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Old 2022-03-18, 22:28   #364
Dr Sardonicus
 
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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...
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Old 2022-03-18, 23:27   #365
tuckerkao
 
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Quote:
Originally Posted by Uncwilly View Post
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 View Post
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
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Old 2022-03-19, 01:38   #366
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Quote:
Originally Posted by tuckerkao View Post
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.
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Old 2022-03-19, 02:07   #367
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Quote:
Originally Posted by slandrum View Post
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
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Old 2022-03-19, 04:48   #368
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Quote:
Originally Posted by tuckerkao View Post
<snip>
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?
<snip>
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).
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Old 2022-03-19, 13:24   #369
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Quote:
Originally Posted by Dr Sardonicus View Post
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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