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1 solution
No of digits is given by lo g N + 1
Here N = 3 2 2 0 1 6 .
Therefore,
no of digits = ( lo g 3 2 2 0 1 6 ) + 1 = ( 2 0 1 6 lo g 3 2 ) + 1 = ( 2 0 1 6 × 1 . 5 0 5 1 4 ) + 1 = 3 0 3 4 + 1 = 3 0 3 5