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Actually we can do any prime numbers if we have first 25 numbers. And also prime numbers minus 1 using Euler's totient function. And also Fibonacci numbers. And also Lucas numbers. And blahuhuhuh
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
⌊sin(3∘)⌋=0
sgn(3)=1 (sign function)
ϕ(3)=2 (Euler's totient phi function)
3
σ(3)=4 (divisor sum)
Σ(3)=5 (prime sum)
3!=6
⌊csc(3)⌋=7
⌈csc(3)⌉=8
⌊antilog(tan(3))⌋=9 (antilog(x)=10x)
Σ(Σ(3))=10
pΣ(3)=11 (n-th prime number)
σ(3!)=12
F⌈csc(3)⌉=13 (Fibonacci function F1=0,F2=1,Fn+2=Fn+Fn+1)
σ(F⌈csc(3)⌉)=14
σ(⌈csc(3)⌉)=15
ϕ(s(σ(σ(σ(3)))))=16
s(σ(σ(σ(3))))=17 (aliquot sum s(n)=σ(n)−n)
σ(s(σ(σ(σ(3)))))=18
p⌈csc(3)⌉=19
σ(p⌈csc(3)⌉)=20
⌊tan(tan(cos(sin(3∘))))⌋=21
⌈tan(tan(cos(sin(3∘))))⌉=22
p⌊antilog(tan(3))⌋=23
(Σ(3))!=24
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Actually we can do any prime numbers if we have first 25 numbers. And also prime numbers minus 1 using Euler's totient function. And also Fibonacci numbers. And also Lucas numbers. And blahuhuhuh
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Well, dxd(3)=0.
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oops yeah...its 0 mis-typo...
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I've commented but you've created...
so credits goes to Kenny Lau and Vinay Sipani
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Ty... But the credit must go to you..