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 2022-03-18, 22:28 #364 Dr Sardonicus     Feb 2017 Nowhere 22·1,459 Posts Not one, but two correct answers to my exercise! Well, at least a little bit of good mathematical thinking came of the latest shambles of a prediction for M52...
2022-03-18, 23:27   #365
tuckerkao

"Tucker Kao"
Jan 2020
Head Base M168202123

74410 Posts

Quote:
 Originally Posted by Uncwilly This is the list of currently active guesses (those stricken through were previous guesses that have been replaced under the rules).
Just checked the quoted post above, I'll use up the final guess on M198626023 with the date of June 23rd, 2022. "Final Answer" like in the "Who Wants to be the Millionaire Show", so UncWilly doesn't have to ask me again.

Quote:
 Originally Posted by slandrum In an even base, if the last digit is even, then the number is even. In an odd base, if the sum of the digits is even, then the number is even. In any base b, the sum of the digits of the number is congruent to the number mod (b-1)
In a trine base, if the last digit is trine(divisible by 3), then the number is trine.
In a (trine + 1) base, if the sum of the digits is trine, then the number is trine.

What's the rule for (trine + 2) base? (3, 6, 118, 148, 178, 228, 258, 308)

Last fiddled with by tuckerkao on 2022-03-18 at 23:56

2022-03-19, 01:38   #366
slandrum

Jan 2021
California

3·11·13 Posts

Quote:
 Originally Posted by tuckerkao Just checked the quoted post above, I'll use up the final guess on M198626023 with the date of June 23rd, 2022. "Final Answer" like in the "Who Wants to be the Millionaire Show", so UncWilly doesn't have to ask me again. In a trine base, if the last digit is trine(divisible by 3), then the number is trine. In a (trine + 1) base, if the sum of the digits is trine, then the number is trine. What's the rule for (trine + 2) base? (3, 6, 118, 148, 178, 228, 258, 308)
There's no one rule, there's a lot of ways to do it.

A simple way is to break the number into two digit pairs, if there's an odd number of digits, the first digit is by itself. Sum those numbers up, and the result will be the same mod 3 as the original number. You can repeat the process until you have a 2 digit number, then add twice the first digit to the 2nd digit, and that will also be the same mod 3 as the original number. Basically you are looking at the number in the base b2, b2 will be congruent to 1 mod 3.

2022-03-19, 02:07   #367
tuckerkao

"Tucker Kao"
Jan 2020
Head Base M168202123

23×3×31 Posts

Quote:
 Originally Posted by slandrum There's no one rule, there's a lot of ways to do it. A simple way is to break the number into two digit pairs, if there's an odd number of digits, the first digit is by itself. Sum those numbers up, and the result will be the same mod 3 as the original number. You can repeat the process until you have a 2 digit number, then add twice the first digit to the 2nd digit, and that will also be the same mod 3 as the original number. Basically you are looking at the number in the base b2, b2 will be congruent to 1 mod 3.
I think it's called the "Alternating sums of the numerical blocks of size 1", must go from right to left if odd digits total, given 118 or 11 of any other (trine + 2) base have the trine character.

Dozenal divisibility of 5: Alternating sums of the numerical blocks of size 2. [Dozenal]25 * 5 = 101
Dozenal divisibility of 7: Alternating sums of the numerical blocks of size 3. [Dozenal]187 * 7 = 1001

Last fiddled with by tuckerkao on 2022-03-19 at 02:19

2022-03-19, 04:48   #368
Dr Sardonicus

Feb 2017
Nowhere

22·1,459 Posts

Quote:
 Originally Posted by tuckerkao In a trine base, if the last digit is trine(divisible by 3), then the number is trine. In a (trine + 1) base, if the sum of the digits is trine, then the number is trine. What's the rule for (trine + 2) base?
Nil sapientiae odiosius obscuritate nimia.

There should be a Forum rule against deliberately using gratuitously obscure lingo like "trine" when standard terminology (e.g. divisible by 3, congruent to 0 (mod 3)) exists.

Note that b^k == (-1)^k mod (b+1). Let dk be the bk base-b digit of n. Then

n == d0 - d1 + ... (mod b+1),

the alternating sum of base-b digits starting with the units digit.

For the decimal base 10 = ten, the alternating digit sum determines whether n is divisible by 11.

If b == 2 (mod 3) then 3|(b+1), so this alternating digit sum is congruent to n (mod 3).

2022-03-19, 13:24   #369
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

659710 Posts

Quote:
 Originally Posted by Dr Sardonicus There should be a Forum rule against deliberately using gratuitously obscure lingo like "trine" when standard terminology (e.g. divisible by 3, congruent to 0 (mod 3)) exists.
Added guidance to https://www.mersenneforum.org/showpo...64&postcount=2, linking back to the quoted post.

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