Insects have 6 legs, and spiders have 8 legs. What is the least number of insects and the least number of spiders possible so that each group has the same number of legs?
6 x 1 = 6 8 x 1 = 8
6 x 2 = 12 8 x 2 = 16
6 x 3 = 18 8 x 3 = 24
6 x 4 = 24
Therefore you need 3 spiders and 4 insects.
The LCM, (Least Common Multiple) of 6 and 8 is 24
so you need 24/6 or 4 insects and 24/8 or 3 spiders.
To find the least number of insects and spiders that have the same number of legs, we need to find a common multiple of 6 and 8.
The first step is to find the least common multiple (LCM) of 6 and 8. To do this, we can find the prime factorization of both numbers.
Prime factorization of 6: 2 * 3
Prime factorization of 8: 2 * 2 * 2
Next, we take the highest power of each prime factor that appears in either prime factorization. In this case, we have 2^3 and 3^1.
Multiplying these factors together gives us the LCM: 2^3 * 3^1 = 24.
Therefore, the smallest number of legs that both insects and spiders could have in equal numbers is 24.
Now, since insects have 6 legs and spiders have 8 legs, we need to determine the number of each type of animal that will result in a total of 24 legs.
Dividing 24 legs by 6 legs per insect gives us 4 insects.
Dividing 24 legs by 8 legs per spider gives us 3 spiders.
So, in order to have the same number of legs, you would need a minimum of 4 insects and 3 spiders.