# Express numbers 1 - 100 using only 1, 2, 3, and 4

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...)

• Is concatenation allowed? Sep 25, 2020 at 15:21
• 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. Sep 25, 2020 at 15:46
• @KentoHarmel so this is a homework assignment? Sep 25, 2020 at 17:14
• 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. Sep 25, 2020 at 19:57
• Sorry guys for the confusion, there are no concatenations allowed. Sep 26, 2020 at 18:58

91 is also possible without concatenation

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

• What does "confirm yours" mean? Also, please put any math in MathJax. Sep 25, 2020 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. Sep 25, 2020 at 21:00

Here comes $$76$$:

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

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

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

• 76: nice one, and in addition you upped the minimum. Sep 26, 2020 at 7:15
• Can you please explain 76 for me, I keep getting a number with decimals, or am I just doing something wrong? Sep 26, 2020 at 19:46
• @KentoHarmel Sure: $3!=6$, $6!=720$ $\frac{720}{2}=360$ $360+1=361$ $\sqrt{361}=19$ $4\times 19=76$. Sep 26, 2020 at 23:04
• Thanks a lot @Albert.Lang Sep 26, 2020 at 23:06

Following Albert Lang's second solution, if we extend to double factorials, we can get $$86$$ and $$93$$

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

• Wow, thank you, that's impressive Sep 27, 2020 at 0:04

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

• Let $A=I(5, m)$. How come $A!^{A-N}=AN$? Something in your notation is unclear. Nov 2, 2022 at 21:36
• @RosieF $!^{(k)}$ (specifically with parentheses) denotes the $k$-th multifactorial, not a normal exponent. Not a great notation, but somewhat standard, if uncommon. Nov 3, 2022 at 1:19