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# If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima

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Joined: 02 Sep 2009
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If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima  [#permalink]

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12 Nov 2017, 00:06
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Difficulty:

65% (hard)

Question Stats:

58% (02:01) correct 42% (01:56) wrong based on 158 sessions

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If $$n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}$$ then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

(A) 2

(B) 3

(C) 5

(D) 7

(E) 9

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If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima  [#permalink]

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12 Nov 2017, 00:24
1
Bunuel wrote:
If $$n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}$$ then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

(A) 2

(B) 3

(C) 5

(D) 7

(E) 9

$$n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}$$
$$n =\frac{2^{-9} (1}{2+1)/(7^{-1})(5^9)}$$
$$n =\frac{(21}{2)/(2^{9})(5^9)}$$
$$n =\frac{(10.5)}{(10^{9})}$$
$$n =\frac{(0.105)}{(10^{7})}$$
7 Zeroes => 0.0000000105
D
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Re: If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima  [#permalink]

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17 Nov 2017, 03:19
2 ^-10 + 2 ^-9/7^-1 * 5^9
=> Multiply numerator and denominator by 2^9 => 2^-1 + 1/7^-1 * 5^9*2^9
= > 2^-1 + 1/7^-1 * 5 ^ 9 * 2 ^9
=> 1.5 * 7/2^9 * 5 ^ 9 = > 10.5 / 10 ^9
= >Divide numerator and denominator by 10^-7
=> 0.105/10^7 and hence answer is D. 7
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Re: If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima  [#permalink]

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20 Nov 2017, 11:15
1
Bunuel wrote:
If $$n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}$$ then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

(A) 2

(B) 3

(C) 5

(D) 7

(E) 9

Let’s simplify the given equation by first factoring out the common factor of 2^-10 from the two terms in the numerator:

[2^-10 + 2^-9]/(7^-1 x 5^9)

2^-10(1 + 2)/(7^-1 x 5^9)

We re-express the two factors containing negative exponents (i.e., 1/7^-1 = 7 and 2^-10 = 1/2^10):

7(1 + 2)/(2^10 x 5^9)

21(5)/(2^10 x 5^10)

105/10^10

105 x 10^-10

We see that 105 x 10^-10 is the number 105 with the decimal point to the right of the digit 5 moved 10 places to the left. After the decimal point is moved 10 places to the left, 105 x 10^-10 will have 7 zeros between the decimal point and the first nonzero digit 1.

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Joined: 30 Jul 2017
Posts: 3
Re: If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima  [#permalink]

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18 Dec 2017, 08:08
Hi,

I don't understand the given explanations. Bunuel, would you be so kind as to help me?

Kind regards,

Sandra
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Joined: 25 Feb 2013
Posts: 1217
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Re: If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima  [#permalink]

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18 Dec 2017, 08:44
Sassie wrote:
Hi,

I don't understand the given explanations. Bunuel, would you be so kind as to help me?

Kind regards,

Sandra

Hi Sassie

Can you explain in detail what exactly you did not understand in this question / solution ?

you simply need to simplify $$n$$ and count the number of 0s before a non digit number.

For eg. in 0.0000001 we have 6 zeros before 1

Now simplification of $$n$$ is what you need to practice through power / exponent section.

If you are still facing difficulty, kindly let me know
Re: If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima   [#permalink] 18 Dec 2017, 08:44
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