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Inspired by this question:

What is the smallest positive integer that you cannot make with 5 ones and five operators?

Rules:

  • You must use all five ones and all five operators
  • Allowed operators are: +, -, *, /, ^, !, Knuth up-arrows. Each up-arrow counts as one operation.
  • Concatenation is not allowed. e.g. you cannot use two 1s to make 11.
  • You may use parentheses freely
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11
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    $\begingroup$ By "smallest number", I assume you mean "smallest positive integer"? $\endgroup$
    – user3294068
    Oct 9, 2015 at 19:34
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    $\begingroup$ By small would -1 be smaller than 0? $\endgroup$
    – Maxqueue
    Oct 9, 2015 at 19:34
  • $\begingroup$ Is concatenating 1s to make 11 allowed? (It's prohibited in the linked question, but not mentioned here.) $\endgroup$
    – f''
    Oct 9, 2015 at 21:23
  • $\begingroup$ Are you counting $\uparrow\uparrow$ as two operations, like in the linked question? $\endgroup$
    – Julian Rosen
    Oct 9, 2015 at 21:29
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    $\begingroup$ I can't imagine this being answered any way except a brute force computer search, making it a boring puzzle. $\endgroup$
    – xnor
    Oct 9, 2015 at 22:21

2 Answers 2

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I could generate 0-9 but I'm stuck at 10:

$0=1/1/1/1-1!$
$1=1/1/1/1/1!$
$2=1+1/1/1/1!$
$3=1+1+1/1/1!$
$4=1+1+1+1/1!$
$5=1+1+1+1+1!$
$6=(1+1+1)!/1/1$
$7=(1+1+1)!+1/1$
$8=(1+1+1)!+1+1$
$9=(1+1+1)\uparrow(1+1!)$
$10=...$
$11=...$
$12=(1+1+1)!*(1+1)$

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  • $\begingroup$ Can you make 5 with just threes 1s and 3 operators? $\endgroup$
    – warspyking
    Oct 10, 2015 at 4:18
  • $\begingroup$ Even disregarding the number of operators, since 5 is not a factorial, the last operation was a sum, product, difference, etc. But the two operands can only be 0, 1 or 2 (since there are up to 2 ones in each side), so this never yields 5. * Assuming you meant three ones $\endgroup$
    – MathET
    Oct 10, 2015 at 16:13
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Sleafar got the right answer - 10 - by forcing it. I got same by the following observation: in order to get numbers above 6 one have to make a group of (1+1+1) to get 3 to further operate with. as the remaining we have two "1" and 3 operations available. Among these 3 operations one should be reserved for connection to the first group. And to get number over 7 one must use + operator to get 2. That leads us to solve the problem having 3 and 2 and two remaining operators. Having these assumptions, 8 and 9 are possible to produce while 10 is not. BTW : the max positive integer is ((1+1+1)↑(1+1))! = 9! = 362880

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  • $\begingroup$ Because you are limited by 5 operators $\endgroup$
    – Yavpolnom Shoke
    Oct 10, 2015 at 5:00

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