Kevin Dorner

Dunno. The original post says "Using only 4 zeros (0) and any mathematical operations..."

and then says "P.S. By mathematical operations I mean only operators, the only numbers are the 4 zeros and the 24."

Are functions "operations"? Didn't introduce any new values (ie constants) into it.

stretch

After figuring out the answer, I posted the same question on a Catholic forum, with a thread title: Are theists as smart as atheists?

So far, nobody has even come close. Oh, well.

InfinityPlusOne

I prefer the solution using the cosine rather than all of the factorials. I've always thought about factorials being the integer values of the gamma function. This is true for all other non-negative integers. Of course Gamma(0) goes to infinity and that's why I think zero factorial has an element of black magic involved. I think it's more for convenience, like when writing infinite series.

I must say, though, that the definition for the factorial given in the link early on (as the number of permutations of a set) is interesting. It may stand up mathematically but I personally don't like that you can permute nothing and one single element the same number of times.

And since we obviously weren't given any clear-cut rules about what we can and cannot use, mathematically, why not use limits? Take the limit of 0/0 and you can get anything. so why not (lim) (0/0)+(0/0). Or subtract. Or multiply. Or divide. I guess I WOULD have to declare how I was getting these zeros, which might break these vague rules; the only numbers added would be equivalent to zero, though.

I could be totally wrong about all of this. After all, I am merely a physics grad student, a group not known for their rigorous math techniques.

Kevin Dorner

I don't think putting anything (even zero) over zero is allowed in any circumstances. A limit denominator can tend to zero but anything over a constant zero is meaningless, like taking the square root of blue.

InfinityPlusOne

Agreed, but it's standard (in my experience, at least) to say that the limit "is" some value.

The derivative in calculus is defined as a limit and I don't say that the derivative of the function tends towards some other function. I admit, though, that I should.

This is probably why I infuriated my math profs during my undergrad years.

Autonemesis

Yeah, but can you use only five zeros to come up with 120?

Loren Pechtel

Originally posted by Farren
Loren

~0 may yield -1 in some languages, but the language is innacurate, not Ipetrichs math.

In pure binary math, the truth table for ~ is

A | ~A
--------
0 | 1
1 | 0

This is invariant and pure math. It is as certain as 1+1=2.

But you can't do 24 in binary math. Thus you can't use binary. You have to use a system that allows you to express 24. At that point you have the 1/-1 problem.

Loren Pechtel

I don't think this is valid. Where did you get the 8 and the 2?

Kevin Dorner

Here's one that only needs three 0's, and doesn't use the factorial:

HeatherD

I found this at a physics forum and oddly enough no one needed any explanation. The p.s. part was mine because you guys are like cats. I knew if I wasn't specific enough you guys would snipe about all the possible options. That's why I like you guys.

Anyway, I would accept cos(0) because cos is a unary operator, operating on the operand "0" Factorial is a mathematical operator just like cos.

The original answer, where I found it was:

[(0!)+(0!)+(0!)+(0!)]!=24

This should also be right, as was mentioned:
[cos(0)+cos(0)+cos(0)+cos(0)]!=24

Kevin's is very creative, I haven't tested it but if it works then that would be another answer

As for the Boolean one, I'm not sure. Certainly ~0 means the bitwise not operation on zero, at least in C, IIRC. That is dealing with binary math and wouldn't work.

I don't know if you can mix Boolean Algebra and "regular math" but a true Boolean operation on 0 looks like this:

!0 - which is NOT 0 which equals 1, even if you stick to the argument that the 0 and the 1 are binary, they still 0 and 1 in base 10.

So I guess you could do this:

[(!0)+(!0)+(!0)+(!0)]!=24

it would work, assuming it is legal to mix Boolean operators with the factorial.

Anyone know!