Make 2019 with single digits

Question

This puzzle is inspired by a puzzle by @Andrew, where the goal is to make 1998 from a minimal amount of 8s. This is a variation of the puzzle, where the goal is to make 2019 using a minimal amount of single digits (0s, 1s, 2s, etc.)

More formal description (to remove any disambiguations and loopholes):
You need to make a (so-called) valid $i$-expression ($i$ is a digit from 0 to 9) which evaluates to 2019 under the common sense of the characters being included into it, and has the minimum number of digits (not counting all other characters) in it (the count is separate for each $i$), according to the following rules (which determine what an $i$-expression is):

• A string of one or more identical digits $ii\dots i$ is a valid $i$-expression (i.e. $2$, $22$ and $22222$ are valid $2$-expressions).
• If $A$ and $B$ are valid $i$-expressions, so are $A+B$, $A-B$, $A\times B$ , $\frac A B$, $A^B$ and $\sqrt[A]B$.
• If A is valid $i$-expression, so are $(A)$, $(A)!$ (factorial - note the parentheses, to explicitly disallow the multifactorials) and $\sqrt A$ (here, 2 is implied).

Example (a "5-expression"): $$(5\times(5+5)-5)^{\frac{5+5}{5}}-5-\frac55=(5\times10-5)^\frac{10}5-5-1=45^2-6=2025-6=2019\ (10\ \mathrm{fives})$$

Clarification: The ultimate aim is to find a minimum number for each kind of digit (i.e. a solution with minumum number of zeroes, ones, etc.), with a total of 10 different solutions. Of course, this does not invalidate the existing answers - providing a single solution is already great.

Answer 1

This is an agglomeration of all existing answers.

If you have a more optimal solution than presented above, please replace the answer with your one, change the number of digits used, and remember to change the link to your profile's.

If your solution alters the status of concatenating, please change that too (i.e. with -> without, or without -> with).

Anyone can edit as this is now a community wiki.

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Zero

• Using thirteen 0s without concatenation (@Bananenkopp)

$(0! + 0! + 0!)!^{0! + 0! + 0! + 0!} + ((0! + 0! + 0!)!)! + 0! + 0! + 0!$

One

• Using eleven 1s with concatenation (@TheSimpliFire, modified @iDriveSidewayz)

$(1+1)^{11}-(1+1+1)(11-1)+1$

Two

• Using nine 2s with concatenation (@TheSimpliFire, inspiration @athin)

$\sqrt{2^{22}}-2^2!-2-2-\dfrac22$

Three

• Using five 3s without concatenation (@TheSimpliFire)

$3!\sqrt{3!^{3!}}+(3!)!+3$

Four

• Using seven 4s without concatenation (@Konstantin)

$\dfrac{\sqrt{\sqrt{4^{4!}}}}{\sqrt{4}} - 4! - 4 - \dfrac{4}{4}$

Five

• Using eight 5s with concatenation (@steffen)

$5 - \frac{5555}{5} + 5^5$

Six

• Using eight 6s without concatenation (@TheSimpliFire, inspiration @OmegaKrypton)

$6\sqrt{6^6}+6!+\dfrac{6+6+6}6$

Seven

• Using ten 7s with concatenation (@TheSimpliFire)

$7\left(\dfrac{7!}{7+7}-77+7\right)-\dfrac{77}7$

Eight

• Using eight 8s without concatenation (@Bananenkopp)

$\dfrac{8!}{(\sqrt{8+8})!-\sqrt{8+8}}+\dfrac{(\sqrt{8+8})!}{8}$

Nine

• Using five 9s without concatenation (@TheSimpliFire)

$(\sqrt9\,!)\sqrt{(\sqrt9\,!)^{\sqrt9\,!}}+(\sqrt9\,!)!+\sqrt9$

Answer 2

My own (very trivial) solution with 0s (using 21 of them - so expecting to be beaten by somebody):

$$(0!+0!)^{0!+0!+0!+0!+0!+0!+0!+0!+0!+0!+0!}-(0!+0!+0!)^{0!+0!+0!}-0!-0!\\=2^{11}-3^3-1-1\\=2048-27-1-1\\=2019$$

Answer 3

After an exhaustive search (by hand) I've found a really good solution for 0 using 14 digits!

