Math Is Fun Forum

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

"Profs Duke It Out in Big Number Duel"

http://tech.mit.edu/V126/N64/64largenumber.html

We could have a similar contest!

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

namealreadychosen

Member

Registered: 2011-07-23

Posts: 16

Re: "Profs Duke It Out in Big Number Duel"

an attempt

x=skewes number
x^x^x....x times
iterate x times,
iterate result x times, call answer y,
replace x with y and repeat lines 2-4 y times
call result z
iterate lines 2-5, replacing x with z, z times
call result w
a=largest number expressible with above notation with z symbols
b=largest number expressible with a symbols
repeat finding largest expressible number b times

Bob

Administrator

Registered: 2010-06-20

Posts: 10,803

Re: "Profs Duke It Out in Big Number Duel"

hi namealreadychosen,

Oh wow!  Couldn't you have started with 1 or something?

MathsIsFun:  Are there rules? 

eg.  What size board?

I notice that the players seemed to take something from the previous play and edit it.  Is that a rule?

Words seem to be allowed so why not "The number he just wrote plus 1".  That would fit easily on most boards.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

SONIC123CDMANIA

Member

Registered: 2023-12-15

Posts: 12

Re: "Profs Duke It Out in Big Number Duel"

The true highest number is every number that isn't mine plus 1.

hypsin_0

Member

From: Earth

Registered: 2025-07-05

Posts: 108

Website

Re: "Profs Duke It Out in Big Number Duel"

I WILL MAKE A LARGE NUMBER
p1(0)=100
pn(a+1)=pn(a)×100
pn(#+pn+1(0))=pn(#+pn(#+...)) (100×pn)
rule3 maybe hard to understand.It means,find the most inside.(right is more inside than left)pn+1(0),and search outside,until the first pn,and copy for 100times.
for example:p1(p2(0)+p2(0))=p1(p2+p1(p2+...))

It seems easy,right?
But TREE(3)＜TREE(TREE(......TREE(3).......))＜......p1(p2(...p100(100)))＜sscg(3)