The 6n Elimination Table - [ last modified: 2007-04-13 | see below ] In 2006, before I was forced by IE to move all I had there offline, I had a major site at ThirdAge, a significant portion of which was dedication to what I named: Number Theory Investigations. That IE made me remove that site/ may have been good. I did discover this year (2007) that at least one person copied my NTI pages there without my knowledge. I hope that person and whoever else became aware of my investigations/ learned some things of value from my endeavors. Being the heuristic mathematician I am, I often present my thoughts as they are occurring, and consequently am liable to change and/or discard those thoughts. - Due to what I began to see this morning, I am next going to place here part of page 12 from my Number Theory Investigations, but first some explanations. # pown = positive odd whole number pewn = positive even whole number E T = Elimination Table tpo = termposition fd = final digit # # 2006-04-19 Wednesday - A pown can be doubled to begin the creation of an Elimination Table which will effectively eliminate every multiple of that pown/ except that pown x 1. Thusly: If c = 6, 6 then can be the pewn constant for an Elimination Table in which neither a (6n - 1) term nor a (6n + 1) term could ever be a multiple of "3". # # 2007-04-12 comment: The "except that pown x 1" is wrong. Onward. Here is the beginning of a 6n Elimination Table: (note: the first "n" = "1")
Column 1 Column 2 Column 3 - (6n - 1) 6n (6n + 1) tpo1 5 6 7 - tpo2 11 12 13 - tpo3 17 18 19 - tpo4 23 24 25 - tpo5 29 30 31 - end of cycle 1 - tpo6 35 36_____________37 - tpo7 41 42_____________43 - tpo8 47 48_____________49 - tpo9 53 54_____________55 - tpo10 59 60_____________61 - end of cycle 2 - -
This is a what-I-started-to-see pause. Knowing that a number whose fd = 5 is a number divisible by 5 is easy, but knowing that a number (whatever its fd is) is divisible by 7 is not easy unless a way can be found which will make it easy. I think I may have found such a way. Important primary point: The power inherent in a pown (let 7 be our example) regarding those numbers it divides into equally/ does not begin until the square of that pown has been reached. In other words, while a given pown does divide equally into a number or numbers which are less than the value of that pown's square, it is the lesser number it is partnered with which rules. Two divides equally into every greater pewn. Important secondary point: In the table being used here, no (6n - 1) or (6n + 1) pown is divisible by 3. Okay, what about 7? 1 x 7 = 7. 3 x 7 = 21. 5 x 7 = 35. 7 x 7 = 49. 21 is not in this table. 35 is, but 5 rules it. 7 is, but 1 rules. As to 49, this: 48/6 = 8. 48 + 1 = 49. 6 + 8 = 14. 7 + 7 = 14. We are forced to skip 63 because it is divisible by 3, and neither 62 nor 64 is divisible by 6. Let's go to 77. 78/6 = 13. 78 - 1 = 77. 6 + 13 = 19. 7 + 11 = 18. What's with this? Let's go to 91. 90/6 = 15. 90 + 1 = 01. 91/7 = 13. 6 + 15 = 21. 7 + 13 = 20. What's with this? Let's go to 119. 120/6 = 20. 119/7 = 17. 6 + 20 = 26. 7 + 17 = 24. Now what's up? Would you like to venture a guess about the next divisible by 7 pown? Let's try 35. 36/6 = 6. 6 + 6 = 12. 5 + 7 = 12. I think there is a pairs thing going on here, wherein 36 and 48 are pair one; 78 and 90 are pair two; 120 and 132 are pair three. Also, 35 and 49 are pair one; 77 and 91 are pair two; 119 and 133 are pair three. I'll be back later. (It is 5:05 PM.) - 6:36 PM Among other things while I was away, I've been doing some calculating. Two worthy results came of it. The first is that the (6n - 1) and (6n + 1) numbers are multiplicands as well as being multipliers. This is simple enough. The other is: If you stay with the 6n E T so as to avoid multiples of "3", 6 times any (6n - 1) number or 6 times any (6n+ 1) number will give you tpo1 of a simple arithmetic sequence based on that tpo/ from which multiples of the pown you have chosen can bedetermined. 6 x 7 = 42. 42 -7 = 35. 42 + 7 = 49. 7's number withinthe parameters of the 6n E T is 42. 42 is tpo1. 84 is tpo2. 126 is tpo3. 168 is tpo4. This means/ 42 times any positive whole number = a product from which two numbers divisible by "7" can be derived. 41 x 42 = 1722. 1722 - 7 = 1715. 1715/7 = 245. 1722 + 7 = 1729. 1729/7 = 247. | 66 is the number for "11"; 78 is the number for "13". - Normally that pown which is two less than the square of a pown is a prime number. There are exceptions. "119" is an exception. 11 x 11 = 121, but 7 x 17 = 119. We already know 6 x 20 = 120. Point of possible interest. 6 + 20 = 26; 7 + 17 = 24; 11 + 11 = 22. In just looking at "119" it does not seem there is any clue pertaining to its divisibility by "7". Yes, 42 x 3 = 126, and 126 - 7 = 119; but what would lead one to even try that? Regarding "3": Any pwn that digit sums to 3, 6, or 9 is divisible by "3". Just a something you likely know. "119" digit sums to 11, or 2. 17's 6n E T parameter number is "102", and 102 + 17 = 119. Isn't this just sweet petunias. 126 - 102 = 24, and 24 x 5 = 120--for whatever that's worth. I wonder where the next 7n = a number which is two less than the square of a pown? - 9:14 PM Something I had forgotten about which solves the "119" question quicker than a blink. It involves "10". 10 times any pwn or any multiple of any pwn will encounter a greater pwn that is divisible by the base pwn. 7 x 7 = 49. 7 x 10 = 70. 49 + 70 = 119. This also proves why "17" has to be the multiplicand: 7 + 10 = 17. I know this isn't entirely satisfactory, but it is a light. # 2007-04-13 8:13 AM - One could call it the "times ten rule". Here is run-through using "7": 1 x 7 = 7; 7 x 10 = 70; 7 + 70 = 77; 1 + 10 = 11. 2 x 7 = 14; 14 + 70 = 84; 2 + 10 = 12; 7 x 12 = 84. 3 x 7 = 21; 21 + 70 = 91; 3 + 10 = 13; 7 x 13 = 91. 4 x 7 = 28; 28 + 70 = 98; 4 + 10 = 14; 7 x 14 = 98. 5 x 7 = 35; 35 + 70 = 105; 5 + 10 =15; 7 x 15 = 105. 6 x 7 = 42; 42 + 70 = 112; 6 + 10 = 16; 7 x 16 = 112. 7 x 7 = 49; 49 + 70 = 119; 7 + 10 = 17; 7 x 17 = 119. 8 x 7 = 56; 56 + 70 = 126; 8 + 10 = 18; 7 x 18 = 126. 9 x 7 = 63; 63 + 70 = 133; 9 + 10 = 19; 7 x 19 = 133. I ignored "0" but I ought to have begun with it. 0 x 7 = 0; 0 + 70 = 70; 0 + 10 = 10; 7x 10 = 70. The finaldigit (fd) cycle for "7" is: 0, 7, 4, 1, 8, 5, 2, 9, 6, 3 This fd cycle is a retrograde cycle because "7" is > "5". Without the leading zero, the fd following the "3" fd would be"0", but that "0" would be the zero in "10". I opted for the leading "0" in part because I did not want to enter what I call "the one set", that set which comprises the teens. I wanted to stay within what I call "the zero set", that set which comprises the single-digit positive whole numbers (pwn's). # Brian A. J. Salchert
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