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If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho

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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 02 Mar 2015, 02:09
Isn't it easier in the particular problem to just calculate the powers?

The ones we are given, 2^9 and 5^3 are pretty much known powers. So, the only time consuming calculation would be this one: 125*512 = 64000.

Then 1/64000 is quite easy to calculate, and gives 0,00001..... So, 4 zeros. ANS B.

Otherwise, I guess we would indeed need to create a 10, in order to ease the calculations. What I have found out in the GMAT Club is that in these questions, 2s and 5s are our friends!
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 10 Apr 2015, 08:10
We have 1/ 5^3 x 2^9

==> we have 1 / 125x512 ===> 1/ 1.25 x 10^2 x 5.12x10^2

If you divide 1 by 1.25 and then by 5.12 you will get something ilke 0.15 (0 digits between the decimal point and the first non zero digit - 1 in that case. But you still have 10^2 and 10^2, which means that you now have 4 zeros. Hence, B
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 17 May 2015, 13:13
Hi, here is an alternative approach for this type of questions:

So we have this expression after some manipulations --> 1/64000, there are 5 digits in the denominator so you must have 5 zero's in total, incl. one zero before decimal --> 0,0000(something that's not a zero) (B)

If you just try to divide 1 by 64000 manualy, you'll understand this approach.

Hope this helps.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 19 Jun 2015, 18:15
TomB 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 fist nonzero digit to the right of the decimal point?

A. Three
B. Four
C. Five
D. Six
E. Nine


We got
T=1/2^9+5^3
2^9=(2*2*2)^3
5^3
T=1/8^3*5^3=1/40^3=1/4*10^4=0.25*10^4=0.000025

So the solution will be B
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 20 Jun 2015, 03:03
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kakal0t29 wrote:
TomB 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 fist nonzero digit to the right of the decimal point?

A. Three
B. Four
C. Five
D. Six
E. Nine


We got
T=1/2^9+5^3
2^9=(2*2*2)^3
5^3
T=1/8^3*5^3=1/40^3=1/4*10^4=0.25*10^4=0.000025

So the solution will be B


Please mark your mistake in Highlighted part

\(\frac{1}{40^3} = \frac{1}{(4^3 * 10^3)} = (\frac{1}{64})*10^{-3} = 0.015625 *10^{-3} = 0.000015625\)

Your answer however is correct but that seems to just by chance. :wink:
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 21 Jun 2015, 00:41
GMATinsight wrote:
kakal0t29 wrote:
TomB 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 fist nonzero digit to the right of the decimal point?

A. Three
B. Four
C. Five
D. Six
E. Nine


We got
T=1/2^9+5^3
2^9=(2*2*2)^3
5^3
T=1/8^3*5^3=1/40^3=1/4*10^4=0.25*10^4=0.000025

So the solution will be B


Please mark your mistake in Highlighted part

\(\frac{1}{40^3} = \frac{1}{(4^3 * 10^3)} = (\frac{1}{64})*10^{-3} = 0.015625 *10^{-3} = 0.000015625\)

Your answer however is correct but that seems to just by chance. :wink:


Yeap!
I've got that mistake ( 1/4^3=1/64). Thank you for your correct. Reserve 1 kudo :-D :thumbup: :)
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 24 Sep 2015, 14:06
TomB 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 fist nonzero digit to the right of the decimal point?

A. Three
B. Four
C. Five
D. Six
E. Nine



2^6 = 64 and 2^3 * 5^3 = 1000


1/64 = 0.01xyz (0.015625 to be more precise but we care only till the first non zero integer to the right of decimal.)
1/64 * 1/1000 = 0.01 * 10^ -3 = 0.00001 ( that means we got four zeros before 1 , and that's the answer.)
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 24 Nov 2015, 02:56
\(\frac{1}{2^9*5^3}\) = \(\frac{1}{2^6*10^3}\)

\(\frac{1}{64}\) = \(0.01...\) ==> add 3 additional zeros to it from the \(10^3\) = 4 zeros
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 02 Mar 2016, 04:07
hi Bunuel,
with your first approach, would I need to multiply 625 x 25 quickly or is there a way to directly know number of digits in 5^6 or for the matter any number x^y.

thx
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 02 Aug 2016, 02:12
Hi,

I got the right answer, but maybe my method is incorrect, I would like to get feedback on that! Thanks a ton.

So I know that 2^9 = 512 and 5^3 = 125

So I just changed the numbers to 500 and 110 which then turns out to be 55,000 as their product.

So 1/55,000 will give you four zeros. I thought this would save my time. Was this approach correct?

Posted from my mobile device
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 02 Aug 2016, 05:48
good solution bunuel...

one small variation to the second approach.

After simplification it is, (1/1000*64) = (1/1000)*(1/64) = (1/1000)*(1/100)*(100/64) = (1/10^5)*(100/64)

100/64 will give the first integer value.

1/10^5 will convert to four decimal point.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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TomB wrote:
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 fist nonzero digit to the right of the decimal point?

A. Three
B. Four
C. Five
D. Six
E. Nine


Here is my take:

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/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 02 Aug 2016, 23:43
Jeff's approach is rock solid.

I'm so thrilled I learned about it this week and was able to answer the question in less than 2 mins. It's hard to believe it this is a 700-level question.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 01 Apr 2017, 05:49
t= 1/(2^9*5^3)
t= 1/(10^3)(2^6)
t= 1/10^3*(0.5)^6
t=(1/10^3)625*25=0.001*0.015625
=0.000015625

Four zeroes
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 16 Aug 2017, 21:56
1/2⁹X5³
1/2⁶X2³X5³
Now1/2=(0.5)
(0.5)⁶X10⁻3
remove decimal for calculation
0.5X0.5=0.25
0.5X0.5=0.25
0.5X0.5=0.2525
25x25x25=0.015625
0.015625X10⁻³
Ans 4 zeros
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 17 Aug 2017, 04:25
Between decimal point and 1st non-zero digit to the right of decimal point:
1 divided by a number greater than 10 will yield 1 Zero
1 divided by a number greater than 100 will yield 2 Zeros
1 divided by a number greater than 1000 will yield 3 Zeros

Hence, answer is 3 Zeros + 1 Zero = 4 Zeros
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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New post 17 Aug 2017, 06:51
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multiply denominator and numerator by 5^6
denominator will become power of 10, and numerator will be 5^6

15625/10^9 = 0.000015625 hence 4 zeroes

B is answer.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho   [#permalink] 17 Aug 2017, 06:51

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