Melody

Thanks Heureka,

But the curved graph does not go through those 2 points.

Here is a bigger version of the graph.

I think it goes through the points (2.75,2.75) (1.75,5) and (5,1.75)

How do I find a graph that will go through these points ?? 

(6x^3 + 4xy^2 + 3y^2 + 13x^2y) ÷ (2x+3y)

This cannot be simplified:

I shall try your method:)

14^15 = 155568095557812224155568095557812224mod15 = 155568095557812224*mod15

As you can see this calculator does not work that way.

I shall try putting it in properly - and yes this calculator is capable of calculating it straight up

(I had thought the number , i mean 14^15, might be too big but it is not)

mod(14^15,15) = 14       

I result is directly from the web2 calc on this posting page :)

Very nice Asad,

Can you do some graphs like that ? :)

Here is another way

14^1=14=    -1mod15

14^2=196= +1 mod15

14^3=2744 = mod(2744,15) = 14 = -1

114^4 = 38416 = mod(38416,15) = 1

We have a pattern

\(For\;\;n\in \text{Positive integers}\\ 14^{2n}=+1\\ 14^{2n-1}=-1 \\ so\\ 14^{15}\;\;mod\;15=-1\;\;or\;\;14\)

Thanks for letting me know Asad, I appreciate that.      

I did unlock it but it must have locked itself again automatically.

If it relocks again could you please let me know again.  Thank you.  

Helperid bought my attention back to this question.

He has not even read my response to him, from a couple of days ago, but I am curious myself.

How can I get a graph to look like that curved graph in his profile pic?

(I hope this is not a doh moment - but too bad if it is)

I think it goes through the points (2.75,2.75) (1.75,5) and (5,1.75)

I assumed it was half a hyperbola and got to xy=1.75*5=8.75 but that does not go through (2.75,2.75)

So it is not correct.     

Can  someone offer a better attack method and answer?