Suppose the 435 members of the house of representatives are placed on committees consisting of more than 2 members but fewer than 30 member of members and each member is to be on one committees. Each committee is to have an equal number of members and each member is to be on one committee. What size committees are possible?
factors of 435: 5 (87) so five folks one 27 committees. (or 87 folks [not allowed] one 5 committees).
factors of 435: 15 (29) would that work? 15 folks on 29 committees?
Any others?
Suppose the 435 members of the House of Representatives
are placed on committees consisting of more than
2 members but fewer than 30 members. Each committee
is to have an equal number of members and each member
is to be on only one committee.
a. What size committees are possible?
b. How many committees are there of each size?
M = Members.
C = Committees.
3M/C * 145C = 435 Members.
5M/C * 87C = 435.
15M/C * 29C = 435.
29M/C * 15C = 435.
a. 3,5,15,and 29M/C are possible.
b. 145,87,29, and 15C, respectively.
To find the possible sizes of committees, we need to consider the factors of 435 (the total number of members in the House of Representatives), excluding 1 and 435 itself.
Here's how you can find the factors of 435:
1. Start by finding the prime factorization of 435.
- Divide 435 by the smallest prime number, 2. 435 divided by 2 is 217.5, which is not a whole number. So try the next prime number, 3.
- Divide 435 by 3. The result is 145, which is a whole number.
- Continue dividing 145 by prime numbers until you cannot divide anymore. 145 divided by 5 is 29, which is also a whole number.
- The prime factorization of 435 is 3 * 5 * 29.
2. Now, you can find the factors of 435 by combining these prime factors. To do that, you can use the formula:
- If the prime factorization is in the form of a^m * b^n * c^o * ..., where a, b, c, ... are prime numbers, and m, n, o, ... are positive integers, then the number of factors is (m+1) * (n+1) * (o+1) * ...
So, for 435: (1+1) * (1+1) * (1+1) = 2 * 2 * 2 = 8.
3. Now, we have 8 factors of 435: 1, 3, 5, 15, 29, 87, 145, and 435.
To find the possible sizes of committees, we need to consider the factors that satisfy the given conditions: more than 2 members but fewer than 30 members.
The sizes of committees that satisfy the conditions are: 3, 5, 15, and 29.
Therefore, the possible sizes of committees are 3, 5, 15, and 29.