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how many zeros

by uptowngirl92 » Thu Oct 01, 2009 4:28 pm
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 nonzero digit to the right of the decimal point?

a) 3
b) 4
c) 5
d) 6
e) 9
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by ssmiles08 » Thu Oct 01, 2009 5:08 pm
1 / (2^9 * 5^3) can also be re-written as 1 / (2^6 *2^3 * 5^3)

1 / [2^6*(2^3 * 5^3)] = 1/2^6 * 1/10^3

1/1000 = .001 = 2 zeros.

1/2^6 = 1/64 (quick division tells you that the quotient is .01xyz)

so it 1/64 = .01 = 1/100

1/100 * 1/1000 = 1/100000 = .00001 = 4 zeros

IMO B.
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by uptowngirl92 » Thu Oct 01, 2009 5:35 pm
I did'nt understand the last part//
Got it till here...1/2^6 x 10^-3
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by ssmiles08 » Fri Oct 02, 2009 5:30 am
uptowngirl92 wrote:I did'nt understand the last part//
Got it till here...1/2^6 x 10^-3
If you got it till 1/2^6 x 10^-3, then all you need to do is multiplication and division.

2^6 = 64

1/64 ~ .015 round it up to .02 if you want...the numerator's exact number does not matter.

.02 = 2/100
10^-3 = 1/1000

2/100 * 1/1000 = 2/100000 (when you multiply you add the zeros in the denominators.)

2/100000 = .00002 (there are 4 zeros between the decimal point and 2)
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