The sum can be divisible by both $3$ and $4$, so the sum is the common multiple of $3$ and $4$.  If the sum is $12$, $12-5=7$, so the sum of the other numbers should be $7$. The other two numbers can be $1$ and $6$ or $3$ and $4$.  If the sum is $24$, $24-5=19$, the other two numbers can only be $10$ and $9$.  If the sum is $36$, it is impossible, because $36-5=31$ and the largest sum of two numbers from $1$ to $10$ is $19$. 
Ten balls numbered from $1$ to $10$ are put into a bag. $3$ balls are taken out and numbers on them are added up. The ball with number $5$ is included in the balls that are taken out, and the sum of numbers on the $3$ balls can be divisible by both $3$ and $4$. Which of the following balls is definitely not taken out? 
Options:
A) $$3$$ 
B) $$4$$ 
C) $$7$$ 
D) $$6$$ 
E) $$9$$ 