Dr. Smiles wants to group the number of monsters that come in the morning and evening in a way that all groups are equal and no monsters are left behind. What is the possible number of monsters in each group?

The correct option is C $9$ and $10$
Monsters that come in the morning $=72$
Monsters that come in the evening $=70$

Check the monsters that come in the morning.

• For a number to be divisible by $5$, the number should end with $0$ or $5$.
• The last digit of $72$ is $2$. Thus, $72$ is not divisible by $5$.
• The group cannot have $5$ monsters in each group.

• For a number to be divisible by $9$, the sum of digits should be divisible by $9$.
• The sum of all the digits of $72$ is $7+2=9,$ which is divisible by $9$.
• The group can have $9$ monsters.

• For a number to be divisible by $10$, it should end with $0$.
• The last digit of $72$ is $2$. Any number whose last digit is zero is divisible by $10$.
• Thus, $72$ is not divisible by $10$.
• The group cannot have $10$ monsters.

Hence, the monsters that come in the morning can be divided into a group of $9$ monsters without leaving any remainder.

Check the monsters that come in the evening.

• The number formed by the last two digits of $70$ are $70$.
• $70$ is not divisible by $4$.
• The group cannot have $4$ monsters.

• The sum of all the digits of $70$ is $7+0=7$.
• $7$ is not divisible by $9$.
• The group cannot have $9$ monsters.

• The last digit of $70$ is $0$.
• Any number whose last digit is zero is divisible by $10$.
• Thus, $70$ is divisible by $10$.
• The group can have $10$ monsters.

Hence, the monsters that come in the evening can be divided into a group of $10$ monsters without leaving any remainder.

Hence, the monsters that come in the morning and evening can be divided into groups of $9$ and $10$, respectively.