ATMC: I think this is a 3-dimensional problem. It can only be solved using matrixes for one linear equation with 3 variables cannot be solved alone therefore 2 more linear equations are needed.

kponke denge:
Stop ur rambling n solve it so we can share d money 2geda. I nor go skol if not I for don solve am tay tay collect d moni.

ATMC: ...a constant would have made this a bit easier *thinking*

Pipsland78: The offer still stands...this is a cool $1 million dollars begging to be earned. Where are our maths professors and straight A students?

Pipsland78: ^^^ At last we have someone smart enough to crack the case and solve the problem. But remember 2% is mine.

Pipsland78:
Please remember I gave you the info when you solve it ooo, so 10% is mine.

digitite:

When we win the prize, I will give you 10%. Make sure it gets to the right source and follow up the news. Keep me posted on any new development. One Love .

digitite:
I came up with this...

A ^x + B^y = C^z

If A = 1, B= 2, C = 7
then x = 3, y = 5, z = 6

therefore;
A^x + B^y = C^z
1³+ 2⁵= 7⁶

That's my 10cent.

digitite:
I came up with this...

A ^x + B^y = C^z

If A = 1, B= 2, C = 7
then x = 3, y = 5, z = 6

therefore;
A^x + B^y = C^z
1³+ 2⁵= 7⁶

That's my 10cent.

digitite:
I came up with this...

A ^x + B^y = C^z

If A = 1, B= 2, C = 7
then x = 3, y = 5, z = 6

therefore;
A^x + B^y = C^z
1³+ 2⁵= 7⁶

That's my 10cent.

panafrican:

I will make less effort than you and get more money
Here is how :
Let's set A=B=C = 0 and suppose x, y, z are exponents.
then any (x, y, z) is a solution provided none of x, y, z is less than or equal to zero.

Amujale:


On my mind you get 10 x 10 per-cent for your effort.

I don’t know for certain if your answer is what the moguls are seeking however , for me, your mathematics is gorgeous and transparent.