47) wrong based on 71 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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C
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Bunuel
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Re M31-09 [#permalink] New post   07 Jun 2015, 09:20
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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


To transform the denominator into a base 10 format, which subsequently enables us to convert the fraction into a decimal of the form 0.xxxx, we should multiply the fraction \(\frac{3^4}{2^3*5^6}\) 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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DJ1986
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Re: M31-09 [#permalink] 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] 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] 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] New post   18 May 2023, 18:55
Sure so, here is a trick that I came up with by running a few cases (Experts pls point to any mistakes, if present):

If you have 2 or 5 power n, the number of non-zero decimals will be n.

If you have a combination of 2 and 5, say 2^a and 5^b: Then the no of decimal digits will be a or b, depending on which is greater. I would like to caveat that some digits here could be zero so if you want non-zero digits might just be easier to divide by separating powers of 10 out

Example:
81/2 = 40.5 (2^1)
81/5 = 16.2 (5^1)
81/50 = 1.62 (5^2, 2^1)
81/400 = 0.2025 (5^2, 2^4)
81/4000 = 0.02025 (I would have divided here so just captured this case)

Hope this helps for anyone who visits this post in the future:)
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Re: M31-09 [#permalink] New post   19 May 2023, 00:48
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M31-09 [#permalink]
19 May 2023, 00:48
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