# Largest number with five 1's and five numeric operations [closed]

You have five 1's at your disposal, together with five arithmetic operations of your choice. However, as you only have five operations, you should choose them wisely.

Question: What is the largest integer that you can generate this way?

Rules:

• Numbers can not be infinite. No dividing by 0.
• You cannot concatenate the 1's (i.e. you cannot use two 1's to make 11)
• You cannot use any other numbers in any other form: no Greek alternatives, no constants such as $e$ or $\pi$.
• Parentheses come for free; you may use as many as you like.
• You may use two or more operations in a row
• You may use any notation you would like. One solution below uses "Knuth's Up Arrow Notation". Each arrow uses one operation of the five allowed operations.

Examples:

  1+1+1+1++1 = 5

((1+1+1)↑↑(1+1)) = 27  <-- Uses Knuth's Up Arrow Notation

(1+1)^((1+1+1)!) = 64

((1+1+1)!)^(1+1) = 81


I have posted my solution below, let's see if you can beat me!

• Hmm, according to Wikipedia 3^^2=27. – Sleafar Oct 9 '15 at 16:18
• @Sleafar I was using the wrong function in Wolfram when I posted my solution. Forgot to update the question. Thanks. – dberm22 Oct 9 '15 at 16:20
• Is it required to use all five $1$'s and all five operations? – Julian Rosen Oct 9 '15 at 16:21
• I was thinking something like $(((1+1+1)!)!)!$ is quite large, but only uses three $1$'s. – Julian Rosen Oct 9 '15 at 16:24
• @dberm22: It is not easy to express this clearly in your question, without explicitly listing all the allowed operators. – Gamow Oct 9 '15 at 16:30

You may use any notation you would like.

Browsing Wikipedia I found the Steinhaus–Moser notation.

(1+1+1)^(1+1)=9


With one operator left we can put the number in a circle:

According to Wikipedia already ② is too big to be displayed. If less than 5 ones can be used, we can make the number even larger.

Update:

Instead of a circle we can use any n-sided polygon to make the number arbitrarily large. See for example the definition of Moser's number in the article linked above.

• Brilliant. Maybe change 1+1+1+1+1 to ((1+1+1)^(1+1)) ? – dberm22 Oct 9 '15 at 17:21
• @dberm22 Thanks, luckily there are still enough UTF8 chars for this. – Sleafar Oct 9 '15 at 17:26
• in the vein of my answer below, if you don't need to use all 1's: circle(circle(circle(circle(1+1)))) would be stupidly stupidly insanely larger – Kevan St. John Oct 9 '15 at 18:02

$$((1+1+1)^{(1+1)})! = 362,880$$

or, ((1+1+1)^(1+1))!

Uses 3 +, 1 ^, and 1 !

Try it on Wolfram Alpha.

• With only a single up arrow, shouldn't your value be 9! or 362880? – JonTheMon Oct 9 '15 at 16:00
• @JonTheMon not sure why Wolfram mislead me. Changed. Thanks for the catch! – dberm22 Oct 9 '15 at 16:12

Must I use all 1's?

$$(((1+1+1)!)!)! = 3!!! = 6!! = 720! = 2.601 × 10^{1746}$$

Using the same notation as Graham's Number and assuming that we don't have to use all five ones:

$$g_{g_{g_{g_{g_{1}}}}} = Something Absurdly Large$$

Note that gg1 makes Graham's Number (g64) look absolutely infinitesimal.

This is much smaller, but uses all five ones:

$$g_{(1+1+1)^{(1+1)}}$$

In general, $$g_n = 3\uparrow^{g_{n-1}}3$$ where $$g_1 = 3\uparrow\uparrow\uparrow\uparrow3$$

S(1+1, (1+1+1)!) = S(2, 6) is a large number, it has not yet been computed but is known to be > 7.4 × 10^36534

Source