Question 64

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:

Solution

Number of factors of $$10^{29} = 2^{29} \times 5^{29}$$

= $$30 \times 30 = 900$$

Factors of $$10^{29}$$ which are multiple of $$10^{23}$$

= $$10^6 = 2^6 \times 5^6$$

= $$7 \times 7 = 49$$

=> Required probability = $$\frac{49}{900} = \frac{a^2}{b^2}$$

=> $$\frac{a}{b} = \frac{7}{30}$$

$$\therefore b - a = 30 - 7 = 23$$

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