The House of Representatives has 435 members. If a committee has a prime number of members, and that number is a factor of 435, then how many members can be on the committee?
The factors of 435 are 3, 5, 29
Are any of those prime numbers?
http://www.mathsisfun.com/prime_numbers.html
To find the number of members that can be on the committee, we need to determine the prime factors of 435.
Prime factors are the prime numbers that divide a given number exactly without leaving any remainder.
To find the prime factors of 435, we can start by dividing it by the smallest prime number, which is 2.
435 ÷ 2 = 217.5
Since 217.5 is not an integer, we move on to the next prime number, which is 3.
435 ÷ 3 = 145
145 is an integer, so 3 is a prime factor of 435.
Now, we divide 145 by 3 again:
145 ÷ 3 = 48.333
Again, 48.333 is not an integer. We continue dividing by the next prime number, which is 5.
145 ÷ 5 = 29
29 is an integer, so 5 is also a prime factor of 435.
At this point, we can no longer divide 29 by any prime numbers since it is a prime number itself.
Therefore, the prime factors of 435 are 3 and 5.
To find the number of members that can be on the committee, we can consider all the factors of 435. Since the problem states that the committee's size must be a factor of 435, we are only interested in the factors that we found.
The factors of 435 are: 1, 3, 5, 15, 29, 87, 145, and 435.
Out of these factors, the ones that are prime numbers are: 3, 5, and 29.
Therefore, the number of members that can be on the committee is either 3, 5, or 29.