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by GMATGuruNY » Fri Jul 05, 2013 9:33 am
If t= 1/(2^9 * 5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first non zero digit to the right of the decimal point?

A - THREE

B - FOUR

C - FIVE

D - SIX

E - NINE

Determine how many powers of 10 are in the denominator.

1/(2�*5³) = 1 / (2�)(2³*5³) = 1/(64*10³) = 1/64,000.

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There are four 0's to the right of the decimal.

The correct answer is B.
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by Brent@GMATPrepNow » Fri Jul 05, 2013 9:56 am
If t = 1/(2^9 x 5^3) 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
(e) Five
(0) Six
(E) Nine
Here's another approach.

First let's evaluate the denominator: (2^9)(5^3)
Notice that we can combine some 2's and 5's here.
(2^9)(5^3) = (2^6)(2^3)(5^3)
= (2^6)(10^3)
= (64)(1,000)
= 64,000
So, t = 1/64,000

Now observe that:
1/100 = 0.01 [1 zero between the decimal point and the first nonzero digit]
1/1,000 = 0.001 [2 zeros between the decimal point and the first nonzero digit]
1/10,000 = 0.0001 [3 zeros between the decimal point and the first nonzero digit]
1/100,000 = 0.00001 [4 zeros between the decimal point and the first nonzero digit]

Since 64,000 is between 10,000 and 100,000, we know that 1/64,000 will be
between 0.00001 and 0.0001

So, there will be 4 zeros between the decimal point and the first nonzero
digit.

Cheers,
Brent
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