# Can anyone solve this?

• Apr 27th 2005, 01:44 AM
Chuck_3000
Can anyone solve this?
:confused: I need help with this maths question and no one can help me!!!

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 the numbers in T

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

Chuck :confused:
• Apr 27th 2005, 03:49 AM
ticbol
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 the numbers in T.

If S = T, then the sum S+T must be an even number, because dividing an odd number by 2 will give non-whole-number halves.
Also, there should be even number of numbers in the set S+T. So that S and T shall have same number of numbers, or S and T be same in size.

Next to 4, n might be 6. Test that:
{1,2,3,4,5,6}
1+2+3+4+5+6 = 21 ...an odd number, so cannot be.

Next n might be 8. Test:
{1,2,3,4,5,6,7,8}
1+2+3+4+5+6+7+8 = 36 ...an even number, so possible set.
For S = T here, then S = T = 36/2 = 18 each subset.

There are 8 numbers in it.
So S and T will have 4 numbers each.

Let us take the first 4 numbers----1,2,3,4.
1 would be in S
2 would be in T. So now T is 1 more than S. So we put 4 in S and the 3 in T---so that S=T for the first 4 numbers.
4 be in S
3 be in T
Now, S=T=5 in sums.

Then the other 4 numbers---5,6,7,8.
5 be in S
6 be in T. Here T is 1 more than S again. So we put the 8 in S and the 7 in T, to balance again.
8 be in S
7 be in T

That would give:
S = {1,4,5,8} = 1+4+5+8 = 18
T = {2,3,6,7} = 2+3+6+7 = 18
S and T have same size and same sum, so, OK.

Therefore, the next n is 8. ----answer.

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

Next to 4, n might be 6. Test that:
{1,2,3,4,5,6}
1^2 +2^2 +3^2 +4^2 +5^2 +6^2
1 +4 +9 +16 +25 +36 = 91 ...an odd number, so cannot be again.

Next n might be 8. Test:
{1,2,3,4,5,6,7,8}
1^2 +2^2 +3^2 +4^2 +5^2 +6^2 +7^2 +8^2
1 +4 +9 +16 +25 +36 +49 +64 = 204 ...an even number, possible.

If S=T here, then S = T = 204/2 = 102 each subset.

By trial and error, I arrived at:
S = {1,4,6,7} = 1 +16 +36 +49 = 102
T = {2,3,5,8} = 4 +9 +25 +64 = 102

Therefore, next n is 8 also. ----answer.
• Apr 27th 2005, 12:27 PM
Chuck_3000
Chuck_3000
thank you so muich!
• Apr 30th 2005, 11:35 PM
Chuck_3000
:confused:

hold on, wouldn't the groups be: {1,2,7,8} {3,4,5,6}

if not then can you please explain to me how the groups are formed!!! because the example was 1,4 & 2,3

or can the groups be anything?

thanks
• May 1st 2005, 12:14 AM
ticbol
You mean for part b), where "...the sum of the squares of the numbers in S equals the sum of the squares of the numbers T." ?

Let us test your {1,2,7,8} {3,4,5,6}

In set {1,2,7,8}, the sum of the squares of the numbers is
1^2 +2^2 +7^2 +8^2
= 1 +4 +49 +64
= 118

In set {3,4,5,6}, the sum of the squares of the numbers is
3^2 +4^2 +5^2 +6^2
= 9 +16 +25 +36
= 86

118 is not equal to 86, so the the two subsets you gave don't satisfy the condition. Hence, your two substs are not it.

Explain how my two subsets were formed?
It will be very long. The long and short of it is that I did it by trial and error. I tested different groupings, using common sense, until S = T per the condition as mentioned.
• May 2nd 2005, 01:14 AM
Chuck_3000
Quote:

Originally Posted by Chuck_3000
b)Find all partitions in a

is there only one?
• May 2nd 2005, 03:03 AM
ticbol
I can find no more.

Try this,
{1,2,3,4,5,6,7,8}
1^2 +2^2 +3^2 +4^2 +5^2 +6^2 +7^2 +8^2
1 +4 +9 +16 +25 +36 +49 +64 = 204 ...an even number, possible.

If S=T here, then S = T = 204/2 = 102 each subset.

Place the largest square, 64, into one subset. Say in S.
So,
S = {8,..,..,..} = 64
We need 3 numbers more to complete the 4-number subset.
The sum of all the squares in the subset shall be 102.
So, 102 -64 = 38.
What 3 numbers whose squares will total 38 can we get from 1,2,3,4,5,6,7 ?

Try that.

-----------
If the question "Find all partitions in a..." is for part "a" instead, where
"...the sum of the numbers in S equal to the sum of the numbers in T.",
then there is another pair of S and T.

Following my explanation in my original answer, you should find this other pair:
S = {1,4,6,7} = 1+4+6+7 = 18
T = {2,3,5,8} = 2+3+5+8 = 18

In my original answer, I found
S = {1,4,5,8} = 1+4+5+8 = 18
T = {2,3,6,7} = 2+3+6+7 = 18
• May 5th 2005, 01:32 AM
Chuck_3000
But wait, there's more...
Terry says she 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 numbers of the cubes of the numbers in T

c show terry is right

Willy says he can partition the set {1,2,...,8} into two subsets S and T, not necessarily of the same size so that
- The sum of the numbers in S and T are equal
- The sum of the squares are equal
- The sum of the cubes are equal

d explain why you do not belive Willy

thanks
• May 5th 2005, 02:34 AM
ticbol
No. I don't say I don't believe Terry and Willy, whoever they are.

I will not explain their findings.
I just don't meddle with others re their answers. I do not even comment on wrong answers.
I only explain my own answers.
• May 5th 2005, 12:38 PM
Chuck_3000
:confused:

this was part c and d of the original question...why did you answer the first question but not the second??? :confused:
• May 5th 2005, 09:58 PM
ticbol
The original question is posted at the head or start of this thread. Look at the start or your question at the start of this thread below. The lowest posting in this tread.

Do you find any part c and d?