there`s a quid riding on this!

 

make a 3 by 3 grid a bit like a Xand O`s game.

inthem are to be Whole numbers that along the lines must add up to 55, the same down the colums and on the diagonal, a bit like those magic squares.

 

so far I can do 54.

 

they MUST be whole numbers, but negative numbers or Zero are allowed too and you may use the same numbers twice if you want.

 

HELP!!!!!

no takers???

 

well here`s what I have so far, I`ve used this a key and was trying to use offsets from this "Key" to make all be 55

 

Key:

 

8 1 6

3 5 7

4 9 2

 

 

all of these in any direction will add up to 15.

I used an offset of 13 to make 54 the sum like so:

 

21 14 19

16 18 20

17 22 15

 

now if each had .333.. reoccuring it would be easy as would filling each square with 18.333...

 

so I tried a different tactic, lets make one that starts with all making Zero:

 

3 -4 1

-2 0 2

-1 4 -3

 

all makes zero, so realy then we need to make each line add up to ONE and then use an offset of 54 (that we established in the 1`st equasion)

 

here`s where I got:

 

4 -5 2

-3 1 3

-2 5 -4

 

all works great untill the bottom line (

 

NOW can anyone take it from here. I`m truly stuck!

What is it u are looking for? A magic square with the numbers total 55 on the columns, rows and the two diagonals?

 

 

 

Edit: What is the maximum number of times you can use a no.? Two?

"What is it u are looking for? A magic square with the numbers total 55 on the columns, rows and the two diagonals?"

 

yup that`s exactly it

 

"Edit: What is the maximum number of times you can use a no.? Two?" there is no maximum, you can use the same number as many times as you like

 

the only restriction is that it must be a Whole number, so non of this 0.1579 stuff

Did un manage to find it? Is it solveable? cause i had a go at it, and i don' think it is

nope not yet, and I`m not sure if it is or not tbh?

cause, i was having a look up on the net, i found 15, and it can't be multiplied to make 55 without getting a horrible decimal

well, something that may help. This puzzle cannot be solved by using positive integers only. The closest u can get to the answer using positives is 54, then it is 57.

 

Hope that helps!

yup, that`s about as far as I got too, the use of negatives seemed encouraging at 1`st, but the best I can get Zeros, then 3 and 6 etc...

I think whatever the answer is must be either zero or a multiple of 3.

it`s darned annoying! )

where did u find this puzzle?

I didn`t, a mate told me about it, I think he heard it on the radio or something like that, not 100% sure?

Off of the top of my head:

25 5 25

5 25 5

25 5 25

 

TRY TO REARRANGE THEM to make 55 on the 2 rowsthat don in this one.

yeah, those diagonals will get ya every time!

If you put your mindto it, you could probably come up with a simple leniar, or maybe quadratic equation to solve this.

 

 

DeoxyriboNucleicAcid said in post #:
simple

I don't think so!

 

DeoxyriboNucleicAcid said in post #:
leniar, or maybe quadratic equation to solve this.

I think so

Sol'n: Not Possible.

if you can solve for 55 you can solve for 1 = [55-(3*18)]

So lets solve for 1

 

start with the middle number a and let 0 denote unknown

0 0 0

0 a 0

0 0 0

 

now for 4 of the lines we have 0 a 0 = 1. Thus all four are of the form x a (1-x-a) = 1...let us choose x,y,z,w to represent each of

the four lines. Therefore:

 

(x) (y) (z)

(1-w-a) (a) (w)

(1-z-a) (1-y-a) (1-x-a)

 

Thus the remaining four lines represent the edges of the box.

 

[1] x+y+z = 1, [2] 3(1-a)-(x-y-z) = 1

[3] ............ [4] ...........

 

now insert [1] into [2] :

[2] --> 3(1-a)=2 ---> 1-a = 2/3---> a= 1/3

but a must be whole...UNSOLVABLE.

it could be more than one number, but that about sums it up

That's what i tohught as well, i tried to to it with 5, and then, multiply everything by 11 (is this the right approach?), and it couldn't work. I just couldn't get 11! You can get 15, but, as i said before, it gives u a horrible decimal!

you keep on using this pattern

8 1 6

3 5 7

4 9 2

 

which makes it impossible to make 55 with whole numbers. if it is possible to make another magic square that adds up to 1 without the pattern

2ndhigh lowest 4thhighest

3rdlowest middle 3rdhighest

4thlowest highest 2ndhighest

 

hey that's pretty neat looking at it with words instead of numbers.

 

anyways, if you can make a magic square that adds to 1 without using the

8 1 6

3 5 7

4 9 2

pattern, you offset every number by 18 and that gives you 55. I'm not familiar with magic squares, but if there's only one specific type of square that generates a specific number, then it would be impossible.

 

to neurocomp:

"Thus all four are of the form x a (1-x-a) = 1...let us choose x,y,z,w to represent each of..." are you adding or multiplying?

psi20

"Thus all four are of the form x a (1-x-a) = 1...let us choose x,y,z,w to represent each of..." are you adding or multiplying?

 

Take a guess.