phi(phi(floor(exp(3!))))=96

This note's credits goes to a comment by Deepak Gowda in this post.

Generate 0 to 100 with only one digit once: the digit 3. You may use any other functions, like the one in the title.

Please enter as much as you can! :)

#Creativity #Brainteaser

Note by Kenny Lau
6 years, 11 months ago

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Comments

sin(3)=0\lfloor \sin(3^{\circ})\rfloor = 0

sgn(3)=1\text{sgn}(3) = 1 (sign function)

ϕ(3)=2\phi(3) = 2 (Euler's totient phi function)

33

σ(3)=4\sigma(3) = 4 (divisor sum)

Σ(3)=5\Sigma(3) = 5 (prime sum)

3!=63! = 6

csc(3)=7\lfloor \csc(3)\rfloor = 7

csc(3)=8\lceil \csc(3)\rceil = 8

antilog(tan(3))=9\lfloor\text{antilog}(\tan(3)) \rfloor = 9 (antilog(x)=10x\text{antilog}(x) = 10^{x})

Σ(Σ(3))=10\Sigma(\Sigma(3)) = 10

pΣ(3)=11p_{\Sigma(3)} = 11 (n-th prime number)

σ(3!)=12\sigma(3!) = 12

Fcsc(3)=13F_{\lceil \csc(3)\rceil} = 13 (Fibonacci function F1=0,F2=1,Fn+2=Fn+Fn+1F_{1} = 0, F_{2} = 1, F_{n+2} = F_{n} + F_{n+1})

σ(Fcsc(3))=14\sigma(F_{\lceil \csc(3)\rceil}) = 14

σ(csc(3))=15\sigma(\lceil \csc(3)\rceil) = 15

ϕ(s(σ(σ(σ(3)))))=16\phi(s(\sigma(\sigma(\sigma(3))))) = 16

s(σ(σ(σ(3))))=17s(\sigma(\sigma(\sigma(3)))) = 17 (aliquot sum s(n)=σ(n)ns(n) = \sigma(n) - n)

σ(s(σ(σ(σ(3)))))=18\sigma(s(\sigma(\sigma(\sigma(3))))) = 18

pcsc(3)=19p_{\lceil \csc(3)\rceil} = 19

σ(pcsc(3))=20\sigma(p_{\lceil \csc(3)\rceil}) = 20

tan(tan(cos(sin(3))))=21\lfloor \tan(\tan(\cos(\sin(3^{\circ})))) \rfloor = 21

tan(tan(cos(sin(3))))=22\lceil \tan(\tan(\cos(\sin(3^{\circ})))) \rceil = 22

pantilog(tan(3))=23p_{\lfloor\text{antilog}(\tan(3)) \rfloor} = 23

(Σ(3))!=24(\Sigma(3))! = 24

Samuraiwarm Tsunayoshi - 6 years, 11 months ago

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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

Samuraiwarm Tsunayoshi - 6 years, 11 months ago

  • cotan(3°)=20\left\lfloor{cotan(3°)}\right\rfloor=20
  • cotan(3°)=19\left\lceil{cotan(3°)}\right\rceil=19
  • μ(μ(3))=1\mu(\mu(3))=1
  • ddx(3)=0\frac{d}{dx}(3)=0
  • σ(3)=4 where σ(n) is sum of positive divisors.

Vinay Sipani - 6 years, 11 months ago

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Well, ddx(3)=0\frac d{dx}(3)=0.

Kenny Lau - 6 years, 11 months ago

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oops yeah...its 0 mis-typo...

Vinay Sipani - 6 years, 11 months ago

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@Vinay Sipani The floor and ceiling is also switched too.

Samuraiwarm Tsunayoshi - 6 years, 11 months ago

I've commented but you've created...

so credits goes to Kenny Lau and Vinay Sipani

Deepak Gowda - 6 years, 11 months ago

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Ty... But the credit must go to you..

Vinay Sipani - 6 years, 11 months ago
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