You are requested to create 3 disjoint
sets such that:
1. Their union is a set of 10 digits (i.e. integers from 0 to 9 inclusive).
2. The average value of the members of set
A is
3.5.
3. The number of members in
B is less than the number of members in
C.
How many distinct solutions are there?
Rem: No empty sets.
Somehow I forgot about group C in this "solution".
Set A must have an even number of members so that the total of its elements is an integer. The number must be larger than 5 so that set B has a smaller cardinality. So set A has 6, 8 or 10 members, while set B has 4, 2 or 0, respectively. The sum of all the digits (the sum of the two sums) must be 45.
Determining the sum of the elements in set A as 3.5 times the number of elements, and of set B as 45 minus the sum of set A:
Set A Set B
elements sum elements sum
6 21 4 24
8 28 2 17
10 35 0 10
It's easier to describe the smaller set, B:
Zero elements can't add up to 10, so naturally there is no empty set.
Two elements can add up to 17 in only one way: 9+8.
Four elements adding to 24:
9+8+6+1
9+8+5+2
9+8+4+3
9+7+6+2
9+7+5+3
9+6+5+4
8+7+6+3
8+7+5+4
That's 9 possibilities in all.
Edited on August 15, 2014, 4:33 pm

Posted by Charlie
on 20140815 13:06:09 