Question 49

Let $$p_1, p_2, p_3$$ be prime numbers and $$\alpha, \beta, \gamma$$ be positive integers. If $$p_1^\alpha p_2^\beta p_3^\gamma$$ is a divisor of 34864764 lying between 100 and 200, then ($$p_1 + p_2 + p_3$$)($$\alpha + \beta + \gamma$$) =

Solution

$$34864764=2^2\times3\times11\times264127.$$

So,$$p_1,p_2,p_3\ could\ have\ values\ 2,3,11.$$ and $$\alpha\ ,\beta\ ,\gamma\ \ could\ have\ values\ of\ 2,1,1.$$

So,$$\left(p_1+p_2+p_3\right)\left(\alpha\ +\beta\ +\gamma\ \right)=\left(2+3+11\right)\left(2+1+1\right)=16\times4=64.$$

A is correct choice.

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