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If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero [#permalink]

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23 Feb 2011, 08:52

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If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

What fluke did above is great. Let me just add here that if you are stuck with how to proceed, don't shy away from quick and easy calculations.

\(\frac{1}{2^9*5^3} = \frac{1}{2^6*1000}\)

Now, I can divide 1 by 64 to get the decimal point: .01

If I divide this further by 1000, the decimal moves 3 places to the left and I get four 0s before the 1.

Sometimes, under pressure in the exam, Math will fail you. Go with your instincts and use logic. (Except if your instincts tell you to multiply a four digit number with a five digit number - then you are definitely missing the point!)
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Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero [#permalink]

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15 Aug 2011, 15:24

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If \(t= 1/(2^9x5^3)\) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero [#permalink]

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28 Mar 2016, 05:07

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If \(t= 1/(2^9x5^3)\) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

(a) 3 (b) 4 (c) 5 (d) 6 (e) 9

Solution:

We use the term "leading zeros" to describe the zeros between the decimal point and the first nonzero decimal digit. To complete this problem we can use the following rule to determine the number of leading zeros in a fraction when it is converted to a decimal:

If X is an integer with k digits, then 1/X will have k – 1 leading zeros unless X is a perfect power of 10, in which case there will be k – 2 leading zeros.

We see that t is in the form 1/X. Because the denominator X has more twos than fives, we know X is not a perfect power of 10. Before considering the fraction as a whole, we first must determine the number of digits in the denominator.

Rewriting the denominator, we get 2^9 x 5^3 = (2^6 x 2^3) x 5^3 = 2^6 x (2^3 x 5^3) = 64 x (1,000) = 64,000, which is a 5-digit integer. Thus, k = 5.

Using our rule, we see that the fraction t has 5 - 1 = 4 leading zeros.

Answer B.
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Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero [#permalink]

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07 Jan 2017, 01:59

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero [#permalink]

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18 Mar 2017, 06:44

Baten80 wrote:

If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

a. three b. four c. five d. six c. nine

I like the explanation from Mike McGarry on Magoosh's site and I'm reposting the explanation here:

t=1/(2^9*5^3) t=1/(10^3*2^6) Now multiply the numerator and denominator by 5^6 so that we can convert the denominator completely in powers of 10 t=5^6/(10^3*2^6*5^6) t=5^6/(10^9) Now 5^3=125, 5^6 = (125)^2. If (100)^2 = 10,000, then (125)^2 will also have 5 digits So fill in the gap in this 9 digit number with the last 5 digits being 125^2 t=_ _ _ _ X X X X X. The first four blanks will obviously have zeros filled

If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?

a. three b. four c. five d. six c. nine

Given: \(t=\frac{1}{2^9*5^3}\).

Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.

Answer: B.

Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).

Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).

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