XAT 2014 — QA & DI Question 12

10 = 5 × 2

10²⁹ = 5²⁹ × 2²⁹

∴ Number of divisors = (29 + 1)(29 + 1) = 30 × 30

We need to find all the divisors of K such that 10²⁹

= K × 10²³

K = 10⁶ = 5⁶ × 2⁶

∴ Number of divisors = (6 + 1)(6 + 1) = 7 × 7

The probability that a randomly chosen positive divisor of 10²⁹ is an integer multiple of 10^{23​​​​​​​}

$=\frac{7\times 7}{30\times 30}=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}$

∴ a = 7 and b = 30

∴ b – a = 30 – 7 = 23

Hence, option (d).