Official TSR Mathematical Society

2/Actually, 2+2=4 is true, which was easy for me to realise as true since i don't know any computer science. Therefore, out of the remaining people, one must be telling the truth.

You have made an error in assuming it is true. Do you know what another base is?

so am i right in assuming my answer is correct and before everyone else has assumed 2+2 = 4 is true they are wrong?

i only know because in year 9 my teacher asked us to tell her one thing which we knew to be 100% true so i said 1+1 = 2 and she showed me using a clock as an example that i was wrong ^_^

2/Actually, 2+2=4 is true, which was easy for me to realise as true since i don't know any computer science. Therefore, out of the remaining people, one must be telling the truth.

You have made an error in assuming it is true. Do you know what another base is?

how does doing computer science make you realise that 2 + 2 = 4 is false?

That is the one time those papers have been disspointing in terms of answers. Ofcourse it would have been ambiguous but having two possibilities for that question is clearly valid.

How about ?

And if you manage that one, then just change the 2 to a 3.

Comment: I can think of a few ways of evaluating this integral, but none of them are particularly pleasant or simple.

Anyone going to have a crack at my previous question?

people seem to be ignoring this, here's my quick stab at it

I know little of modular arithmetic, but there are more solutions.

What is your method... (please)

Btw are you sure this converges?

3(b+1)(c+1) = 4bc -> bc - 3b - 3c = 0. Try factorizing, and you can find b and c. Rinse and repeat for different values of a.

I am very very tired but I do not think this is true.

3(c+1)(b+1) = 4bc --> bc - 3c - 3b - 3 = 0

Yes, you're right, it's late. But the method works.

Sorry, the post referred to the expansion of 3(b+1)(c+1) which is incorrect in your spoiler.

However, I am not convinced there is a solution for a=2;

Yes, you're right, it's late. But the method works.