The arithmetic mean of five natural numbers is 40 . The largest exceeds the smallest number by 10 . If…

The arithmetic mean of five natural numbers is 40 . The largest exceeds the smallest number by 10 . If $\alpha$ is the maximum possible value for the largest of these 5 numbers, then the number of positive integral divisors of $\alpha$ is
  1. 12
  2. 10
  3. 9
  4. 5

Solution

Let five natural numbers be $\alpha-10, x, y, z, \alpha$. $ \begin{aligned} \text { Given, } & \frac{\alpha-10+x+y+z+\alpha}{5}=40 \\ \Rightarrow & \alpha-10+x+y+z+\alpha=200 \\ & x+y+z=210-2 \alpha \end{aligned} $ Now, $\alpha-10 < \frac{x+y+z}{3} < \alpha$ $ \begin{aligned} & \Rightarrow \quad 3 \alpha-30 < x+y+z < 3 \alpha \\ & \Rightarrow \quad 3 \alpha-30 < 210-2 \alpha < 3 \alpha \\ & \Rightarrow \quad \alpha < 48 \text { or } \alpha>42 \end{aligned} $ $\therefore \quad$ Maximum value of $\alpha=48$ $ 48=2^4 \times 3^1 $ $\therefore \quad$ Positive divisors of $48=(4+1)(1+1)$ $ =5 \times 2=10 $

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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