CF1985B Maximum Multiple Sum

Given an integer $ n $ , find an integer $ x $ such that: - $ 2 \leq x \leq n $ . - The sum of multiples of $ x $ that are less than or equal to $ n $ is maximized. Formally, $ x + 2x + 3x + \dots + kx $ where $ kx \leq n $ is maximized over all possible values of $ x $ .

The first line contains $ t $ ( $ 1 \leq t \leq 100 $ ) — the number of test cases. Each test case contains a single integer $ n $ ( $ 2 \leq n \leq 100 $ ).

For each test case, output an integer, the optimal value of $ x $ . It can be shown there is only one unique answer.

For $ n = 3 $ , the possible values of $ x $ are $ 2 $ and $ 3 $ . The sum of all multiples of $ 2 $ less than or equal to $ n $ is just $ 2 $ , and the sum of all multiples of $ 3 $ less than or equal to $ n $ is $ 3 $ . Therefore, $ 3 $ is the optimal value of $ x $ . For $ n = 15 $ , the optimal value of $ x $ is $ 2 $ . The sum of all multiples of $ 2 $ less than or equal to $ n $ is $ 2 + 4 + 6 + 8 + 10 + 12 + 14 = 56 $ , which can be proven to be the maximal over all other possible values of $ x $ .