Testbook Test Series

Question

Answer 1

option C i.e 5.

let b be the radix.

hence

$(312)_{b} = (20)_{b} * (13.1)_{b}$

$3b^{2} + b + 2 = 2b * (b + 3 + b^{-1})$

$3b^{2} + b + 2 = 2b^{2} + 6b + 2$

$b^{2} – 5b = 0$

Now b >=4 since the max digit used is 3. Hence b ≠ 0.

Hence b = 5