I Need some help with a math puzzle!

[YT2095]

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!!!!!

[YT2095]

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!

[aommaster]

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?

[YT2095]

"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

[aommaster]

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

[YT2095]

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

[aommaster]

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

[aommaster]

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!

[YT2095]

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! )

[aommaster]

where did u find this puzzle?

[YT2095]

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

[DeoxyriboNucleicAcid]

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.

[YT2095]

yeah, those diagonals will get ya every time!

[DeoxyriboNucleicAcid]

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

[aommaster]

 

 

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

[Neurocomp2003]

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.

[DeoxyriboNucleicAcid]

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

[aommaster]

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!

[psi20]

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?

[Neurocomp2003]

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.