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If d = 1/(2^k*5^m), m > k (m and k are positive integers) is a termina

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If d = 1/(2^k*5^m), m > k (m and k are positive integers) is a termina  [#permalink]

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New post 15 Nov 2019, 02:30
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If d = 1/(2^k*5^m), m > k (m and k are positive integers) is a termina  [#permalink]

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New post 15 Nov 2019, 06:24
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Bunuel wrote:
If \(d = \frac{1}{(2^k*5^m)}\), m > k (m and k are positive integers) is a terminating decimal, then there are how many non-zero digits in its decimal notation?


(1) \(m + k =10\)

(2) \(3 < m – k < 7\)



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\(d = \frac{1}{(2^k*5^m)}=\frac{1}{(2^k*5^k*5^{m-k})}=\frac{1}{10^k*5^{m-k}}\)
Now \(\frac{1}{10^k}\) will always give ONE non-zero terminating decimal, and that too does not effect overall non-zero digits as it is 1... 0.1 or 0.01 or 0.001 etc
So, our answer depends on \(\frac{1}{5^{m-k}}\)=\(\frac{2^{m-k}}{10^{m-k}}\), which depends on \(2^{m-k}\)..

(1) \(m + k =10\)
Insuff

(2) \(3 < m – k < 7\)
m-k can be 4, 5 or 6..
\(\frac{1}{5^{m-k}}\)=\(\frac{1}{5^{4}}=\frac{2^4}{10^4}=\frac{16}{10000}=0.0016\)..
\(\frac{1}{5^{m-k}}\)=\(\frac{1}{5^{5}}=\frac{2^5}{10^5}=\frac{32}{100000}=0.00032\)..
\(\frac{1}{5^{m-k}}\)=\(\frac{1}{5^{6}}=\frac{2^6}{10^6}=\frac{64}{100000}=0.000064\)
So, in each case, the answer is 2
Suff

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If d = 1/(2^k*5^m), m > k (m and k are positive integers) is a termina   [#permalink] 15 Nov 2019, 06:24
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