The sum of $49$ different positive integers is $2016$. What is the minimum number of these integers which are odd?
If we start from $0$ odd numbers we can consider the equation to be $2(n_1+n_2+n_3+\cdots+n_{49})=2016$. The minimum values of this expression is $>2016$, hence no possible solution for $0$ odd numbers. Similarly, we can continue taking number of odd numbers $=1,2,3,4…$ and get the answer which is 6 but isn't there a shorter method?
Best Answer
Notice that even + odd = odd, hence we can have only an even number of odd numbers.
The minimum sum that can be made out of the 49 positive numbers, with $2k$ odd numbers and $(49-2k)$ even numbers is given by $$S = (49-2k)(50-2k) + 4k^2$$ We then need to solve the inequality $$S \le 2016$$ Solving this gives us $$8k^2 - 198k + 434 \le 0$$ The solution set of this inequality is approximately $[2.43,22.32]$. Since $k$ can only assume integer values, the smallest value of $k$ is 3, hence the minimum number of odd numbers is 6. $\blacksquare$