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

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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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New post 12 Nov 2017, 00:06
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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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New post 12 Nov 2017, 00:24
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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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New post 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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New post 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.

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

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New post 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 :-)
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Re: If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima &nbs [#permalink] 18 Dec 2017, 08:44
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