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