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Terminating Decimal

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by neelgandham » Wed Nov 23, 2011 1:45 am
1/(2^11 * 5^17)
= 1/(2^11 * 5^11 * 5^6)----Since (a^(m+n) = (a^m) * (a^n))
= 1/(10^11 * 5^6)----Since ((a^m) * (b^m) = (ab)^m)
= 2^6/(10^11 * 5^6 * 2^6)----Multiply the numerator and denominator by 2^6
= 64/(10^11 * 10^6)
= 64/(10^17)
= 0.0000..64 (17 0's after the decimal)
Non-zero numbers are 6 and 4. Hence 2, Option B

In simple terms, multiply the numerator and denominator by a number which can convert the denominator into a multiple of 10(in this case)
i.e. 1/(2^11 * 5^17) = 2^6/10^17 = 64/10^17 = 0.000....64, Option B
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by shankar.ashwin » Wed Nov 23, 2011 2:10 am
1/[(2^11) * (5*11)] = 1/10^11 - which is nothing but a bunch of zeros followed by 1.

So the question basically asks you what 1/5^6 is.

1/5 = 0.2, so (0.2)^6 = 0.000XX 64 - 2 numbers B
Last edited by shankar.ashwin on Wed Nov 23, 2011 8:12 am, edited 1 time in total.
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by GMATGuruNY » Wed Nov 23, 2011 2:45 am
MBA.Aspirant wrote:If 1/(2^11 * 5^17) is expressed as a terminating decimal, how many non-zero digits will the decimal have?

A) 1
B) 2
C) 4
D) 6
E) 11
1/(2¹¹*5¹�) = 2�/(2¹�*5¹�) = 64/10¹� = 64*10ˉ¹�.

Thus, there will be two non-zero digits: 6 and 4.

The correct answer is B.
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by ariz » Wed Nov 23, 2011 11:47 am
Great approaches, thank you all!
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