# Sixteen Two's to get Two Thousand and Twenty-Two

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?

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

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)

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

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 $$\times$$.

• oh wow thanks i didnt know that I learned something today Commented Aug 3, 2023 at 22:31

Because roots are cool:

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

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

For the follow-up:

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

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