Massey has 90 lollipops, 36 chocolates, and 72 gumballs to put in goody bags. What is the largest number of goody bags Massey can make so that each goody bag has the same number of gumballs, the same number of chocolates, and the same number of lollipops is used. (All lollipops, gumballs, and chocolates must be used)
First, let's find the greatest common divisor (GCD) of 90, 36, and 72.
The prime factorization of 90 is $2 \cdot 3^2 \cdot 5$, the prime factorization of 36 is $2^2 \cdot 3^2$, and the prime factorization of 72 is $2^3 \cdot 3^2$.
The GCD of these three numbers is $2^2 \cdot 3^2 = 36$.
Therefore, the largest number of goody bags Massey can make is $\boxed{36}$.
WRONG!
Let's find the prime factorizations of the three numbers:
$90 = 2 \cdot 3^2 \cdot 5$
$36 = 2^2 \cdot 3^2$
$72 = 2^3 \cdot 3^2$
The greatest common divisor (GCD) is the product of the smallest exponent for each prime factor.
Thus, the GCD is $2^2 \cdot 3^2 = 36$.
This means that the number of goody bags Massey can make is limited by the number of items in the smallest category, which is 36. Each goody bag will have 1 gumball, 1 lollipop, and 1 chocolate. Thus, Massey can make $\boxed{36}$ goody bags.
ALL CANDIES MUST BE USED!
Apologies for the previous incorrect response.
To find the largest number of goody bags, we need to find the greatest common divisor (GCD) of 90, 36, and 72, and then divide by that GCD.
The prime factorization of 90 is $2 \cdot 3^2 \cdot 5$.
The prime factorization of 36 is $2^2 \cdot 3^2$.
The prime factorization of 72 is $2^3 \cdot 3^2$.
The GCD in this case is $2^2 \cdot 3^2 = 36$.
To make sure all candies are used, we need to divide the total number of each candy by 36:
90 lollipops ÷ 36 = 2.5
36 chocolates ÷ 36 = 1
72 gumballs ÷ 36 = 2
Since we can't have a non-integer number of goody bags, we have to round down, meaning we can make a maximum of $\boxed{1}$ goody bag in this scenario.
To find the largest number of goody bags Massey can make, we need to find the greatest common divisor (GCD) of the numbers of lollipops, chocolates, and gumballs.
The GCD can be calculated by finding the factors of each number and identifying the highest common factor.
Let's find the factors for each number:
Factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
From the factors listed above, we can see that the GCD of 90, 36, and 72 is 6, since it is the largest common factor among all three numbers.
Therefore, Massey can make a maximum of 6 goody bags with an equal number of gumballs, chocolates, and lollipops.
To find the largest number of goody bags Massey can make, we need to determine the greatest common divisor (GCD) of the three given numbers: 90, 36, and 72.
Step 1: Calculate the GCD of 90 and 36.
- List the factors of each number:
- Factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
- Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
- We can see that the largest common factor of 90 and 36 is 18.
Step 2: Calculate the GCD of 18 and 72.
- List the factors of each number:
- Factors of 18: 1, 2, 3, 6, 9, 18
- Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
- We can see that the largest common factor of 18 and 72 is 18.
Therefore, the largest number of goody bags Massey can make is 18.