The art department had a party at the teacher's house. There are 50 brownies and 40 celery sticks served as refreshments. Each art student had the same number of brownies and the same number of whole celery sticks, & nothing was left over. The art teacher didn't eat. What is the largest number of art members that might have been at the party?
Here we look for the highest number of members N. Since each member eats the same number of celery (c) and brownies (b), then
Nb=50 and Nc=40
We need to find the highest common (HCF) denominator of 50 and 40 to find N.
There are two ways to find the HCF.
The traditional way taught in schools is
to factorize the numbers and extract common factors. These factors are then multiplied to give the HCF.
50=2*5*5
40=2*2*2*5
So we see that the common factor is 2*5=10, which is the HCF.
The other way is using Euclid's algorithm, which is suited for mental calculations.
Take the difference between the two numbers, and repeat using the smaller number and the difference until the difference divides the smaller number.
50-40=10. Since 10 divides 40, 10 is the HCF.
To determine the largest number of art members that might have been at the party, we need to find the common factors of the number of brownies and celery sticks. Since we know that each art student had the same number of brownies and celery sticks, the number of art members must be a divisor of both 50 and 40.
First, let's find the factors of 50: 1, 2, 5, 10, 25, and 50.
Next, let's find the factors of 40: 1, 2, 4, 5, 8, 10, 20, and 40.
Now, let's find the common factors of 50 and 40: 1, 2, 5, and 10.
Therefore, the largest number of art members that might have been at the party is 10. Each art student would have received 5 brownies and 4 celery sticks, resulting in a total of 50 brownies and 40 celery sticks.