$(0!+0!+0!)\times(((0!+0!+0!)!)!-((0!+0!+0!)!+0!))-((0!+0!+0!)!-0!)!$
$=3\times((3!)!-(3!+1))-(3!-1)!$
$=3\times(6!-(6+1))-(6-1)!$
$=3\times(720-7)-5!$
$=3\times(713)-120$
$=2139-120$
$=2019$

Proof for optimisation

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To make this proof less convoluted (and to help in future similar problems) this is a dictionary of the construction of all numbers from 1-10. [Number ($i$ count): Expression]

1 (1): $0!$
2 (2): $1+1$
3 (3): $2+1$
4 (4): $3+1$ or $2^2$
5 (4): $3!-1$
6 (3): $3!$
7 (4): $6+1$
8 (5): $2^3$ or $6+2$
9 (5): $3^2$
10 (6): $9+1$ or $2\times5$

Proof that it isn't possible to reach 2019 using two (1-10) number $i$-expressions, so at least more than 8 digits are required

The only way to reach 2019 using two $i$-expressions is to use an expression either of the form $A \times B$, $A^B$ or $A-B$ ($\frac{A}{B}$ and $A+B$ are redundant). The only factors of 2019 are 3 and 673 and no factorial can equal 673 so $A \times B$ is ruled out and $A^B$ is ruled out by default. For $A-B$ to bear a solution, $A \ge 2019$ however no large number $B$ obtainable by a 1-10 number expression exists to constrain possible large $A$.

Following off of that proof, we now know the approach to take to get 2019

$A^B$ and $\frac{A}{B}$ are completely out of the question. From this it is safe to say that either $3\times673=2019$, $A+B$, $A-B$ or $A=2019^B$

On the subject of factorials...

No perfect power factorial exists above $1!$. Also no expression containing only factorials (above $1!$) can equal a prime number due to common factors.

This helps to drastically rule out many expressions

$A!=2019^B$ is out. $673$ is a prime number so the only way to reach it is by $A \pm B$. In the case of $673$, $A \pm B$ is disprovable by exhaustion for 1-10 numbers since nested factorials are disproved. In a more general case, $A=2019^B$ cannot be done if $C$ is added either. For it to even bear results it would have to be $A \pm B = 2019^C$ or $A \times B = 2019^C$. Since $2019^C=3^C673^C$ it clearly isn't possible using nested factorials so this can be disproved by trivial exhaustion. A common factor of $3^C$ can be brought out of $A!\pm B!$ for $A!,B! > 3!$ leaving two other factorials which must sum or difference to $673^C$ which isn't possible because it is many of the same prime. This is now trivially disprovable by exhaustion! Thus this case generally requires at least 12+ digits since more than three (1-10) number $i$-expressions are required. I doubt this will lead anywhere so it is fair to say that it has been disproved! {of course this isn't 100% irrefutable proof since it is possible for $D$ to enter the equation however the numbers (1-10) would be highly restrained and I don't want to write a full paper on this}

On the subject of $3\times673=2019$ we already know that $3$ takes 3 digits so all I have to do is show that $673$ cannot be made in a reasonable amount of digits. We know that $A,B$ expressions that contain only factorials cannot be prime and the other solutions are trivially disprovable by exhaustion ($A \pm B$). This brings it up to 11+ digits. Adding in $C$, the cases to be examined are $A \times B \pm C$, $A-B \times C$ and $A^B-C$. The unlisted cases are either clearly logically wrong, already listed and/or trivially disproved by exhaustion. $A \times B \pm C$ requires a nested factorial somewhere in order for the numbers to be large enough for a solution. If $A$ or $B$ is a factorial $>1!$ then $C$ must be an odd number for the result to be prime. Therefore either all numbers are nested factorials (disproved by previous logic) or only one of them are a nested factorial. Since only one of them can be a nested factorial, this is disprovable by exhaustion (gee this is getting exhausting!). This is extended to $A-B \times C$ (which is now very obvious without proof by "..."). For $A^B-C$, $A$ and $C$ cannot both be nested factorials so logically only $B,C$ should be examined. $A$ must then be odd. Looking at $A^{B!}-C!$, the cases where $A \le C$ - have a common factor thus disproved. $A>C$ restricts nested factorial $C$ to $A$'s upper bound thus is disprovable by "...". All of this proof comfortably disproves $3\times673=2019$ since $673$ requires more than three (1-10) numbers and at worse case scenario (unlike my $A=2019^B$ proof) is +12 digits but should generally require +15 digits.

