Mark:
GROUPS
How about...
cubes: (1,2,4,8,9,12),
(3,5,6,7,10,11)
squares: (1,4,8,10,12), (2,3,5,6,7,9,11) moved (2,9,10)
integers: (4,5,8,10,12), (1,2,3,6,7,9,11) moved (1,5)
Yeah, how about it! Let me consult my notes
Each group of cubes must equal 3042
Each group of squares must equal 325
Each group of integers must equal 39
So now the difficult bit. Finding the two groups of cubed integers that add up to 3042. For a start, we know that 1728 + 1331 > 3042, therefore, those two numbers are in separate groups. The 1000 can't go with 1728, because that leaves 314 to make from the remaining integers and it is not possible. And with a bit of trial and error, I managed to get these figures to work. One group is 1728, 729, 512, 64, 8, 1 which uses the integers 12, 9, 8, 4, 2, 1. The second group is 1331, 1000, 343, 216, 125, 27 which uses the integers 11, 10, 7, 6, 5, 3.
Next up is to move as few integers as possible so that the squares of the integers in each group are equal. First off, with the existing groups, take a look at what the sum of the squares are. The first group adds up to 310, and the second to 340. The number we are looking for them to equal is 325. You can't make 15 from any 2 numbers, so what we are most likely looking for is to move a minimum of 3 integers. 2 integers from one side and one from the other would be ideal. So we look at the squares, and for each square, we see if a number that is 15 less can be made up from 2 other integers. And it can. We can use integer 10 as 100, and 15 less is 85, which can be made from integers 9 and 2 which are 81 and 4 respectively. So we can move the 10 over to group 1 and move the 9 and the 2 over to group 2. So we now have:
Group 1: 1, 4, 8, 10, 12
Group 2: 2, 3, 5, 6, 7, 9, 11
Both of those groups add up to 325 using squares of the integers.
Next job is again move as few integers as possible so that the sums of the integers themselves in each group are equal, in this case 39. If we take a look at what each group currently adds up to. Group 1 is 35 and group 2 is 43. So we take a look at the two groups and see if there is a number in group 2 that is exactly 4 higher than a number in group 1. There is, the integer 5, which we can swap with integer 1 from group 1, which leaves us finally with this:
Group 1: 4, 5, 8, 10, 12
Group 2: 1, 2, 3, 6, 7, 9, 11
And so to answer the original question is not necessary, as Mark has done that.
Sister Act wrote,
"Prove the sum of (X^4) minus (Y^4) is not equal to 0 for all integers X,Y between 1 and 12 by inference?"
How you been keeping toots, slow day at the office?
Let x be one number.
Let y be the second #.
2x+3y=-14(if its -14)
x-3=y
Now, you substitute the value of y, x-3, into 2x+3y=-14
y=x-3
2x+ 3(x-3)=-14
2x+3x-9=-14
5x=-5
x=-1
y=-1-3
y=-4
Value of x is -1
Value of y, the second # is -4. Q.E.D. :wink:
Now, do you wanna strut yo stuff, how about a twirl round my pole?
I have a good feeling about today's offering:
GBRATRMAAE
(Three words)
AFAPLALRGT
