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

Question

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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chetan2u · Answer

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

B

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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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