mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2022-03-13, 03:41   #12
tuckerkao
 
"Tucker Kao"
Jan 2020
Head Base M168202123

2·389 Posts
Default

Quote:
Originally Posted by VBCurtis View Post
As usual, you reply to a post with incorrect mathematical content.

Or, please tell us how the last digit allows you to identify whether a number is divisible by 3.

Here are numbers not divisible by 3: 10, 11, 22, 23, 34, 35, 46, 47, 58, 59. There's all ten last digits. So, explain.
Sorry I was thinking the math dozenally which the trine numbers always end in 0, 3, 6, 9 and non-trine numbers end in 1, 2, 4, 5, 7, 8, Ӿ, Ɛ.

When the thread title showed up, I immediately thought all the multiples of 6 were ended in the dozenal 0 or 6. 6n+1 end in 1 or 7, 6n+5 end in 5 or Ɛ.

Last fiddled with by tuckerkao on 2022-03-13 at 03:50
tuckerkao is offline   Reply With Quote
Old 2022-03-13, 14:15   #13
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

10111011011112 Posts
Default

If memory serves, the Pari-GP function nextprime(n) merely insures that candidates ≥ n are in one of the residue classes relatively prime to the modulus 210 = 2*3*5*7 before doing a "brutal" BPSW pseudoprime test. (The function is written to make sure it doesn't overlook the primes 2, 3, 5, and 7. It may have other provisions for "small" inputs.) The function returns the first number ≥ n that "passes" the test.

Obviously, this function is not intended for very large inputs.

As to using the binomial theorem to determine 2n - 1 (mod 3), it might be clearer to note that 2 = 3 + (-1), so that 2n = (3 + (-1))n.

In the binomial expansion of (3 + (-1))n, every term except (-1)n is divisible by 3. Thus 2n == (-1)n (mod 3).

Of course, one can skip the binomial theorem altogether and just say, 2 == -1 (mod 3) whence 2n == (-1)n (mod 3).
Dr Sardonicus is offline   Reply With Quote
Old 2022-03-13, 18:18   #14
Dobri
 
"Καλός"
May 2018

5·73 Posts
Default

Quote:
Originally Posted by Dr Sardonicus View Post
Of course, one can skip the binomial theorem altogether and just say, 2 == -1 (mod 3) whence 2n == (-1)n (mod 3).
This general statement could be a bit confusing for the OP if not being a mathematician.
The case considered in their thread is concerned with the quoted positive integer n being an even number and 2n = 3 × Quotient + (-1)n with (-1)n = 1 so that
2n = 3 × Quotient + 1,
2(2n) = 2(3 × Quotient + 1),
2n+1 = 6 × Quotient + 2, and finally
2n+1 - 1 = 6 × Quotient + 1,
where the odd number p in my previous post is p = n + 1.
Dobri is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
How do I calculate chances of finding a prime, while testing multiple exponents? drkirkby Math 4 2021-04-13 12:58
Linux and multiple GPU bgbeuning GPU Computing 52 2016-06-09 05:26
Using multiple PCs numbercruncher Information & Answers 18 2014-04-17 00:17
Are multiple gpu's necessary for CUDA... WraithX GPU Computing 16 2012-03-22 10:44
Multiple systems/multiple CPUs. Best configuration? BillW Software 1 2003-01-21 20:11

All times are UTC. The time now is 18:08.


Thu Oct 6 18:08:22 UTC 2022 up 49 days, 15:36, 0 users, load averages: 0.85, 1.39, 1.54

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