# Interesting ways to write 2023 [closed]

The year 2023 is near and today I found this nice way to write that number:

$$\displaystyle\color{blue}{\pi}\left(\frac{(\pi !)!-\lceil\pi\rceil\pi !}{\pi^{\sqrt{\pi}}-\pi !}\right)+\lfloor\pi\rfloor=2023$$

where $$\color{blue}{\pi}$$ is the counting function of prime numbers.

My question is, do you know any other interesting way to write 2023? By the way, happy new year everyone

• How do you define factorial of $\pi$?
– RFZ
Dec 31, 2022 at 6:05
• @ZFR $\Gamma(1+\pi)$ Dec 31, 2022 at 6:12
• Perhaps it is worth mentioning that we have $$2023\mid (-20!+23!)$$ Dec 31, 2022 at 15:56
• $2023 = 7 \times p_{7}^{2}$ where $p_{7}$ is the $7$th prime.
– Vue
Jan 2, 2023 at 4:43
• $\int_{0}^{\infty}\left(\sec\left(\frac{\pi x}{2}\right)\frac{\sin\left(2023\pi x\right)}{\pi x}\right)^{2}dx=2023$ Jun 1, 2023 at 20:40

## 19 Answers

$$(2+0+2+3)(2^2 + 0^2 + 2^2 + 3^2)^2 = 2023$$

update:

$$(20+24) + (20+24)(20+24) + (20+24) = 2024$$

• @Angelo Thanks; I'm just so excited about the prospect of living through a square year when $2025 = 45^2$ arrives! Dec 31, 2022 at 7:46
• @BenjaminDickman, damn so you weren't around when it was $1936$?
– user1059904
Dec 31, 2022 at 7:59

$$2023 = ((( 9\times 8\times 7) +2 ) \times 4 ) – ( 5+3) + ( 6 + 1 + 0)$$

(i.e., using all digits exactly once)

Update :- $$2024 = 22^2 + 20^2 +18^2 +16^2 +14^2 +12^2 +10^2 +8^2 +6^2 +4^2 +2^2$$

• No need for subtraction! 2023 = 1 + 2 + (3 + 7) x (4 x (5 x 8 + 9) + 6) Dec 31, 2023 at 20:01
• @mathlander great! btw, check my update! Jan 1 at 4:58

$$\text{2022}$$+$$\text{1}$$=$$\text{2023}$$

• @Peter, it's the way that lends itself most easily to generalization! It also has some significance to those to whom 2022 wasn't a great year. Dec 31, 2022 at 20:25
• @Peter I think the point is that there is absolutely nothing special apriori about a given year and coming up with ways to write it, while a fun game mathematicians might sometimes enjoy, isn't very mathematical and thus this question isn't very mathematical. Jan 1, 2023 at 11:31
• Wow, it's simple to write .............+1 Jan 8, 2023 at 23:01

A palindromic hexadecimal number:

$$2023_{10} = 7e7_{16}$$

You can write $$2023$$ as the sum of four squares ($$61$$ ways) using any row of the following table.

