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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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Joined: 02 Sep 2009
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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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15 Nov 2019, 02:30
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95% (hard)

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18% (03:44) correct 82% (03:03) wrong based on 27 sessions

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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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Joined: 02 Aug 2009
Posts: 8322
If d = 1/(2^k*5^m), m > k (m and k are positive integers) is a termina  [#permalink]

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15 Nov 2019, 06:24
1
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) $$3 < m – k < 7$$

Are You Up For the Challenge: 700 Level Questions

$$d = \frac{1}{(2^k*5^m)}=\frac{1}{(2^k*5^k*5^{m-k})}=\frac{1}{10^k*5^{m-k}}$$
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   [#permalink] 15 Nov 2019, 06:24
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