So, I've been having trouble with a problem from a Mock AMC 10.
The question is as follows:
$58$ candies are distributed among five boxes. Let $a_n$ be the number of candies in the $n$th box, starting from $1$. It is known that $3 \leq a_1 \leq 8$, $6 \leq a_2 \leq 8$, $9 \leq a_3 \leq 12$, $12 \leq a_4 \leq 20$, and $a_5=15$. Aniwey is allowed to open two of the five candy boxes. Assuming he chooses optimally, what is the maximum number of candies that he is guaranteed to obtain?
My solution:
We know that $3 \leq a_1, a_2, a_3 \leq 12$. So, $a_4$ and $a_5$ are greater than these three no matter what. And that means that the minimum is $12+15=27$, and the maximum is $20+15=35$. So, since the guaranteed is the minimum, we get $\boxed{27}$.
The answer is 30. BTW, this is a mock so yes, I am suspecting that they are wrong.
Please clarify, and thanks in advance.
Best Answer
Note that $a_1+a_2+a_3\le8+8+12=28$, while $a_1+\cdots+a_5=58$, so $a_4+a_5\ge30$.