$$\left( \begin{array}{cccc} 1 & 2 & 13 & 43 \\ 1 & 5 & 29 & 34 \\ 1 & 7 & 23 & 38 \\ 1 & 10 & 31 & 31 \\ 1 & 11 & 26 & 35 \\ 1 & 13 & 22 & 37 \\ 1 & 17 & 17 & 38 \\ 2 & 5 & 25 & 37 \\ 2 & 7 & 11 & 43 \\ 2 & 7 & 17 & 41 \\ 2 & 11 & 23 & 37 \\ 2 & 13 & 13 & 41 \\ 2 & 13 & 25 & 35 \\ 2 & 17 & 19 & 37 \\ 2 & 23 & 23 & 31 \\ 3 & 3 & 18 & 41 \\ 3 & 3 & 22 & 39 \\ 3 & 5 & 15 & 42 \\ 3 & 5 & 30 & 33 \\ 3 & 9 & 13 & 42 \\ 3 & 13 & 18 & 39 \\ 3 & 14 & 27 & 33 \\ 3 & 18 & 27 & 31 \\ 3 & 21 & 22 & 33 \\ 5 & 5 & 23 & 38 \\ 5 & 6 & 21 & 39 \\ 5 & 7 & 10 & 43 \\ 5 & 10 & 23 & 37 \\ 5 & 11 & 14 & 41 \\ 5 & 14 & 29 & 31 \\ 5 & 17 & 22 & 35 \\ 5 & 19 & 26 & 31 \\ 6 & 9 & 15 & 41 \\ 6 & 13 & 27 & 33 \\ 6 & 23 & 27 & 27 \\ 7 & 11 & 22 & 37 \\ 7 & 13 & 19 & 38 \\ 7 & 17 & 23 & 34 \\ 7 & 22 & 23 & 31 \\ 9 & 9 & 30 & 31 \\ 9 & 14 & 15 & 39 \\ 9 & 18 & 23 & 33 \\ 9 & 22 & 27 & 27 \\ 10 & 11 & 11 & 41 \\ 10 & 11 & 29 & 31 \\ 10 & 13 & 23 & 35 \\ 11 & 11 & 25 & 34 \\ 11 & 13 & 17 & 38 \\ 13 & 13 & 23 & 34 \\ 13 & 14 & 17 & 37 \\ 13 & 15 & 27 & 30 \\ 13 & 18 & 21 & 33 \\ 13 & 22 & 23 & 29 \\ 14 & 19 & 25 & 29 \\ 15 & 15 & 22 & 33 \\ 17 & 17 & 17 & 34 \\ 17 & 17 & 22 & 31 \\ 17 & 22 & 25 & 25 \\ 17 & 23 & 23 & 26 \\ 18 & 21 & 23 & 27 \\ 19 & 19 & 25 & 26 \\ \end{array} \right)$$

For example, using the first row, we have

$$2023 = 1^2 + 2^2 + 13^2 + 43^2$$

Another

How many solutions does the following have?

$$20 x_1 +23 x_2 =2023$$

We have

$$(x_1, x_2) = (8, 81), (31, 61), (54, 41), (77, 21), (100, 1)$$

Another

$$2023$$ as the sum of five cubes (I think it is the only one)

$$2023 = 2^3+5^3+6^3+7^3+11^3$$

Another

Write $$2023$$ as the sum of Fibonacci numbers ($$18$$ ways)

$$\begin{array}{l} 1597+377+34+13+2 \\ 1597+377+34+8+5+2 \\ 1597+233+144+34+13+2 \\ 987+610+377+34+13+2 \\ 1597+377+21+13+8+5+2 \\ 1597+233+144+34+8+5+2 \\ 1597+233+89+55+34+13+2 \\ 987+610+377+34+8+5+2 \\ 987+610+233+144+34+13+2 \\ 1597+233+144+21+13+8+5+2 \\ 1597+233+89+55+34+8+5+2 \\ 987+610+377+21+13+8+5+2 \\ 987+610+233+144+34+8+5+2 \\ 987+610+233+89+55+34+13+2 \\ 1597+233+89+55+21+13+8+5+2 \\ 987+610+233+144+21+13+8+5+2 \\ 987+610+233+89+55+34+8+5+2 \\ 987+610+233+89+55+21+13+8+5+2 \\ \end{array}$$

Another

$$2023 = MMXXIII$$

• Do you allow negatives cubes? Otherwise $2023=2^3+(-9)^3+14^3+0^3+0^3$ is the sum of five cubes. Jan 1, 2023 at 20:58
• @DietrichBurde: That would also make a good answer and I would include them! Happy New Year!
– Moo
Jan 1, 2023 at 21:29

$$2023$$ can be written as

$$\text{the year you were born}+\text{how many years old you are}+1$$

This works $$100\%$$ of the time when this calculation is performed at the very beginning of $$2023$$.

