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

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

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21 Mar 2012, 09:04
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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
[Reveal] Spoiler: OA
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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21 Mar 2012, 10:46
40
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Expert's post
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TomB wrote:
If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it 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

bunnel , can you please explain this problem

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 $$0.0000xxxx$$.

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

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21 Mar 2012, 10:56
Bunuel: Good approach!
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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23 Mar 2012, 04:36
nice approach Bunuel. It shortens the long process and makes it less error prone.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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15 Apr 2012, 19:43
1
This post was
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Bunuel wrote:
TomB wrote:
If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it 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

bunnel , can you please explain this problem

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 $$0.0000xxxx$$.

Can someone please explain how is multiplying 5^6 to the denominator (2^9 * 5^3) get 10^9?
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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15 Apr 2012, 19:50
4
KUDOS
catty2004 wrote:
Bunuel wrote:
TomB wrote:
If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it 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

bunnel , can you please explain this problem

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 $$0.0000xxxx$$.

Can someone please explain how is multiplying 5^6 to the denominator (2^9 * 5^3) get 10^9?

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

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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15 Apr 2012, 21:00
Bunuel wrote:
TomB wrote:
If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it 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

bunnel , can you please explain this problem

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 $$0.0000xxxx$$.

1st method is really awasome to follow...thanks Bunuel
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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16 Apr 2012, 06:14
Wow Bunuel, that was really good!
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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29 Apr 2012, 10:34
Please I would like to know why you multiplied by 5^6. and I did not understand your second approach.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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29 Apr 2012, 13:08
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Expert's post
olwan wrote:
Please I would like to know why you multiplied by 5^6. and I did not understand your second approach.

We are multiplying by 5^6/5^6 to convert denominator to the base of 10, so to simplify it to the decimal form: 0.xxxx.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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20 Oct 2012, 02:25
1
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t= 1 / (2^9 * 5^3)
or t=1/(2^3*5^3)*2^6

t=1/(10^3)*64

1/64 will be 0.01 and shifting the decimal point by three places to account for 1/10^3..

we get 4 zeros followed by 1..
Ans: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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20 Oct 2012, 02:58
With a question like this always try to convert the numbers to 10^(x) times something so that you can see the shift of the decimal point. As stated above, the answer is 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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07 Jan 2013, 21:58
Hi, I still dont understand why we have to multiply by 5^6/5^6 i understand that this equals one but what is the general rule for this? how did you know to pick 5^6?

Thanks

Bunuel wrote:
TomB wrote:
If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it 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

bunnel , can you please explain this problem

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 $$0.0000xxxx$$.

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

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08 Jan 2013, 03:16
shahir16 wrote:
Hi, I still dont understand why we have to multiply by 5^6/5^6 i understand that this equals one but what is the general rule for this? how did you know to pick 5^6?

Thanks

Bunuel wrote:
TomB wrote:
If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it 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

bunnel , can you please explain this problem

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 $$0.0000xxxx$$.

Welcome to GMAT Club shahir16.

We want the denominator of the fraction to be written as some power of 10. We need that in order to transform the fraction into decimal easily.

Now, the denominator = 2^9 * 5^3, hence we need to multiply it by 5^6 to get 10^9.

Hope it's clear.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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08 Jan 2013, 06:49
Can you please show me step-by-step how to convert 2^9 * 5^3 to a power of 10? I am unclear on the concept of converting an expression with exponents to a power of 10. I appreciate your help

Welcome to GMAT Club shahir16.

We want the denominator of the fraction to be written as some power of 10. We need that in order to transform the fraction into decimal easily.

Now, the denominator = 2^9 * 5^3, hence we need to multiply it by 5^6 to get 10^9.

Hope it's clear.[/quote]
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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08 Jan 2013, 10:15
shahir16 wrote:
Can you please show me step-by-step how to convert 2^9 * 5^3 to a power of 10? I am unclear on the concept of converting an expression with exponents to a power of 10. I appreciate your help

Here it goes: $$(2^9 * 5^3)*5^6=2^9 * (5^3*5^6)=2^9*5^9=(2*5)^9=10^9$$.
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Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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

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23 Jun 2014, 02:59
$$\frac{1}{2^9 5^3}$$

$$= \frac{5^6}{2^9 5^3 5^6}$$

$$= \frac{125^2}{10^9}$$

Count the numbers in numerator as equivalent to zero's in denominator

$$= \frac{15625}{100000 . 10^4}$$

$$10^4$$ remains in denominator

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

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11 Sep 2014, 11:35
1
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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

t = 1/(2^9*5^3)

2^9 = 8^3

8^3 *5^3 = 40^3

1/64000 = there will be at least 3 zeros

1/64=0.0something ,,, adding another 3 zeros then 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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11 Sep 2014, 18:37
Bunuel wrote:
TomB wrote:
If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it 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

bunnel , can you please explain this problem

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 $$0.0000xxxx$$.

Awesome approach Bunuel...
Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho   [#permalink] 11 Sep 2014, 18:37

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