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Happy New Year!

Here's a nice puzzle for you guys.

2021 has started and can be written as $$ (-9+8)*7-6*(5-(4+3)^{2+1})+0=2021 $$ Ofcourse many other solutions are possible!

So, the obvious puzzle is: "How can you write 2022 with all digits in descending order (so 9 to 0)?"
All classical mathematical defined operators and concatenations are allowed. Eg. $*$,$/$,$+$,$-$,$!$, $\sqrt{}$, $\mod{}$, $\log{}$, $\int{}$ ...
EDIT: The shortest answer wins. Non-defined operators are not allowed.
Order of operations are as you would type them into LaTeX/classical calculators. So $\sqrt[a]{b}$ is a before b.

KUDOs

  • @Retudin for finding $(-9+8)*7-6*(5-(4+3)^{2+1})+0!=2022$
  • @Deusovi for the mentioning of non-defined operators
  • @tsh for the mentioning of operands like $\sqrt[a]{b}$.
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    $\begingroup$ ...+0! seems a bit too obvious to make an answer $\endgroup$
    – Retudin
    Jan 6, 2021 at 8:09
  • $\begingroup$ Which exact operators are allowed? Without specifying that, I could just define the operator $⋄$ as $a⋄b = 2022$, and then take $9⋄8⋄7⋄6⋄5⋄4⋄3⋄2⋄1$. And if you don't allow that, but still don't list exactly which operators are allowed, the question will likely be subjective. (In any case, questions with many possible equally "good" answers are discouraged here; they do not fit the format of this site, which looks for one definitive answer.) $\endgroup$
    – Deusovi ♦
    Jan 6, 2021 at 8:12
  • $\begingroup$ Also, the order of operand should also be defined. Is $\sqrt[a]{b}$ defined as a goes before b or otherwise? $\endgroup$
    – tsh
    Jan 6, 2021 at 8:32
  • $\begingroup$ Thanks for feedback. I've edited my question. Please not that this was my first ever question on PSE. So downvoting like that is a bit harsh. $\endgroup$
    – IT M
    Jan 6, 2021 at 8:40
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    $\begingroup$ How do you define shorter? Is more concatenation better? $\endgroup$
    – hexomino
    Jan 6, 2021 at 11:26

2 Answers 2

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I'm not sure how "shorter" is defined but if it means more concatenation then we can get the following

$\lceil 9876/5 + 4 + 32 + 10 \rceil = 2022$
$\lfloor 9876/5 + 43 + 2 + 1 + 0! \rfloor = 2022$

However, if the floor and ceiling functions are disallowed then my offering is the following

$(987+65-43+2)\times(1+0!)$

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  • $\begingroup$ floor and ceiling functions count, so both solutions work fine $\endgroup$
    – IT M
    Jan 6, 2021 at 11:57
  • $\begingroup$ Also nice solution with the better known operators. I'll mark your answer as the best solution for now, since the ceiling function contains the least amount of characters (which in essence is shorter). $\endgroup$
    – IT M
    Jan 6, 2021 at 11:59
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Using $+,-,\times,(,)$:

$9+8\times7\times6\times(5+4-3)-2-1+0=2022$

And the good news is we can also make it works in a few years...

$9+8\times7\times6\times(5+4-3)-2-1\times0=2023$
$9+8\times7\times6\times(5+4-3)-2+1+0=2024$
$9+8\times7\times6\times(5+4-3)-2\times1\times0=2025$
$9+8\times7\times6\times(5+4-3)+2-1+0=2026$
$9+8\times7\times6\times(5+4-3)+2-1\times0=2027$
$9+8\times7\times6\times(5+4-3)+2+1+0=2028$

Some more:

$9\times(8+7\times6\times5+4+3)-2-1+0=2022$
$9\times8\times7\times(6-5)\times4+3+2+1+0=2022$
$9\times8\times7\times(6-5)\times4+3\times2+1\times0=2022$
$(9\times8\times7+6-5)\times4+3-2+1+0=2022$

And with $\div$:

$9\times(8\times7\times6+5-4)\div3\times2+1\times0=2022$
$(9\times8\times7\times6+5+4)\div3\times2+1\times0=2022$

If concatenation is allowed:

$(9-8+7)\times654-3210=2022$
$9+(87+654)\times3-210=2022$

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  • $\begingroup$ Certainly a nice way of going! $\endgroup$
    – IT M
    Jan 6, 2021 at 11:57

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