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 $\boxed{36}$.
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)
5 answers
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.
$90 = 2 \cdot 3^2 \cdot 5$
$36 = 2^2 \cdot 3^2$
$72 = 2^3 \cdot 3^2$
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, 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.
1 answer