Q68 / 200 · WBJEE 2014 · Mathematics

The number of digits in $20^{301}$ (given, $\log _{10} 2=0.3010$ is

  • A
    602
  • B
    301
  • Correct answer
    C
    392
  • D
    391

Solution

Let $y=20^{301}$ Number of digits $=$ Integral part of $\left(301 \log _{10} 20\right)+1$ Integral part of $\left.\left[301 \log _{10} 10+\log _{10} 2\right)\right]+1$ $=$ Integral part of $[301(1+0.3010)]+1$ $=$ Integral part of $[301 \times 1.3010]+1$ Integral part of $[391.601]+1$ $=391+1=392$

Hence, the correct answer is option: 'C'.

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