- 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. - --- In mathforfun@yahoogroups.com, "mathsman3" <mathsman3@y...> wrote:
> This problem, I think, is a lot easier than it sounds. What is the

Not all mathematics is formula-driven. Thinking is likely to get you

> formula I need? Any clever thinking also welcomed.

>

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

If you don't know how to answer this, make several slips of paper 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

>

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

proceed to your other questions.

Regards, adh