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December 22

2^222222222

What is 2 to the power of 222222222 —Preceding unsigned comment added by 24.15.66.119 (talk) 22:36, 22 December 2007 (UTC)[reply]

Remember to sign next time and the answer is 4938271590617284. I do not remember where to put the places. Rgoodermote 22:39, 22 December 2007 (UTC)[reply]

I don't see what that number has to do with the question. 2^222222222 is a very large number. It is the value of 2 × 2 × 2 × ... × 2, with 222222222 times the number 2. If you write it out as a decimal number, it consists of 66895555 digits. That is an awful lot; if you were to print it out on paper, it would be enough to wallpaper a big room. The number starts with ~~3350427837...~~ 3350427823... and ends on ...4451042304. --Lambiam 23:44, 22 December 2007 (UTC)[reply]

I did it on paper and my math skills are not that great, I would have done it by calculator but mine is broken and I need to send it in. Rgoodermote 23:50, 22 December 2007 (UTC)[reply]

Even by inspection, it is not hard to see that the numbers are far too large to handle using pen and paper in any reasonable amount of time. —Preceding unsigned comment added by 75.12.149.26 (talk) 02:41, 23 December 2007 (UTC)[reply]

I doubt that your calculator, even when fixed, can handle 66895555-digit numbers. --Lambiam 01:26, 23 December 2007 (UTC)[reply]

It is quite obvious that what Rgoodermote has calculated is actually

222222222^{2}

. -- Meni Rosenfeld (talk) 09:35, 23 December 2007 (UTC)[reply]

I agree on the digit count and end but get 3350427823... in the start. PrimeHunter (talk) 00:00, 23 December 2007 (UTC)[reply]

You're right, I must have made an error somewhere. Here are the first 100 digits, in groups of 10: 3350427823 2757955011 2615529851 2161657287 7758144507 2745761854 2197277741 8185959255 1313691825 1073290414. --Lambiam 01:23, 23 December 2007 (UTC)[reply]

(edit conflict, also, someone should remove "Template:Edit, I changed the wording so that it appears in the right context. I meant to reply to the original poster, non someone else" if possible)Hi. Well, I can't do it on my scientific calculator, because mein only goes up to 9.999999999 x 10⁹⁹, and after that, it only displays as "Error 2", which it will also display, for example, if division by zero is performed, which indicates the number is out of the range. Thanks. ~A H 1^(TCU) 01:31, 23 December 2007 (UTC)[reply]

(I have removed [1] the curly braces from the mentioned non-template comment "{{Edit, I changed the wording so that it appears in the right context. I meant to reply to the original poster, not someone else}}" in [2] from December 17).

Use of logarithms enables things like easily computing the number of digits, first digits and last digits of 2^222222222 (but an ordinary PC with bignum software could also compute the complete decimal expansion without problems). PrimeHunter (talk) 02:03, 23 December 2007 (UTC)[reply]

How do logarithms help in finding the last digits? You need modular arithmetic for that. -- Meni Rosenfeld (talk) 09:35, 23 December 2007 (UTC)[reply]

Right. That's what I really used for the last digits. PrimeHunter (talk) 21:23, 23 December 2007 (UTC)[reply]

Unknown Series

x	y
0	1
1	1
2	2
3	6
4	24
5	120
6	720
7	5040
8	40320
9	362880
10	3628800
11	39916800
12	479001600
13	6227020800

Can anyone help me spot the pattern in this number series, I cant i thought it might be in the form of $e^{p(x)}$ where $p(x)$ is some polynomial. But I cant seem to be able to come up with anything. Thanks. ΦΙΛ Κ 23:25, 22 December 2007 (UTC)[reply]

Haha, of course, how did I miss that! Thanks anyway! ΦΙΛ Κ 13:17, 23 December 2007 (UTC)[reply]

See Factorial. For example, 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040. --Lambiam 23:47, 22 December 2007 (UTC)[reply]

For future reference, the easy way to answer such questions is the OEIS. Algebraist 22:43, 23 December 2007 (UTC)[reply]