Math Is Fun Forum

Proof: This proof is by induction on n.

(Base Case) Let n = 25.  Then 25! > 10^25.  So the base case holds.

(Inductive Assumption) Now let n be an arbitrary number, such that n ≥ 25.  Assume that n! > 10^n.

Since n ≥ 25, n+1 > 10.  Also, since we already know n! > 10^n:

(n+1)*n! > 10*10^n, or rather, (n+1)! > 10^(n+1).

∴By the principle of mathematical induction, n! > 10^n for all n ≥ 25.  QED.

Nope, not true.  0^4 = 0^5.

Yea, but can you really call this exponential?  If you do, then just about every single equation is exponential.  Example:

y = x^2 + 2.

Certainly doesn't seem exponential, does it?  But watch:

y = x^2 + 2 * (1^x)

Since for all values, 1^x = 1, these are the same equations.  Now would you call this exponential?  Nah, it makes no sense to.

Only it's been falsified by it's creation.

I'm still getting confused with this "all" thing.  When you say all numbers 1, 2, 3... N, I think you need:

1, 2, 3, .... N-2, N-1, N all in the list.

Apparently, that is not the case (from your examples).  So I take it that N is just the highest number you use.  Is that right?

I modified my program so that it selects from a list of integers from 1 to 100 and creates lists:

Edit: I realized my old program was producing duplicate results, for example:

2 34 47
2 47 34

So I made it eliminate (actually, not even check) these.

Size 3:

2 23 98
2 34 47
3 22 78
4 21 60
5 20 44
6 19 30
6 75 94
10 54 90
12 52 69
14 35 86
15 34 66
16 33 48
26 74 95
29 52 92
30 51 70
48 73 96

Size 4:
1 3 22 99
1 3 33 48
1 3 97 99
1 8 41 80
1 24 97 99
2 7 42 79
3 6 43 78
4 5 44 77

It's currently running, but takes quite a while.  I'll update posts with results as I get them.

How exactly would you say "&&"?

"The sum of all perfect squares is 2n(n+1)/2"

What exactly do you mean by the sum of all perfect squares?  That would be infinity.

Edit:

My program has ruled out all numbers up to a list of size 13.  Of course, with each size increase, the time it takes goes up by a factorial, so I fear it won't finish 20 by tonight.  We'll see.

I worked up my own proof, but krassi beat me to it.  I figure I'd post it anyways.

I've only seen Z refer to integers, but like what krassi said, we need the context as well.

Sort of.  Since the coefficients are all low numbers, you know the factors have to be low as well.  So you can just try all the low integer factors -5 to 5 and see if they work.  If those don't, you can be reasonably sure the factors aren't integers.

Maybe he just has the midas touch.

Well, no one said math should be immediately apparent

If you don't want to try to factor, you could always try the cubic formula:

http://mathworld.wolfram.com/CubicFormula.html

So you don't have to have a calculator, but it is much simpler if you do.  Isn't that why those were invented anyways?