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

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

bunnel , can you please explain this problem
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Re: fractions [#permalink]  21 Mar 2012, 10:46
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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 [#permalink]  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 [#permalink]  23 Mar 2012, 04:36
nice approach Bunuel. It shortens the long process and makes it less error prone.
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Re: fractions [#permalink]  15 Apr 2012, 19:43
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: fractions [#permalink]  15 Apr 2012, 19:50
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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: fractions [#permalink]  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 [#permalink]  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 [#permalink]  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 [#permalink]  29 Apr 2012, 13:08
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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Terminating decimal OG [#permalink]  20 Oct 2012, 01:31
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 first nonzero digit to the right of the decimal
point?
(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine
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Re: Terminating decimal OG [#permalink]  20 Oct 2012, 02:25
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: Terminating decimal OG [#permalink]  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: Terminating decimal OG [#permalink]  20 Oct 2012, 04:19
closed271 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 first nonzero digit to the right of the decimal
point?
(A) Three
(B) Four
(C) Five
(D) Six
(E) Nine

Merging similar topics. Please refer to the solutions above.

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Re: fractions [#permalink]  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: fractions [#permalink]  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: fractions [#permalink]  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: fractions [#permalink]  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: fractions   [#permalink] 08 Jan 2013, 10:15
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