Soldato
- Joined
- 8 Mar 2007
- Posts
- 10,938
MAFFS!
2+4 is 1 if you are in mod 5, not mod 4![]()

I'm expected to solve this sort of stuff at degree level, it's an example from monday's lecture 
But if you use BIDMAS (or BODMAS) should you not do the addition first which makes the sum 6-(0+1) making the answer 5?
Not so simple now? Hmmmm!
Oh bugger, I was being an idiot and counting from zero with 2.
Haven't done modulo arithmetic in a couple of years to be fair!
Some simultaneous equations:
(3x^2)+(2y^4)+z=0
(4x^(1/2))+(3iy^e)+4sinz=23
((4cos2x)/(3cos3x))+4y+(z^2x)=14
Solve!
And good luck
Edit: without Wolfram Alpha if you canI'm expected to solve this sort of stuff at degree level, it's an example from monday's lecture
![]()


African or European?
With all due respect, I am someone who spends all day formulating and solving complex partial differential equations, which involves the numerical solution of millions of simultaneous equations. (I think Duff-Man on here is also someone who develops finite-element / finite-volume codes) I therefore don't feel too enamoured towards solving some simple sets of equations in my spare time - if I wanted to do this then I'd sit in the college admissions again for maths / engineering / physics undergrads and watch candidates try not to mess them up
Once you're not an undergrad / high school student any more then you don't need to prove to yourself you can do it, you fire up Mathematica and get the computer to solve it in a couple of seconds rather than wasting your own time! You reach that stage where simple maths starts to seem complicated again - I had to do a couple of numerical tests for job applications, and I was in that awkward situation of not wanting to mess up currency conversion questions / percentage increases by deeming it below me![]()



I am a professional mathematician, and I can confirm that the correct answer is infact potato.