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

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Question Stats: 55% (01:45) correct 45% (01:38) wrong based on 302 sessions

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-t-1-2-9-5 ... 47-20.html

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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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5
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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5
1
$$t= \frac{1}{2^9 * 5^3}$$

$$t= \frac{1}{2^6 * 2^3 * 5^3}$$

$$t= \frac{1}{2^6 * (2*5)^3}$$

$$t= \frac{1}{64 * (10)^3}$$

Multiplying numerator and denominator by $$10^2$$

$$t= \frac{10^2}{64 * (10)^5}$$

$$t= \frac{1.something}{(10)^5}$$

$$t= 1.something * 10^{-5}$$

To remove $$10^{-5}$$ we need to move decimal point 5 digits to the left

$$t= .00001something$$

4 zeros between decimal and first non-zero digit.

Ans: "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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I think there is no easier way than this.

fluke wrote:
$$t= \frac{1}{2^9 * 5^3}$$

$$t= \frac{1}{2^6 * 2^3 * 5^3}$$

$$t= \frac{1}{2^6 * (2*5)^3}$$

$$t= \frac{1}{64 * (10)^3}$$

Multiplying numerator and denominator by $$10^2$$

$$t= \frac{10^2}{64 * (10)^5}$$

$$t= \frac{1.something}{(10)^5}$$

$$t= 1.something * 10^{-5}$$

To remove $$10^{-5}$$ we need to move decimal point 5 digits to the left

$$t= .00001something$$

4 zeros between decimal and first non-zero digit.

Ans: "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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1
1/1000 * 1/2^6 = 5^6/5^6 * 1/2^6 * 1/1000 = 5^6/10^9 = 5 digits/10^9

=> Answer is B (4 zeroes)
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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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2
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
Director  Joined: 01 Feb 2011
Posts: 517
Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero  [#permalink]

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2
given expression can be re written as

5^6/(2^9*5^3*5^6) = 5^6/10^9

15625/10^9

=> 4 zero's between the decimal point and the first non zero digit to the right of the decimal point.

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Posts: 567
Schools: University of Chicago, Wharton School
Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero  [#permalink]

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This should be the approach. +1.

Spidy001 wrote:
given expression can be re written as

5^6/(2^9*5^3*5^6) = 5^6/10^9

15625/10^9

=> 4 zero's between the decimal point and the first non zero digit to the right of the decimal point.

If the question were only about the terminating decimal, I would solve as under:

t= 1/(2^9 * 5^3)
t= 1/(2^6 * 10^3)
t= (0.5)^6 * (0.1)^3

Since the sum of the powers of the two decimals is 9, the terminating decimal has 9 decimals.
Manager  Joined: 05 Oct 2011
Posts: 148
Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero  [#permalink]

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1
You could also use bench mark values as MGmat strategy guide talks about.
so here
1/100,000 < 1/64,000 <1/10,000
which is
0.00001 < 1/64000< 0.0001

So, t can be a like 0.000011, 0.000012 etc
so there are four zeroes.

Takeaway is to use Benchmark values.
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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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1
restore 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) 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.

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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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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
Math Expert V
Joined: 02 Sep 2009
Posts: 61537
Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero  [#permalink]

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1
1
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

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.

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 $$0.0000xxxx$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-t-1-2-9-5 ... 47-20.html
_________________ Re: If t= 1/(2^9x5^3) is expressed as a terminating decimal, how many zero   [#permalink] 18 Mar 2017, 05:56
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