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

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If n ={2^{-10}+2^{-9}}/{(7^{-1}) (5^9)} then n is a terminat [#permalink] New post 12 Nov 2017, 00:05
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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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Re: If n ={2^{-10}+2^{-9}}/{(7^{-1}) (5^9)} then n is a terminat [#permalink] New post 12 Nov 2017, 23:32
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In questions about the number of zeroes, we have to look for how many times does 10 appear in our expression. Thus, if we rearrange we get \(\frac{2^{-9}(2^{-1}+1)}{7^{-1}5^9} = \frac{7(2^{-1}+1)}{2^95^9} = = \frac{10.5)}{10^9}\).

Now, we have to check how many zeroes are there after the decimal point but before the first non-zero digit. If we rewrite our expression as \(10.5*10^{-9} = 0.105*10^{-7}\) we get that from the situation in which the first digit after the decimal point is a non-zero digit, we have to move the point back 7 digits, so that there will be 7 zero digits before the 1.

Answer D
Re: If n ={2^{-10}+2^{-9}}/{(7^{-1}) (5^9)} then n is a terminat   [#permalink] 12 Nov 2017, 23:32
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If n ={2^{-10}+2^{-9}}/{(7^{-1}) (5^9)} then n is a terminat

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