FREE DIAMOND MEMBERSHIP FOR WHO GETS THIS RIGHT! NOT KIDDING!

Benzo, XOR is a logic operation. XOR entails that if A = B, then output is 0. If A ≠ B, then output is 1. Look up basic logic gates and operations to know more.

The loop I performed was running the logic operation on the 1st and 2nd digits of the four-digit number, then on the 3rd and 4th digits, leaving me with a 2-digit number. Then I performed the operation on the 2-digit number, leaving me with a single digit.

Mate, I know what bitwise AND, NAND, OR, NOR, XOR, XNOR operations/gatters are.

And I also get that you're "looping" on the same number.

However, it doesn't make any sense.

Why don't you do it from the outside-in or from the inside-out? See, it doesn't make any sense!!!

What is the answer?

A plausible answer.  It's most likely that or: 2 + 2 = 4

Would you like me to do the entire proof again from the inside-out and the outside-in? It doesn't make any sense because it's not intended to make any sense, I just decided to break it into binary and perform a bunch of random logic operations and a bit of calculus for the hell of it.

Umm, how did you integrate f(x)? I don't quite understand it.

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LIMIT 2 -- LOLOL

MSC, now that you mention it, I completely screwed up the integral. I took a derivative >.< - haven't been in calc class in a couple of months.

I should have used the a^x integral rule where the integral of a^(x+c) is (a^(x+c))/ln(a).

EDIT: The proof for that integral is a bit overcomplicated, I can't remember exactly how it goes but it's something like throwing in a ln(e) and doing integration by parts.

Without ln I guess. Ln is only when there's 1/x.

So it should be 2^x/x.

Yeah, I feel you. Just had mathematical exam 4 days ago and now I'm checking how much I forgot. :)

EDIT: Whoops, I think I know what you mean and I also sceewed a bit/lot. :)

The proof is as follows (It's actually a u-substitution):

int(a^(x+c)) dx

int((e^ln(a))^(x+c)) dx

int((e^(ln(a)*(x+c)) dx

Set u = ln(a)*(x+c)

Set du = ln(a) dx

= int((e^u)(du/ln(a))

= 1/ln(a) int(e^u) du

= e^u/ln(a)

= a*(x+c)/ln(a)

Alright, attempt number three, the integral is correct this time. :/

Aha, ok, yes. Don't remember if we've done it in school, but I understand it, very nice job!

On the other hand, we could have used WolframAlpha. ;)

No one caught the joke really? This has been done before.... <_<

i've seen this once, I believe it is an equation or 5 or a statement

I Wolframmed it first and then decided to do the proof myself. :P

You didn't give the answer I wanted, but you win because I didn't know that.

Twinchicky, you fail, though you appear to be the smartest.

iTS 3

more like you stole it from me from the future

Crangle drungels!