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

Posted 1 month ago
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## What is your one-liner for getting 2023? Comment below!

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Posted 1 month ago
 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 1 month ago

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 1 month ago
 Going into algebraic number theory -NumberFieldDiscriminant[x /. Solve[x^3 == x^2 + 6 x + 5][[1]]] 
Posted 1 month ago
 = (11/x+1 )^-1 which = 184 184 * 11 - 1 = 2023
Posted 1 month ago
 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%5E2Or in WL: 7 7^^23^2
Posted 1 month ago
 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 1 month ago
 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 1 month ago
 $3^2+7^2+11^2+15^2+19^2+23^2+27^2 = 2023$
Posted 1 month ago
 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 1 month ago
 Looks like we got another interesting postcard post by Fermat's Library on LinkedIn: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 1 month ago
 From a Tweet by Srinivasa Raghava (he also shared a number of other interesting results of 2023): https://twitter.com/SrinivasR1729/status/1609257145798889472
Posted 1 month 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 1 month ago
 [ (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)] = 20232023 within a solution using the next twin primes
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