The arithmetic mean of five natural numbers i | vedclass

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(B) Let the five natural numbers be $n_1, n_2, n_3, n_4, n_5$ such that $n_1 \le n_2 \le n_3 \le n_4 \le n_5 = \alpha$. Given $n_5 - n_1 = 10$,so $n_1

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$10$

Vedclass · Answer

(B) Let the five natural numbers be $n_1, n_2, n_3, n_4, n_5$ such that $n_1 \le n_2 \le n_3 \le n_4 \le n_5 = \alpha$. Given $n_5 - n_1 = 10$,so $n_1 = \alpha - 10$. Since the numbers are natural numbers,$n_1 \ge 1$,so $\alpha \ge 11$. The sum of the five numbers is $5 	imes 40 = 200$. Thus,$(\alpha - 10) + n_2 + n_3 + n_4 + \alpha = 200$,which implies $n_2 + n_3 + n_4 = 210 - 2\alpha$. Since $n_1 \le n_2 \le n_3 \le n_4 \le n_5$,we have $3n_1 \le n_2 + n_3 + n_4 \le 3n_5$. Substituting the values: $3(\alpha - 10) \le 210 - 2\alpha \le 3\alpha$. From $3\alpha - 30 \le 210 - 2\alpha$,we get $5\alpha \le 240$,so $\alpha \le 48$. From $210 - 2\alpha \le 3\alpha$,we get $5\alpha \ge 210$,so $\alpha \ge 42$. The maximum value of $\alpha$ is $48$. The prime factorization of $48$ is $2^4 	imes 3^1$. The number of positive integral divisors is $(4+1)(1+1) = 5 	imes 2 = 10$.

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

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