Question

The probability that a randomly chosen positive divisor of $10^{29}$ is an integer multiple of $10^{23}$ is $a^2 / b^2$, then 'b - a' would be:

Accepted Answer

D. $10^{29} = 2^{29} 	imes 5^{29}$. Total divisors $= (29+1)(29+1) = 900$. Multiple of $10^{23} = 2^{23} 	imes 5^{23} \implies 10^{29} / 10^{23} = 10^6 = 2^6 	imes 5^6$. Divisors that are multiples $= (6+1)(6+1) = 49$. Probability $= 49/900 = (7/30)^2 \implies a=7, b=30$. $b - a = 30 - 7 = 23$.

Divisors Multiple Probability

Explanation

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