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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

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

19 Answers 19

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


update:

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

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    $\begingroup$ @Angelo Thanks; I'm just so excited about the prospect of living through a square year when $2025 = 45^2$ arrives! $\endgroup$ Dec 31, 2022 at 7:46
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    $\begingroup$ @BenjaminDickman, damn so you weren't around when it was $1936$? $\endgroup$
    – user1059904
    Dec 31, 2022 at 7:59
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$$2023 = ((( 9\times 8\times 7) +2 ) \times 4 ) – ( 5+3) + ( 6 + 1 + 0)$$

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

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

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  • $\begingroup$ No need for subtraction! 2023 = 1 + 2 + (3 + 7) x (4 x (5 x 8 + 9) + 6) $\endgroup$
    – mathlander
    Dec 31, 2023 at 20:01
  • $\begingroup$ @mathlander great! btw, check my update! $\endgroup$ Jan 1 at 4:58
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$\text{2022}$+$\text{1}$=$\text{2023}$

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    $\begingroup$ @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. $\endgroup$ Dec 31, 2022 at 20:25
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    $\begingroup$ @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. $\endgroup$ Jan 1, 2023 at 11:31
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    $\begingroup$ Wow, it's simple to write .............+1 $\endgroup$
    – TShiong
    Jan 8, 2023 at 23:01
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A palindromic hexadecimal number:

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

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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

$$2023 = MMXXIII$$

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

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

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$2023$ 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. $$ 2023=11111100111. $$

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

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

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

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

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

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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 $\color{blue}{1}$.

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

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

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$45^{2}-2=2023$

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

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

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

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

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.

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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 $3.42 \times 10^{-28}$.

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

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

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

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Something easy :

$$2023+3202=5225$$

Wich is palindromic too .

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    $\begingroup$ 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. $\endgroup$
    – Peter
    Jan 5, 2023 at 14:34

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