Terminating decimals

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Terminating decimals [#permalink]

27 Nov 2011, 19:34

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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

[Reveal] Spoiler: OA

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Re: Terminating decimals [#permalink]

28 Nov 2011, 04:04

ashiima 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

The question is based on very simple concepts but the application is a little tricky which actually makes it a good question.

First realize that 2^{11} * 5^{17} = 2^{11}*5^{11}*5^6 = 10^{11}*5^6
So \frac{1}{10^{11}*5^6} is just \frac{0.00...001}{5^6}

Now what do you get when you divide .01 by 5? You get .002
You write 0s till you get 10 and then you get a non-zero digit.

Now what do you get when you divide .01 by 125? You get .00008

Do you notice something? The non 0 term is 8 = 2^3

The reason is this: You will only get 1 followed by as many 0s as you want in the dividend. 125 = 5^3 so you will need 2^3 i.e. you will need 10^3 as the dividend and then 125 will be able to divide it completely (i.e. the decimal will terminate)

Now, using the same logic, what will be the non zero digits if you are dividing .00001 by 625?
625 = 5^4. You will need 2^4 = 16 to get 10^4 and that will end the terminating decimal. So you will have two non 0 digits: 16

What will you get when you divide by 5^6? Your non zero digits will be 2^6 = 64 i.e. you will have 2 non-zero digits.

Try doing some calculations to better understand the concept used.
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Re: Terminating decimals [#permalink]

11 Dec 2011, 08:30

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VeritasPrepKarishma wrote:

ashiima 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

Better way:

\frac{1}{2^{11}*5^{17}}=\frac{1}{(2^{11}*5^{11})*5^6}=\frac{1}{10^{11}*5^6}. Multiply both nominator and denominator by \frac{2^6}{2^6} so that to have only power of 10 in denominator: \frac{1}{10^{11}*5^6}*\frac{2^6}{2^6}=\frac{2^6}{10^{17}}=\frac{64}{10^{17}}, so the decimal will have two non-zero digits - 64.

Correct answer B.

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Re: Terminating decimals [#permalink]

11 Dec 2011, 23:18

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we have in denominator 2^11 * 5^17
as we know 2*5 =10 which is a number very easy to be handle
here we have 11 powers of 2 and 17 powers of 5
multiply both numerator and denominator by 2^6 so that we now have exp in denominator as
2^17 * 5^17 = 10^17

in numerator we have only 2^6 = 64

so we have only 2 non zero digits in this decimal

Re: Terminating decimals [#permalink] 11 Dec 2011, 23:18

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Terminating decimals

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