Represent these numbers as
1. If u consider 2 raised to 222 is correct, type 2p222
2.If u consider 22raised to 22 is correct, type 22p22
3.There are many more options, those may be correct as well
4.Factorials or other such craps are not allowed...try powers of 2 only
2006-11-27 01:27:47 · 12 answers · asked by shricooldude92 2 in Science & Mathematics ➔ Mathematics
Since 2p22>222 then
2p(2p22) is the greatest number
My calculator could not handle that high a number.
So I figured , using an inductive method, that the result
is a 1 followed by 1,200,000 zeros.
Anyone care to confirm or contradict.
Here's how
2p4,000,000=10p(x)
x=4mil(log2)= 1.2 mil
2006-11-27 01:45:22 · answer #1 · answered by albert 5 · 1⤊ 0⤋
should be the one that many have already answered, 2 ^ (2 ^ 22)
2006-11-27 20:08:34 · answer #2 · answered by pravkas 2 · 0⤊ 0⤋
It's 2p(2p22), unless you're allowed to create infinity by dividing by zero e.g. 2+2 / (2-2)
2006-11-27 01:50:29 · answer #3 · answered by Anonymous · 0⤊ 0⤋
2^(2^22)
2006-11-27 01:30:43 · answer #4 · answered by mathman241 6 · 3⤊ 0⤋
2^2^22
which is 1 followed by at least 1.3 million zeroes.
2006-11-27 01:31:43 · answer #5 · answered by Anonymous · 0⤊ 0⤋
ac to me it is
2p2p22
reason being it is of the order
~10p10p6
wheras others do not stand anywhere
i did this using log3=0.301
which everyone knows
2006-11-30 20:05:42 · answer #6 · answered by sidharth 2 · 0⤊ 0⤋
You can always find higher then the one you thought about, there is no end to it.
In mathematical language it is call "Many" infinite or one more or one higher.
2006-11-27 01:38:16 · answer #7 · answered by minootoo 7 · 0⤊ 0⤋
2 p(2 p22)
2006-11-30 23:47:08 · answer #8 · answered by Sachin 1 · 0⤊ 0⤋
I think it's 2p222.
Edit: sorry, mathman was right. It's 2p(2p22).
2006-11-27 01:31:23 · answer #9 · answered by Amy F 5 · 0⤊ 0⤋
2p222
2006-11-27 01:54:06 · answer #10 · answered by raj 7 · 0⤊ 0⤋