Xavier Aptitude Test 2014 Paper

(d) $10=5\times 2$
${10}^{29}=5^{29}\times 2^{29}$
$\therefore$ Number of divisors of ${\left(10\right)}^{29}=\left(29+1\right)\left(29+1\right)=30\times 30$
We need to find all the divisors of $\mathit{m}$ such that ${10}^{29}=$
$\mathrm{}\mathit{m}\times {10}^{23}$
$\mathrm{}\mathit{m}={10}^6=5^6\times 2^6$
$\mathrm{therefore}$ Number of divisors of $\mathrm{mathrm}\mathit{m}=\left(6+1\right)\left(6+1\right)=7\times 7$
The probability that a randomly chosen positivedivisor of ${10}^{29}$ is an integer multiple of ${10}^{23}$
$=\frac{7\times 7}{30\times 30}=\mathrm{}\frac{{\mathit{a}}^2}{{\mathit{b}}^2}$
$\therefore \mathrm{}\mathit{a}=7$ and $\mathit{b}=30$
$\therefore \mathrm{}\mathit{b}-\mathit{a}=30-7=23$.