(d) Let there be n number and he missed a number k, then theaverage (which he has calculated =
(
n(n+1)
2
−k)
n
=15 ⇒ n2 − 29n = 2k ⇒ n(n − 29) = 2k Thus at n = 29 or n < 29, the expression is invalid since the value of k is neither zero nor negative, which is actually a natural number. So for the least possible valued of n = 30 k = 15 and for n = 31, k = 31 Again for n > 31, k is beyond the range i.e., greater than n. Since k can not be greater than n. Hence there are only two values of k. So, there is no unique value of n.