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[CHALLENGE] MAKE 2023 with MATH! PowerMod[7, 7, 7!] = 2023. Your turn!

Ed Pegg, Wolfram Research
Posted 2 years ago

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

enter image description here
POSTED BY: Ed Pegg
13 Replies
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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

Also can be checked in Wolfram|Alpha: https://www.wolframalpha.com/input?i=7%5E7+mod+7%21

BTW - Cool idea! Here is something very simple:

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

In[]:= 2^11-5^2
Out[]= 2023

And thanks to John Conway's discovery of 1988 that $e^{\pi }-\pi \approx 20$ we have

$$\text{Round}\left[101 e^{\pi }-100 \pi \right]= 2023$$

In[]:= Round[101*E^Pi-100*Pi]
Out[]= 2023

I am trying to find some exact ones based on Golden Ratio, Pi, E, other irrational and transcendental numbers. Similar to Fibonacci sequence formula, but much neater than this (I hope someone will figure this out):

$$\frac{\phi ^{18}-(1-\phi) ^{18}}{\sqrt{5}}-3 \cdot 11 \cdot 17=2023$$

In[]:= FullSimplify[(GoldenRatio^18-(1-GoldenRatio)^18)/Sqrt[5]-3 11 17]
Out[]= 2023
POSTED BY: Vitaliy Kaurov

Going into algebraic number theory

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

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

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

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

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

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

Looks like we got another interesting postcard post by Fermat's Library on LinkedIn:

https://www.linkedin.com/feed/update/urn:li:activity:7014999420580597760

enter image description here

which is in Wolfram:

In[]:= (10+(9+8*7)*6)*5+4*3*2-1
Out[]= 2023

I wonder is it easy to find such combination? How many numbers we can get out of specific range using a specific set of operations? And how many combinations we have for a single number on average?
POSTED BY: Vitaliy Kaurov
Tianyi Hu
Tianyi Hu
Posted 2 years ago

From a Tweet by Srinivasa Raghava (he also shared a number of other interesting results of 2023):

https://twitter.com/SrinivasR1729/status/1609257145798889472

enter image description here
POSTED BY: Tianyi Hu
Paul Cleary
Paul Cleary
Posted 2 years ago

If we allow concatenation.

1*2*3*456+7-8*9*10
1+2*34*5*6-7+8-9-10
1/2*345*6+78+910
12+3/4*5*67*8-9+10
12/34*5678+9+10
12+345*6-78+9+10
10-9*8-76+5*432+1
10-9*87+65*43+2-1
10*98-7*6+543*2-1
10*987/6+54/3*21
POSTED BY: Paul Cleary

[ (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

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

184 * 11 - 1 = 2023

POSTED BY: Richard Andrews
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