Think Simple!

Algebra Level 4

Given $100$ reals $a_{1}, a_{2},\ldots , a_{100}$ such that $\displaystyle \sum ^{100}_{i=1}a^{16}_{i} = 2^{32}$ , we have maximum value of $\displaystyle \sum ^{100}_{i=1}a^{17}_{i}$ to be of the form $a^{k}$ , where $a$ is a prime and $k$ is a positive integer. Find $(a+k)$ .

The answer is 36.

1 solution

Venkata Karthik Bandaru
Aug 14, 2015

NOTE : The maximum value is unaffected of the number of reals. The solution below considers a more general case of taking $n$ reals, instead of $100$ as mentioned in the problem.

Moderator note:

Good observation that $a^{17} = a \times a^{16}$ , and thus that can give us a bound.

We say that $||l_{17} || \leq || l_1 || \times || l_16 ||$ .

Think Simple!

The answer is 36.

1 solution

Moderator note:

0 pending reports