### Instruction: If $a$ is the smallest positive two-digit multiple of $3$, and $b$ is the smallest positive three-digit multiple of $4$, then find $a+b$.### Response: I need to find the smallest positive two-digit multiple of $3$ and the smallest positive three-digit multiple of $4$.
I can use divisibility rules to help me find them.
A number is divisible by $3$ if the sum of its digits is divisible by $3$.
The smallest two-digit number is $10$, and the sum of its digits is $1+0=1$, which is not divisible by $3$.
The next two-digit number is $11$, and the sum of its digits is $1+1=2$, which is also not divisible by $3$.
The next one is $12$, and the sum of its digits is $1+2=3$, which is divisible by $3$.
So, $12$ is the smallest positive two-digit multiple of $3$.
A number is divisible by $4$ if the last two digits form a number that is divisible by $4$.
The smallest three-digit number is $100$, and the last two digits are $00$, which is divisible by $4$.
So, $100$ is the smallest positive three-digit multiple of $4$.
Now, I just need to add $a$ and $b$.
$a+b=12+100=112$.