The greatest common factor of 120 and 72 is 24 because the common factors of 120 and 72 are 1, 2, 3, 4, 6, and 24 and 24 is the largest of these. The rabbit could be feeding 120/24=<<120/24=5>>5 bunnies carrots
The rabbit could be feeding 72/24=<<72/24=3>>3 bunnies lettuce leaves.
This means that the rabbit must have a total of 5*3=<<5*3=15>>15 baby bunnies. Answer: \boxed{15}.
A rabbit is packing lunches for his children. He has 120 carrots and 72 leaves of lettuce. If he is giving each bunny the same number of carrots and the same number of leaves of lettuce, how many baby bunnies must he have. He will have no extra vegetables after he packs the lunches.
3 answers
15 is not an option
Apologies for the confusion. Let's reassess the problem. Since the rabbit is giving each bunny the same number of carrots and leaves of lettuce, we need to find the common factors of 120 and 72.
The prime factorization of 120 is $2^3 \cdot 3 \cdot 5$.
The prime factorization of 72 is $2^3 \cdot 3^2$.
The common factors of 120 and 72 are $2^3 \cdot 3 = 24$.
Therefore, the rabbit must have 24 baby bunnies. Answer: $\boxed{24}$.
The prime factorization of 120 is $2^3 \cdot 3 \cdot 5$.
The prime factorization of 72 is $2^3 \cdot 3^2$.
The common factors of 120 and 72 are $2^3 \cdot 3 = 24$.
Therefore, the rabbit must have 24 baby bunnies. Answer: $\boxed{24}$.