Maths in bed - Latin square

A diff one I tried -



Using the info I had, I filled in the certain boxes with every number they could possibly have in them...it didn't help

I'd help

But not at this time

What book is this??

Post it in Maths

@NotNotBatman You need a different Latin square.


@Biryani007

Start with the highest and lowest valued equations.

D E 1=11. This covers two squares and the only combination of two numbers is 5 and 6. Implying $D1,E1,\in\{5,6\}$.

We can also infer that 5,6 cannot be in A1,B1,C1,F1.

D 4 5 6 =6. Three integers that are all different that add to 6, can only be 1,2,3 etc.

Final Solution:

Is it even possible? If we have a 6x6 matrix : $\begin{bmatrix}a & b & c & d &e &f \\ b &a &f &e & c &d \\ c& f & \bf b& \bf a & d &e \\ d&c & e &\bf b & f &a \\ e & d & a &\bf f & b &c \\ f & e & d &\bf c &a &b \end{bmatrix}$

And CD3=11,(I'm assuming this means C3 +D3 =11), then $a+b = 11$, since it must be less than 6 then $5\leq a\leq 6$ and $5\leq b\leq 6$, but D456 =6, meaning $b+c+f = 6$, this restricts it so that $b\leq 3$

but, there is no number for:
$b\leq 3$ and $5\leq b\leq 6$

Or maybe I'm reading it wrong.

I think you have be changing the numbers (if you have to) sometimes so that the equation at the top are true.

So the first one is A345=11 so A3, A4, A5 are three numbers that add up to 11.

So you can have 5+5+1, 7+3+1 (etc) in boxes A3, A4, A5.

@NotNotBatman You need a different Latin square.


@Biryani007

Start with the highest and lowest valued equations.

D E 1=11. This covers two squares and the only combination of two numbers is 5 and 6. Implying $D1,E1,\in\{5,6\}$.

We can also infer that 5,6 cannot be in A1,B1,C1,F1.

D 4 5 6 =6. Three integers that are all different that add to 6, can only be 1,2,3 etc.




Final Solution:

Is that how you do it? Using one of those squares with letters filled in?

In my second post I uploaded a pic of one I tried...I did what you did and found all the possible numbers you could have in the boxes mentioned...it doesn't make it any clearer...how would you know what order the numbers go in the squares? Just trial and error after you've narrowed it down or is there a method?

So, would a systematic trial and error method need to be used, since there are many matrices that are orthogonal to the one I wrote? Or is there another method?

There's no formal method that I'm aware of. Just see what's the next step. What can you deduce from your current information.

E.g. in your second post.
D345=6. So, these must be 1,2,3 in some order.

It follows that D1,D2,D6 are each one of 4,5,6.

But DE2=5, So, D2,E2 are at most 4, hence D2=4.

And, D1, D6 are now 5 or 6.

DE2=9. Since D2 is now 4, then E2=5

It's a bit like Sudoku, and a lot of other logic games.

I get that it's done logically...it's placing the initial numbers that's stopping me really...the 1,2,3 do I place in any order and then change up if I need to later on? D3 has an equal chance of being 1, 2 or 3 right? My question was how do I know in what order to place them..

You don't. You put 1,2,3 in each of the three cells. Then as more information becomes available you can start crossing out.

You'll find you have to run through the given clues numerous times. You need to maximise what you can deduce. E.g. If DE3=7, and D3 has {1,3,4} in it, then E3 can only have {6,3,4}

Sometimes the next step is obvious, sometimes it's incredibly difficult.

Eurgh I hate how confused I am

Ive done that for all the boxes I can in post 2...looking at that what would your next step be?

Sorry

Looks as if you don't have any squares finalised there. See my post #14. It had a massive error, so you can't have read it. Corrected now.

Ahh yeah I think I've got it now...so 1,2,3 you wont have finalised yet because you cant...that'll be done with info later on, right?

For other squares, how'd you pick a good place to start...clues that cover the most squares and then work in the column/row around it like you did for this one...coz I'm guessing starting anywhere wouldn't really work?