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Bunuel
If X = 625,000, how many nonzero digits are in the decimal representation of 1/X ?

(A) One
(B) Two
(C) Three
(D) Four
(E) Five
One easy method: multiply both numerator and denominator by \(\frac{a^{b}}{a^{b}}\) to create only powers of 10 in the denominator.
Division by 10 changes the decimal position, not the number of non-zero digits in the numerator.

Find the prime factors of 625,000
\(625,000 = (625 * 1,000) =\)
\((5^4*10^3)=(5^4*5^32^3)\)


\(\frac{1}{X}=\frac{1}{625,000}=\frac{1}{(5^4*5^32^3)}\)

There are seven powers of 5, but only three powers of 2. We want only powers of 10 in the denominator. We need four more 2s

\(\frac{1}{(5^45^32^3)}*(\frac{2^4}{2^4})=(\frac{2^4}{10^7})\)
\(2^4 = 16\)


There will be TWO non-zero digits in the decimal representation.

Answer B

(The number is .0000016)
Edit: Digvijay101 , simultaneous post. No wonder my browser got stuck. :-)
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