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Make 2023 with math: PowerMod[7, 7, 7!] = 2023. Your turn!

Ed Pegg, Wolfram Research
Posted 3 years ago

What is your one-liner for getting 2023? Comment below!

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POSTED BY: Ed Pegg
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[ (2023)^2 - 2023 ] / [6* (2023)] = 337

New year : 337 (6) + 1 = 2023 or [337 (7)* (2023)] / 2023] - 338 = 2023

Last year : 337 (6) = 2022 or 337(7) - 337 = 2022

Moving along : 337 + 1 = 338

Finding the next primes using the formula above :

[ [2023 * ( 338)* (7) - 2023] / 2023] - 338 = 2027 is prime

[ [2023 ( 338) (7) + 2023] / 2023] - 338 = 2029 is prime

[(2029+338) 2023 - 2023 / (2366)] = 2023

[(2027+338) 2023 -+2023 / (2366)] = 2023

2023 within a solution using the next twin primes

POSTED BY: Richard Andrews
Paul Cleary
Paul Cleary
Posted 3 years ago
POSTED BY: Paul Cleary
Tianyi Hu
Tianyi Hu
Posted 3 years ago
POSTED BY: Tianyi Hu
POSTED BY: Vitaliy Kaurov

Not a formal per se but 2023 is a multiple of 7 whose digits also add to 7. In other words 2023 is the numbers congruent to 7 mod 63.

POSTED BY: Salvatore Mangano

$3^2+7^2+11^2+15^2+19^2+23^2+27^2 = 2023$
POSTED BY: Francisco Javier García Capitán

I prefer to look forward to what 2023 might bring to us! 2023 is a very interesting 3-color totalistic cellular automaton:

ArrayPlot[CellularAutomaton[{2023,{3,1}},{{1},0},500],ColorRules->{0->White,1->Gray,2->Red}]
POSTED BY: Keith Stendall

There are some good answers in comments on Reddit Math: https://wolfr.am/1a2XCVsb0

Also there is this link with the properties: https://www.numbersaplenty.com/2023

POSTED BY: Vitaliy Kaurov

With this: https://www.wolframalpha.com/input?i=prime+factorization+of+2023 I thought it would be interesting to see what could be one in base 7 to get to 2023.

https://www.wolframalpha.com/input?i=decimal+form+of+7*23_7%5E2

Or in WL: 7 7^^23^2
POSTED BY: Dorothy Evans

Going into algebraic number theory

-NumberFieldDiscriminant[x /. Solve[x^3 == x^2 + 6 x + 5][[1]]]
POSTED BY: Ed Pegg

= (11/x+1 )^-1 which = 184

184 * 11 - 1 = 2023

POSTED BY: Richard Andrews
POSTED BY: Vitaliy Kaurov

A few more 

LinearRecurrence[{3, -3, 1}, {7, 34, 79}, 15] // Last 
LinearRecurrence[{0, 1, 1, 2}, {0, 2, 3, 10}, 17] // Last 
LinearRecurrence[{4, 0, -5, -2}, {0, 2, 10, 41}, 7] // Last  
N[FromContinuedFraction[Flatten[{44, Table[{1, 43, 1, 88}, 5]}]]^2, 40]
POSTED BY: Ed Pegg
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