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# PS - Exponents - OG 12

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Senior Manager
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PS - Exponents - OG 12 [#permalink]  30 Mar 2009, 19:30
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Question Stats:

100% (01:01) correct 0% (00:00) wrong based on 5 sessions
If $$t = \frac{1}{(2^9 * 5^3)}$$ 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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Re: PS - Exponents - OG 12 [#permalink]  30 Mar 2009, 20:16
B. ?
t = 1/(2^6 * 10^3) => Multiply 5s and 2s to get three 10s.
t = (1/2^6 ) * 10^-3. Now 1/64 = 0.01xxxx
So t will have 3 + 1 = 4 zeroes before 1.

mrsmarthi wrote:
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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Re: PS - Exponents - OG 12 [#permalink]  30 Mar 2009, 23:11
1/10^3 * 1/2^6 = 0.001 * 0.01 = 0.00001

B.
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Re: PS - Exponents - OG 12 [#permalink]  31 Mar 2009, 15:43
OA is B.

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Re: PS - Exponents - OG 12 [#permalink]  01 Oct 2009, 16:29
I did'nt understand the last part//
Got it till here...1/2^6 x 10^-3
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Re: PS - Exponents - OG 12 [#permalink]  02 Oct 2009, 08:27
tejal777 wrote:
I did'nt understand the last part//
Got it till here...1/2^6 x 10^-3

1/2^6 = 1/64 = 0.01XXX ;
10^-3=1/1000=0.001
so the product of these two numbers is 0.00001 so the answer is B
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Re: PS - Exponents - OG 12 [#permalink]  03 Oct 2009, 23:41
Another way to look at this problem is (assuming you know what 2^9 and 5^3 is) (512 and 125)

Both 1/2^9 and 1/5^3 will yield - 0.001xx and 0.00x

Take away : As long as the denominator in 1/xx is more than 101 and less than 1000, you will always have 0.00xx as the resulting decimal

Cheers!

tejal777 wrote:
I did'nt understand the last part//
Got it till here...1/2^6 x 10^-3

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Re: PS - Exponents - OG 12 [#permalink]  09 Oct 2009, 18:07
Another way to look at it is once you reach the poinwhere t = 1/64,000 you can set up a long division diagram with 1 underneath and 64,000 on the outside.

Then, place the decimal on top, and proceed to fill in zeros underneath until you reach a number that 64,000 goes into. In this case, you will get to 100,000 before you place a 1 above the last zero of 100,000, since 64,000 goes into 100,000 one time. Then just fill in the remaining zeroes above the division line and count them.

The procedure is diagrammed in the attached .jpg. The red zeroes get filled in one by one

Attachments

File comment: diagram of long division method

OG prob.JPG [ 2.31 KiB | Viewed 1972 times ]

Re: PS - Exponents - OG 12   [#permalink] 09 Oct 2009, 18:07
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# PS - Exponents - OG 12

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