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

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Math Expert
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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45% (medium)

Question Stats:

57% (01:27) correct 43% (01:07) wrong based on 69 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
[Reveal] Spoiler: OA

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Kudos [?]: 135380 [0], given: 12691

Director
Joined: 18 Aug 2016
Posts: 597

Kudos [?]: 179 [0], given: 136

GMAT 1: 630 Q47 V29
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
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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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Kudos [?]: 179 [0], given: 136

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Joined: 02 Apr 2017
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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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Target Test Prep Representative
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Kudos [?]: 1014 [1], given: 3

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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
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Expert's post
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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Kudos [?]: 1014 [1], given: 3

Re: If n =(2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating decima   [#permalink] 20 Nov 2017, 11:15
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