• Not if you were born on Jan 1 12:00am right? Jan 2, 2023 at 20:23
• @ShukantPal well you'd have to be born exactly then, which has probability $0$. Jan 3, 2023 at 1:31
• That’s true - and the probability that you do this calculation exactly at the beginning of the year is $0$ as well :-) Jan 4, 2023 at 3:21

You writing $$2023$$ using only $$\pi$$ and this video from Presh Talwalkar inspired me to give this representation of $$2023$$ as an answer:

$$2023=\frac{\ln({\frac{\ln(\pi)}{\ln(a)}})}{\ln(-\cos(\pi)-\cos(\pi))},$$

$$a=\pi^{\frac{1}{b}},b=2^{2023}.$$

This discussion on Puzzling Stack Exchange is worth checking out as well.

$$2023$$ is a sum of four squares (but not of fewer ones), e.g., $$2023=10^2+11^2+11^2+41^2.$$ It is also the sum of three cubes, e.g., $$2023=2^3+(-9)^3+14^3.$$ It is sort of close to $$111111111111$$ in the binary system. $$2023=11111100111.$$

$$2023=\lfloor (45-\frac{1}{45})^2\rfloor$$

$$2023=\Big(\frac{2^3+3^2}{2^{10}3^2}\Big)\phi(2022)\phi(2023)$$

where $$\phi$$ is Euler's totient function.

$$9^3+8^3+7^3+6^3+5^3+4^3+3^3+2^3-1^3=2023$$

$$\begin{array}{} 20 \cdot 2^{3} \quad + \quad 23 \cdot 3^{2^{2}}\\ \end{array}$$

There are total $$\left(\dfrac{3\times3\times2\times1}{2!}\right)=9$$ four-digit significant numbers, formed by permuting the four digits $$\color{red}{2},\color{red}{0}, \color{red}{2}, \color{red}{3}$$ without repetition, which can be arranged in ascending order as follows

$$\begin{pmatrix} \color{red}{2023}\\ 2032\\ 2203\\2230\\2302\\2320\\3022\\3202\\3220 \end{pmatrix}$$ The rank of $$\color{red}{2023}$$ in the ascending order in its table is $$\color{blue}{1}$$.

So close to $$2^{10} + 10^3$$.

$$2023 = 2^{10} + 999$$

$$45^{2}-2=2023$$

$$2^{11}-5^2=2023$$

$$2^{8}+12^{3}+39=2023$$

$$3^6+6^4-2=2023$$

$$3+4\cdot5+40\cdot50=2023$$

Let $$P_{n}$$ denotes the nth prime,then:

$$35+\sum_{n=1}^{33} P_{n} =2023$$

$$-104+\sum_{n=1}^{34} P_{n} =2023$$

I wish great happy new year for all people.

Thanks to Wolfram Alpha

$$2023 \sim \frac A{11^2}$$ where $$A=18468 \binom{\pi }{\pi !}+5548 \binom{\pi !}{\pi }+8724 \binom{\pi !}{\log (\pi )}-19734 \binom{\log (\pi )}{\pi !}-$$ $$15884 \binom{\pi }{\log (\pi )}-4309 \binom{\log (\pi )}{\pi }$$

The difference between rhs and lhs is $$3.42 \times 10^{-28}$$.

Just keep it simple $$\int_0^1 x^{2022} \, \mathrm{d}x = \frac{1}{2023}$$

And I know one more cool integral which is $$\int_0^\infty \frac{\tan^{-1}(2022 x)}{x(x^2+1)} \mathrm{d}x = \frac{\pi}{2} \ln(2023)$$

We have $$\require{cancel} \sum_{n=1}^{2+0+2+3}\left(1+\frac{2.\cancel{0}.2.3}{\cancel{0}}[n+1]n\right)=2023$$

Something easy :

$$2023+3202=5225$$

Wich is palindromic too .

• This is however true for every positive integer with digits not exceeding $4$. As long as their are no carry-overs , adding the "reversal number" to a number must result in a palindrome. Jan 5, 2023 at 14:34