18
$\begingroup$

Inspired by this puzzle, I've come up with the following:

Can you find a way to make

$2 \; 0 \; 2 \; 2 \; 2 \; 0 \; 2 \; 2 = 2022$

by only adding any of the following operations or symbols:

$+,\ -,\ \times,\ !,\ /,\, \hat\, ,\ (, \, )$

Additional Rules:

  • Concatenation is not allowed. For example, $\operatorname*{concat}(2 \; 0 \; 2 \; 2) + (2 \times 0 \times 2 \times 2) = 2022$ is disallowed.
  • The same symbol cannot be adjacent (except brackets). For example, $!!$ or $\hat{}\hat{}$ is disallowed.
  • Base conversion is not allowed, that is, both LHS and RHS must use base 10.
  • $=$ strictly denotes numerical equality.
  • The only valid interpretations for the above symbols are those listed here.
  • Rearranging numbers is not allowed.
  • Symbolic manipulation is not allowed (for both symbols and numbers). For example, $\neq$, $\substack{0\\0}$, $!\!=$, etc. is disallowed.
  • Modifications may only be made to LHS.
  • No new symbols or numbers may be added. For example, $\mid$, $[]$ or commas may not be added.

Hint (for one possible solution):

$337 \times 6 = 2022$

Another hint (for more solutions):

You may use $!n$ to denote the subfactorial.

$\endgroup$
4
  • 1
    $\begingroup$ So (2+0!)!! is not allowed, but ((2+0!)!)! is. Is this a loophole? $\endgroup$
    – Florian F
    Jul 19 at 13:26
  • 2
    $\begingroup$ @FlorianF No. $\endgroup$
    – user80184
    Jul 19 at 13:27
  • $\begingroup$ Maybe I am nitpicking but by enumerating allowed "operations or symbols" you equate the two. So each symbol is understood to be an operation. $\endgroup$
    – Florian F
    Jul 19 at 13:35
  • $\begingroup$ @FlorianF IMO the intent is clear, if it is not, ask away! An operation can be represented by one or more symbols. It happens so that all the operations given in the next line are represented by a single symbol. Later it is clarified that operation represented by adjacent symbols, such as !! are not allowed. This makes sense because they are not in the list. If each exclamation mark treated as an operation factorial it is allowed as your example demonstrates. But double factorial is not allowed. $\endgroup$ Jul 21 at 8:32

7 Answers 7

10
$\begingroup$

Here's another one:

$-2-0!+(2(2+2)!-0!-2)^2=-3+(2\times 24 -3)^2=-3+45^2=2025-3=2022$

Using hint 2:

$!(2+0!+2) \times (2 + !(2+0!+2)) - 2 = 44\times 46 -2 = 2024-2 =2022$

Or:

$(2+!0)!+(!(2+2))!/((2+!0)!)!\times 2\times 2=6+9!/6!\times 4=6+504 \times 4 = 2022$

$\endgroup$
3
  • $\begingroup$ @loopyWait You are missing a × in front of the parentheses in 2(2. $\endgroup$
    – lukas.j
    Jul 20 at 7:57
  • 1
    $\begingroup$ No I do not: en.wikipedia.org/wiki/Juxtaposition#Mathematics @lukas.j $\endgroup$
    – loopy walt
    Jul 20 at 8:03
  • 1
    $\begingroup$ Then you could drop the × in your second solution. $\endgroup$
    – lukas.j
    Jul 20 at 8:17
14
$\begingroup$

It was pretty challenging and fun.

$$\left(2+0!\right)\times\left(\left(\left(2+2\right)!+2\right)^{0+2}-2\right) \\= 3 \times (26^2 - 2) = 3 \times 674 = 2022$$

$\endgroup$
7
$\begingroup$

Following the hint, there's also

$$(2+0!)! \left(-2^2! + \frac{((2+0!)!)! +2}{2} \right)$$ $$ = 3! \left( -4! + \frac{(3!)! + 2}{2}\right)$$ $$ = 6 \left( -24 + \frac{720 + 2}{2}\right)$$ $$ = 6(-24 + 361) = 6\times337 = 2022$$

$\endgroup$
6
$\begingroup$

Was fun! Probably took me a bit longer than it should have, had about 4 prime factorization calculators open.

$(2 + 0!) \cdot (-(2 \cdot (2 + 2)!) + ((0! + 2)!)! + 2)$

$\endgroup$
2
  • 1
    $\begingroup$ I thought it's a minor variation of one of existing answers, but it's not. Well done! $\endgroup$
    – Bubbler
    Jul 20 at 6:44
  • 2
    $\begingroup$ Neatest one so far. $\endgroup$
    – WhatsUp
    Jul 20 at 9:14
3
$\begingroup$

Does this one count?

\begin{equation*} \begin{split}((2+0!)!)+((2+2)!)^2+(((0!+2)!)!\cdot2)&=\\ ((2+1)!)+(4!)^2+(((1+2)!)!\cdot2)&=\\ 3!+(4!)^2 (((3)!)!\cdot2)&=\\ 6+24^2+((6!)\cdot2)&=\\ 6+576+(720\cdot2)&=\\ 582+1440&=2022 \end{split} \end{equation*}

$\endgroup$
2
$\begingroup$

Great challenge!

Here's my solution:

$$-\left(2+0!\right)+\left(\left(\left(2+2\right)!\cdot2\right)-\left(0!+2\right)\right)^{2}$$ $$=-(3)+((4!\cdot2)-(3))^{2}$$ $$=-3+(48-3)^{2}$$ $$=-3+2025$$ $$=2022$$

$\endgroup$
0
$\begingroup$

I guess I'll leave in my own answer as well, which is admittedly not as clean as some of the other answers here.

\begin{align}& \quad \,(!((2 + 0!)!) + (2+2)!\times (2 + 0!)) \times (2 + !2)!\\&= (!(3!) + 4! \times 3) \times 3!\\&= (!6 + 24 \times 3) \times 6\\&= (265 + 72) \times 6\\&= 337 \times 6\\&= \boxed{2022}\end{align}

$\endgroup$