Nay wants to divide 153 toothpicks into at least 4 but no more than 11 bags. Can he put them so there are none left over?

Lets look for the factors of 153 that are 4 or greater, and 11 or less. first step is to solve for the prime factors.

153

/ /

9 17

/ / /

3 3 17

So the prime factors of 153 are 3,3, and 17. From this, we now look for two or more numbers among the prime factors that have a product thats 4 or greater, and 11 or less.

3*3=9

4≤9≥11

So the toothpicks can be divided into 9 bags with none left over.

Yes because 9 can go into 153 so he can have 9 groups with, none leftover