-2
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Get the number 2022 by only using the number 2 a maximum of 16 times. There is a solution.

Allowed:

  • $+$
  • $-$
  • $*$
  • "$()$"
  • "$\frac x y$"
  • $x^y$ Note: When you use $2^2$, it counts as using two 2s, not one 2!
  • $\sqrt x$

Follow-Up: Can you do the same with number 3? But without square-roots, without raising to the power, and without using parenthesis?

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    $\begingroup$ Why does the title say 2023, when the question asks for 2022? $\endgroup$
    – fljx
    Commented Jul 31, 2023 at 11:17
  • $\begingroup$ I edited the question (typo), but forgot to change the title $\endgroup$
    – WOWOW
    Commented Jul 31, 2023 at 11:44
  • $\begingroup$ @fljx Thanks, for pointing it out! $\endgroup$
    – WOWOW
    Commented Jul 31, 2023 at 11:44
  • $\begingroup$ I meant one 2, sorry $\endgroup$
    – WOWOW
    Commented Aug 2, 2023 at 11:03

4 Answers 4

3
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The solution (powers down to top):

$\left({\left({2^2}\right)^2}\right)^2 \times 2\times2\times2 - ({2^2})^2\times2 +2+2+2 \times \frac22 = 2022$

Question 2:

Because this one only allows the four main operations, PEMDAS/BODMAS requires that 2022 must be the sum/difference of multiple powers of three. An economical solution would be $3^7 - 3^4 - 3^4 - 3 = 3\times3\times3\times3\times3\times3\times3 - 3\times3\times3\times3 - 3\times3\times3\times3 - 3 = 2022$

Proof that this is the only solution:

$(2022)_3 = 2202220$ (not doable by adding the first 2s)
$(3^7-2022)_3 = 20010$ (or -$2220$, but it's not doable either, so just stick to $20010$ as in using 9 more threes)

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  • $\begingroup$ Not my wanted solution, but correct nontheless! $\endgroup$
    – WOWOW
    Commented Jul 31, 2023 at 11:55
  • $\begingroup$ View my Solution, and view follow up question! $\endgroup$
    – WOWOW
    Commented Jul 31, 2023 at 11:55
  • $\begingroup$ You have the exact same solution, I posted 30 minutes ago, for the follow-up question! But it is possibly the only solution :P $\endgroup$
    – WOWOW
    Commented Jul 31, 2023 at 12:41
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    $\begingroup$ The first solution uses division, which is not in the list of allowed operations. You can replace the x2/2 with +2-2 $\endgroup$
    – fljx
    Commented Jul 31, 2023 at 15:08
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    $\begingroup$ @Nautilus After fljx' warnings. $\endgroup$
    – z100
    Commented Jul 31, 2023 at 17:00
6
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Solution using only 3 operations and brackets because simple is nice:

$(2 \times 2 \times 2 + 2) ^ 2 \times (2 \times 2 \times 2 + 2) \times 2 + (2 \times (2 \times 2 \times 2 + 2)) + 2 = 2022$

How!?!

$(2\times2\times2+2) = 10 \implies (2\times2\times2+2)^2 = 100 \implies \left[(2\times2\times2+2)\times(2\times2\times2+2)^2\right] = 1000 \implies \left[(2\times2\times2+2)\times(2\times2\times2+2)^2\right]\times2 = 2000\text{ }\&\text{ }[(2 \times 2 \times 2 + 2) \times 2 + (2 \times (2 \times 2 \times 2 + 2))]=20 \therefore [(2 \times 2 \times 2 + 2) ^ 2 \times (2 \times 2 \times 2 + 2) \times 2 + (2 \times (2 \times 2 \times 2 + 2)) + 2] = 2022$


Totally unrelated side note

I notice people using the letter x to represent multiplication, but you can actually use \times within MathJax, which results in $\times$.

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  • $\begingroup$ oh wow thanks i didnt know that I learned something today $\endgroup$ Commented Aug 3, 2023 at 22:31
4
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Because roots are cool:

$ [ 2 \sqrt(\sqrt(2)) \times (\sqrt(\sqrt(\sqrt(2))) ] ^{2\times2\times2} -[2\times2-\sqrt(2)]^2 - [2 +2\sqrt(2)]^2 +2^2 = 2022$

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-1
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My answer would have been:

$\left(\left(2^2\right)^2\right)^2 \times \left(2^2 \times 2 \right) - (2^2)^2 - 2^2 \times 2 - 2 = 2022$

 

For the follow-up:

$3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 - 3 \times 3 \times 3 \times 3 - 3 \times 3 \times 3 \times 3 - 3 = 2022$

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    $\begingroup$ Your first answer uses division, which is not in the list of allowed operations $\endgroup$
    – fljx
    Commented Jul 31, 2023 at 15:09
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    $\begingroup$ It is in the list $\endgroup$
    – WOWOW
    Commented Jul 31, 2023 at 15:43
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    $\begingroup$ It's there now. It wasn't when I made that comment. $\endgroup$
    – fljx
    Commented Jul 31, 2023 at 16:59
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    $\begingroup$ Agree with @fljx, if you're trying to stretch the rules, please don't. If it was an honest mistake though, that's fine. $\endgroup$ Commented Jul 31, 2023 at 17:08
  • $\begingroup$ It was an honest mistake $\endgroup$
    – WOWOW
    Commented Aug 1, 2023 at 7:26

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