You can use the digits 1,2,and 3 once only and any mathematical symbols you are aware of, but no symbol is to be used more than once. The challenge is to see if you can make the largest numbers.

Here are some numbers to set the ball rolling: 321, 21to the third power, (3/.1)to the second power.

(levik: I guess this is more of a competition)

(In reply to The challenge taken up.... by Dej Mar)

not sure if anybody is every going to see this, but obviously this number is huge.  I am simply going to show that googleplex pales in comparison to this number

let y={H([1E+(3!)]#)}$
and let S(a,b) be a to the superpower of b

for ease of notation,I will use Log(x) for base 10 log of x, and ln(x) for base 2 log of x

then we want to show S(2,y)>googleplex
take base 10 log of both sides
Log(S(2,y))>log(googleplex)=log(10^google)=google
log(S(2,y))>10^100
now S(2,y)=2^S(2,y-1) thus
s(2,y-1)log(2)>10^100
s(2,y-1)>10^100/log(2)
log(s(2,y-1))>log(10^100/log(2))
s(2,y-2)log(2)>log(10^100)-log(log(2))
s(2,y-2)log(2)>100-log(log(2))
s(2,y-2)>(100-log(log(2)))/log(2)=x
now x is aproximately 333.92483

now lets look at how fast s(2,t) grows
s(2,1)=2
s(2,2)=2^s(2,1)=2^2=4
s(2,3)=2^s(2,2)=2^4=16
s(2,4)=2^16=65536

so we have y-2>3 or y>5 or y>=6

so to sum it all up for s(2,y)>googleplex we simply need
y>=6
now y=H([1E+(3!)]#)$>=6
for x$>=6
x>=3
thus we need H(x)>=3 which gives
x>=2
thus x#>=2 thus x>=3 thus
[1E+(3!)]>=3
10^6>=3
which is true.  Thus we have proven that S(2,y)>googleplex

in fact, if you look at the end result of all the initial transformations
that reduce googleplex to a measily 333.92, yet S(2,y) is simply brought
down to S(2,y-2) you can see that s(2,y) is to googleplex as the universe
is to a subatomic particle, but even that analogy I feel is insufficient.
I guess my question which is beyond my ability to answer, is if you re-arrange
the order of the notations could you get a larger number, or is this maximal?

Posted by Daniel on 2008-09-01 16:28:09