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

Posted 1 year ago

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

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Posted 7 months ago
 (2 + 0 + 2 + 3) × (22 + 02 + 22 + 32)2 = 2023pretty simple one. 2023 is the year when I started learning math at a more advanced level. Now I have math classes three times a week, and I can say that I like it, but it also makes me stressed simultaneously. From time to time, I even take extra lessons with a tutor because I can't understand something. And pretty often, I use the help of https://edubirdie.com/do-my-math-homework because the tasks can be so hard, and I spend a lot of time on it with no results. But I really love math, and I need good math skills for my future, so I will do my best to improve my skills and knowledge.
Posted 11 months 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
Posted 11 months 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 11 months 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 11 months 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 year 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 year ago
 $3^2+7^2+11^2+15^2+19^2+23^2+27^2 = 2023$
Posted 1 year 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 year 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 year 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 year ago
 Going into algebraic number theory -NumberFieldDiscriminant[x /. Solve[x^3 == x^2 + 6 x + 5][[1]]] 
Posted 11 months ago
 = (11/x+1 )^-1 which = 184 184 * 11 - 1 = 2023
Posted 1 year 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 year 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]