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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 parentheses and brackets. No other digits besides one of each of 1, 2, 3, and 4.

(This is basically what user Bernardo Recamán Santos asked 4 years ago but just without using 0. I have all the numbers except 86 and 93...)

Stiv
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MathQuestionMark
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4 Answers4

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91 is also possible without concatenation

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

Retudin
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  • What does "confirm yours" mean? Also, please put any math in MathJax. – bobble Sep 25 '20 at 19:56
  • 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. – 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 $85$:

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

Albert.Lang
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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 $$N - (I(5, m) - N) = 2N - I(5, m) < 0$$

tehtmi
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  • Let $A=I(5, m)$. How come $A!^{A-N}=AN$? Something in your notation is unclear. – Rosie F Nov 02 '22 at 21:36
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    @RosieF $!^{(k)}$ (specifically with parentheses) denotes the $k$-th multifactorial, not a normal exponent. Not a great notation, but somewhat standard, if uncommon. – tehtmi Nov 03 '22 at 01:19
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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 $
$93 = ((3!)!! \times 2) - 4 + 1$

hexomino
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