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

Maybe fx = lxl because of the way the x is positioned is that what he