Jane makes necklaces .she uses one pendant for each necklace.there are 6pendants and 10 ribbons in a pack.what is the least number of necklaces jane can make without having any ribbons or pendants left over
What is the least common multiple of 6 and 10?
To find the least number of necklaces Jane can make without having any ribbons or pendants left over, we need to find the common factor between the number of pendants and the number of ribbons.
In this case, the common factor is 2. To determine this, we can find the greatest common divisor (GCD) of the number of pendants and the number of ribbons, which is 2.
Dividing the number of pendants (6) and the number of ribbons (10) by their GCD (2), we get:
6 ÷ 2 = 3
10 ÷ 2 = 5
This means that Jane can make a maximum of 3 necklaces, each requiring 2 pendants and 2 ribbons, without having any ribbons or pendants left over.