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The least number that cannot be written using the numbers 1, 2, and 3, each exactly once, and any combination of standard arithmetic operations (including factorials) is 41. What is the least such number if the numbers 1, 2, 3, and 4 are allowed?

Allowed operations are addition, subtraction, multiplication, division, factorials, exponents, square roots, intermediate non-integer results, and any amount of parenthesis and brackets. No other digits besides one of each of 1, 2, 3, and 4.

(This is basically what a person asked 4 years ago but just without using 0)

(I have all the numbers except 86, 93,)

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  • $\begingroup$ Is concatenation allowed? $\endgroup$ – hexomino Sep 25 '20 at 15:21
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    $\begingroup$ So, for example, if you can have concatenation, you could just do $86 = 43 \times 2$. Do you think this is okay? Also, do you have a reference for the $1,2,3$ version as this might help. $\endgroup$ – hexomino Sep 25 '20 at 15:46
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    $\begingroup$ @KentoHarmel so this is a homework assignment? $\endgroup$ – Ben Barden Sep 25 '20 at 17:14
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    $\begingroup$ If 19 is possible with just 1, 2 and 3, then 76 is possible by multiplying that result by 4. I'll admit I'm not coming with the former easily. $\endgroup$ – Jeremy Dover Sep 25 '20 at 19:57
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    $\begingroup$ Sorry guys for the confusion, there are no concatenations allowed. $\endgroup$ – MathQuestionMark Sep 26 '20 at 18:58
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91 is also possible without concatenation

$((3!)!)/(2\times4) + 1$

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  • $\begingroup$ What does "confirm yours" mean? Also, please put any math in MathJax. $\endgroup$ – bobble Sep 25 '20 at 19:56
  • $\begingroup$ I meant with that that I also found solutions for the numbers Kento solved. I guess that does not serve a purpose. I'll remove that part to avoid confusion. $\endgroup$ – Retudin Sep 25 '20 at 21:00
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Here comes $76$:

$76 = 4 \sqrt{\frac {(3!)!}{2}+1}$

If we are allowed double factorials we can do $85$:

$85 = \frac{2^{4!!}-1}{3}$

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  • $\begingroup$ 76: nice one, and in addition you upped the minimum. $\endgroup$ – Retudin Sep 26 '20 at 7:15
  • $\begingroup$ Can you please explain 76 for me, I keep getting a number with decimals, or am I just doing something wrong? $\endgroup$ – MathQuestionMark Sep 26 '20 at 19:46
  • $\begingroup$ @KentoHarmel Sure: $3!=6$, $6!=720$ $\frac{720}{2}=360$ $360+1=361$ $\sqrt{361}=19$ $4\times 19=76$. $\endgroup$ – Albert.Lang Sep 26 '20 at 23:04
  • $\begingroup$ Thanks a lot @Albert.Lang $\endgroup$ – MathQuestionMark Sep 26 '20 at 23:06
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Following Albert Lang's second solution, if we extend to double factorials, we can get $86$ and $93$

$86 = ((3!)!! - 4 - 1) \times 2 $
$93 = ((3!)!! \times 2) - 4 + 1$

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  • $\begingroup$ Wow, thank you, that's impressive $\endgroup$ – MathQuestionMark Sep 27 '20 at 0:04
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An answer for $99$:

$(3!+4)^2-1$

Regarding $85$, $86$, $93$:

I think these are impossible without concatenation, double factorial (deemed acceptable by OP's comment), or other operations of unknown status. I could not find answers even with computer aid.

With concatenation:

$85 = 43 \times 2 - 1$ (based on hexomino's comment)

$86 = 43 \times 2 \times 1$ (based on hexomino's comment)

$93 = 12 + 3^4$

If arbitrary multifactorials are allowed, we can obtain any positive integer in a rather silly way:

Suppose we want to reach $N$. Let $I(n, m)$ be the itererated factorial function, applying the ordinary factorial $m$ times starting with $n$. Choose $m$ such that $I(5, m) > 2N$. Then $$N = I(4+1, m)!^{(I(5, m) - N)} / I(2+3, m)$$ because $$I(5, m)!^{(I(5, m) - N)} = I(5, m) \times (I(5, m) - (I(5, m) - N)) = I(5, m) \times N$$ Note, the next term in the multifactorial expansion would be $$N - (I(5, m) - N) = 2N - I(5, m) < 0$$

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