sw00879math--digit-addition-families

26


 Confusion reigns even where there is order.


1, 10, 19, 28, 37, 46, 55, 64, 73, 82   |   91, 100, 109, 118, 127, 136, 145, 154, 163, 172
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2, 11, 20, 29, 38, 47, 56, 65, 74, 83   |   92, 101, 110, 119, 128, 137, 146, 155, 164, 173
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3, 12, 21, 30, 39, 48, 57, 66, 75, 84   |   93, 102, 111, 120, 129, 138, 147, 156, 165, 174
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4, 13, 22, 31, 40, 49, 58, 67, 76, 85   |   94, 103, 112, 121, 130, 139, 148, 157, 166, 175
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5, 14, 23, 32, 41, 50, 59, 68, 77, 86   |   95, 104, 113, 122, 131, 140, 149, 158, 167, 176
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6, 15, 24, 33, 42, 51, 60, 69, 78, 87   |   96, 105, 114, 123, 132, 141, 150, 159, 168, 177
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7, 16, 25, 34, 43, 52, 61, 70, 79, 88   |   97, 106, 115, 124, 133, 142, 151, 160, 169, 178
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8, 17, 26, 35, 44, 53, 62, 71, 80, 89   |   98, 107, 116, 125, 134, 143, 152, 161, 170, 179
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9, 18, 27, 36, 45, 54, 63, 72, 81, 90   |   99, 108, 117, 126, 135, 144, 153, 162, 171, 180


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First:  There cannot be a "0" digit-addition (0da) family.
Second:  In the 9da there is no prime number.
Third:  Each single-digit prime is also the da it is in (2, 3, 5, 7).
Fourth:  The first prime in the 1da is "19".
Fifth:  In the 6da there is no prime number.
Sixth:  The first prime in the 4da is "13".
Seventh:  The first prime in the 8da is "17".
Eighth:  In the 3da the only prime is "3".
Ninth:  Every integer in the 3da, 6da, and 9da is divisible by "3".
Tenth:  Each digit-addition family has a 10-term pattern.
Eleventh:  There are only 9 da patterns 
                   pertaining to powns and their pown multiples,
                   and each of these patterns has 9 terms,
                   and "9" is the 5th term in each pattern.  
                   Why?  The da pattern repeats after every 9 terms
                   because the da for x 1 equals the da for x 19; and,
                   for this same reason, the da pattern repeats after
                   every 18th higher pown.  19's da and 37's da is "1".
                   Also, any integer times 9 equals a 9da integer.

How crazy I am is an unknown; nonetheless, I'm pressing on.

One interest is knowing how a prime relates to its square.  Did a few
calculations and it seems that in a given da family, both the primes and
the squares of those primes, while each in a different da family, are
respectively in the same da family.  Thus, if the prime is in the 1da, its-- 
this is an exception--square is inthe 1da.  
If a prime is in the 2da, its squareis in the 4da.
If a prime is in the 3da (and only one is), its square is in the 9da.  
If a prime is in the 4da, its square is in the 7da (because 4 x 4 = 16).
If a prime is in the 5da, its square is in the 7da (because 5 x 5 = 25).
If a prime is in the 7da, its square is in the 4da.
If a prime is in the 8da, its square is in the 1da.

So?  Sew baritones on your underwear.

Now, 11 (2da) times 17 (8da) equals 187 (7da)
and 29 (2da) times 53 (8da) equals 1537 (7da).
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What about 105, which is 3 x 5 x 7?
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This is heading into combinatorics.
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29 (2da) times 19 (1da) equals 551 (2da),
and 11 times 37 equals 407.
29 times 13 equals 377 (8da).
29 times 23 equals 667 (1da).
29 times 43 equals 1247 (5da).
29 times 47 equals 1363 (4da).
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Conclusion 1:  Every integer product is divisible by "1".
Conclusion 2:  The da of a pown product may be divisible by a pown
                        of any other da.

Had to fix an error in the da chart.  After the correction, I noticed that "11"
and "101" are both 2da, that "13" and "103" are both 4da, that "17" and
"107" are both 8da, and that "19" and "109" are both 1da.  This suggests
that at multiples of 90 there may be other such prime pairs.  In order to
avoid being misunderstood, the pairs here are 11 and 13, 17 and 19,
101 and 103, 107 and 109.  90 up from 105 is 195 (3 x 5 x 13), and
neither 7 nor 11 divide without remainders into 191, 193, 197, or 199.
A prime does not assert its power until it attains its square.  Primes > 13
do not, therefore, matter.  17 x 17 = 289. 






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Brian A. J. Salchert