Free Practice Question

PS · Problem Solving

705-805

Divisibility/Multiples/Factors

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?

Answer Choices

Correct answer marked below

A

2

B

3

C

5

7

Correct

E

9

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Explanation

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Official Explanation

Official

Simplifying the numerator:

2^{-10} + 2^{-9} = 2^{-10} \cdot (1 + 2) = 3 \cdot 2^{-10}

Thus, we have:

\frac{3 \cdot 2^{-10}}{7^{-1} \cdot 5^9} = \frac{3 \cdot 7^1}{2^{10} \cdot 5^9} = \frac{21}{2^{10} \cdot 5^9}

…

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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?

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If n=2−10+2−9(7−1)(59)n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}n=(7−1)(59)2−10+2−9​ then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

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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?