Eleven Positive Integers

Average of 11 distinct positive integers is 12, this means their sum is $12 \times 11 \Rightarrow 132$ .

Now we need one maximum possible number, which means rest of the ten integers should be minimum but distinct, we know 1 is the smallest positive integer, so the sum of minimum possible ten positive integers is $1 + 2 + 3 + 4 + \cdots +10 = 55$ .

Largest possible number $\Rightarrow 132 - 55 = \boxed{77}$

Make sure you include "distinct integers" or otherwise it is wrong. Anyway, the sum of the integers is 132. Based off the answers, the numbers are distinct. For maximization of the last one, we pick 1 to 10. Hence 132-55=77.