Sets and subsets

Expand Messages
• This problem, I think, is a lot easier than it sounds. What is the formula I need? Any clever thinking also welcomed. ... The set {1,2,3,4} can be partitioned
Message 1 of 2 , May 2, 2005
• 0 Attachment
This problem, I think, is a lot easier than it sounds. What is the
formula I need? Any clever thinking also welcomed.

--->>

The set {1,2,3,4} can be partitioned into two subsets {1,4} and {2,3}
of the same size. Notice that 1+4 = 2+3

(a) Find the next whole number n, above 4, for which the set
{1,2 ...n} can be partitioned into two subsets S and T of the same
size, with the sum of the numbers in S equal to the sum of numbers in
T

(b) Find all partitions in (a) with the additional property that the
sum of the square of the numbers in S equals the sum of the squares
of the numbers in T

Tim says he can partition the set {1,2 ... 16} into two subsets S and
T of the same size so that:

- the sum of the numbers in S equals the sum of the numbers in T
- the sum of the squares of the numbers in S equals the sum of the
squares of the numbers in T
AND
- the sum of the cubes of the numbers in S equals the sum of the
cubes of the numbers in T

Show that Tim is right.

Tim then claims he can partition the set {1,2 ... 8} into two subsets
S and T, not neccesarily of the same size, so that:

- the sum of the numbers in S equals the sum of the numbers in T
- the sum of the squares of the numbers in S equals the sum of the
squares of the numbers in T
AND
- the sum of the cubes of the numbers in S equals the sum of the
cubes of the numbers in T

Explain why you don't believe Tim this time.
• ... Not all mathematics is formula-driven. Thinking is likely to get you farther with these questions. However, it may be helpful to know that the sum of the
Message 2 of 2 , May 2, 2005
• 0 Attachment
--- In mathforfun@yahoogroups.com, "mathsman3" <mathsman3@y...> wrote:
> This problem, I think, is a lot easier than it sounds. What is the
> formula I need? Any clever thinking also welcomed.
>
Not all mathematics is formula-driven. Thinking is likely to get you
farther with these questions. However, it may be helpful to know that
the sum of the first n positive integers, the sum of their squares,
and the sum of their cubes are given by

1+2+...+n = n*(n+1)/2,

1+4+...+n^2 = n*(n+1)*(2n+1)/6,

1+8+...+n^3 = [n*(n+1)/2]^2 = [1+2+...+n]^2

(The last is a remarkable coincidence.)

> The set {1,2,3,4} can be partitioned into two subsets {1,4} and
> {2,3} of the same size. Notice that 1+4 = 2+3
>
> (a) Find the next whole number n, above 4, for which the set
> {1,2 ...n} can be partitioned into two subsets S and T of the
> same size, with the sum of the numbers in S equal to the sum
> of numbers in T
>
If you don't know how to answer this, make several slips of paper and
number them consecutively starting at 1. Then try to partition the
slips into sets having the same sum. You'll soon recognize patterns.
It seems you'll need a thorough understanding of this question to