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!


closed as too broad by Deusovi, Julian Rosen, Tyler Seacrest, Engineer Toast, Gordon K Oct 9 '15 at 19:54

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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

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 '15 at 17:21
  • $\begingroup$ @dberm22 Thanks, luckily there are still enough UTF8 chars for this. $\endgroup$ – Sleafar Oct 9 '15 at 17:26
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    $\begingroup$ 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 $\endgroup$ – 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.

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


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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