Answer 4

The best solution I found so far (painful to say with concatenation...):
With 6 copies of 3

$( (3!)! - 3!3! ) \times 3 - 33$

$ = ( 6! - 6*6 ) \times 3 - 33$
$ = ( 720 - 36 ) \times 3 - 33$
$ = ( 684 ) \times 3 - 33$
$ = 2052 - 33$
$ = 2019$

Answer 5

Update:

The first offering for (sixteen) $7$s based on the $1$s solution

$(\frac{7777 - 777 + 77 - 7}{7}) (\frac{7 + 7}{7}) - (\frac{7}{7}) = 1010 \times 2 - 1$

I have fourteen $1$s

$(1 + 1)(1111 - 111 + 11) - 1 - 1 - 1 = 2 \times 1011 - 3$

I have eight $4$s

$44(44 + \sqrt{4}) - 4 - \frac{4}{4} = 44 \times 46 - 5$

Answer 6

With six nines:

$((\sqrt 9)!)!\times\sqrt 9 -((9+(\sqrt 9)!)\times9) -(\sqrt 9)!$
$=720\times3-(15\times9)-6$
$=2160-135-6$
$=2019$

Answer 7

For $4$, we can use seven digits:

$\sqrt{\sqrt{\sqrt{4^{44}}}} - ((4-4)! + 4! + 4)$

Answer 8

Ten 5s (different from @TheSimpliFire and OP)

$(5*5-5)*((5*5-5)*5+\frac{5}5)-\frac{5}5=2019$

Eight 6s (thanks @TheSimpliFire)

$6\sqrt{6^6}+6!+\frac{6+6+6}6=2019$

Ten 6s

$(666+6+\frac{6}{6})\times\frac{6+6+6}{6}=2019$

Ten 8s

$\frac{\sqrt{8^8}}{\sqrt{\sqrt{8+8}}} - \sqrt{(8+8)}! - \sqrt{8+8} - \frac{8}{8} = 2019 $
$\sqrt{8^8}=4096$
$\sqrt{\sqrt{8+8}}=2$
$\sqrt{(8+8)}!=24$

Answer 9

Seven 4's no concatenation

\begin{align}\frac{\sqrt{\sqrt{4^{4!}}}}{\sqrt{4}} - 4! - 4 - \frac{4}{4}&= \frac{4096}{2} - 4! - 4 - \frac{4}{4}\\&= 2048 - 24 - 4 - 1\\&= 2019\end{align}

Answer 10

I have found a solution with 7 (threes)

$$3!^3*3!+(3*3)^3-3! = 6^3*6*(9)^3-6=1296 + 729 - 6 = 2019$$

Answer 11

All solutions without concatenation.

A solution for $0$ using 13 digits

$(0! + 0! + 0!)!^{0! + 0! + 0! + 0!} + ((0! + 0! + 0!)!)! + 0! + 0! + 0! = 1296 + 720 + 3 = 2019$

A solution for $7$ using 10 digits

$\frac{(\frac{7! + 7 + 7}{7} - (7\times7)) \times (7+7+7)}{7} = \frac{673 * 21}{7} = 2019$

A solution for $8$ using 8 digits

$\dfrac{8!}{(\sqrt{8+8})!-\sqrt{8+8}}+\dfrac{(\sqrt{8+8})!}{8} = 2016 + 3 = 2019$

Old solutions:

A solution for $8$ using 9 digits:

$ \frac{8! + 8*8 - \sqrt{8+8}}{8+8+\sqrt{8+8}} = \frac{40320 + 64 - 4}{16 + 4} = 2019 $

Answer 12

a "3-expression" instead of 5 in example:

$((3!^3 \times 3)+((3! \times (3+ \frac 33))+ \frac 33)) \times 3$
$((6^3 \times 3) + ((6 \times 4)+1)) \times 3$
$((216 \times 3) + (24+1)) \times 3$
$(648+25) \times 3$
$673\times3 = 2019$
10 threes

Answer 13

A solution for 7 using 18 7's

$7\times7\times7\times7 - \frac{777}{7} - \frac{777}{7} - \frac{777}{7} - 7\times7 = 2019$
$2401 - 111 -111 -111 -49$

Answer 14

12 1s:

$(1+1)^{11} - (1+1+1)^{1+1+1} - (1+1)$

Answer 15

Nine 3s without concatenation.

$3(3\times3)^3-3!\left(3^3+\dfrac{3}{3}\right)$

Answer 16

Nine ones with concatenation and decimal point:

$(1+1)*\frac{1111}{1.1}-1$

Eight fives with concatenation:

$5 - \frac{5555}{5} + 5^5$