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Q.

The number $N = 6 \log_{10} 2 + \log_{10} 31$ lies between two successive integers whose sum is equal to

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a

7

b

10

c

9

d

5

answer is B.

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Detailed Solution

$\begin{array}{l} N = \log_{10} 64 + \log_{10} 31 = \log_{10} 1984 . Therefore, \\ 3 < N < 4 \Rightarrow 7 \end{array}$ 1000<1984<10000

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The number N=6log10⁡2+log10⁡31 lies between two successive integers whose sum is equal to