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Three subsets (Posted on 2014-08-15) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

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computer solution | Comment 1 of 13
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 2014-08-15 13:06:09

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