All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
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:

That's 9 possibilities in all.

Edited on August 15, 2014, 4:33 pm
  Posted by Charlie on 2014-08-15 13:06:09

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information