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?


  • 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.


  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!

  • $\begingroup$ Hmm, according to Wikipedia 3^^2=27. $\endgroup$
    – Sleafar
    Oct 9, 2015 at 16:18
  • $\begingroup$ @Sleafar I was using the wrong function in Wolfram when I posted my solution. Forgot to update the question. Thanks. $\endgroup$
    – dberm22
    Oct 9, 2015 at 16:20
  • $\begingroup$ Is it required to use all five $1$'s and all five operations? $\endgroup$ Oct 9, 2015 at 16:21
  • 2
    $\begingroup$ I was thinking something like $(((1+1+1)!)!)!$ is quite large, but only uses three $1$'s. $\endgroup$ Oct 9, 2015 at 16:24
  • 1
    $\begingroup$ @dberm22: It is not easy to express this clearly in your question, without explicitly listing all the allowed operators. $\endgroup$
    – Gamow
    Oct 9, 2015 at 16:30

5 Answers 5


You may use any notation you would like.

Browsing Wikipedia I found the Steinhaus–Moser notation.

If all ones must be used we can start with:


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.


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.

  • $\begingroup$ Brilliant. Maybe change 1+1+1+1+1 to ((1+1+1)^(1+1)) ? $\endgroup$
    – dberm22
    Oct 9, 2015 at 17:21
  • $\begingroup$ @dberm22 Thanks, luckily there are still enough UTF8 chars for this. $\endgroup$
    – Sleafar
    Oct 9, 2015 at 17:26
  • 1
    $\begingroup$ 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 $\endgroup$ Oct 9, 2015 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.

  • $\begingroup$ With only a single up arrow, shouldn't your value be 9! or 362880? $\endgroup$
    – JonTheMon
    Oct 9, 2015 at 16:00
  • $\begingroup$ @JonTheMon not sure why Wolfram mislead me. Changed. Thanks for the catch! $\endgroup$
    – dberm22
    Oct 9, 2015 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:


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



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