User talk:Bubba73

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Help:Archiving a talk page

DeadeyeDave73,"meprabot@gmail"You sure do get to see a lot of American history Bubba73.I sure would've liked to see the guidestones w/ya b4 their demise.Sure would like to know what you think about things!?"><"

I don't have much of an opinion about them. After they were demolished, they said that they were going to rebuild them exactly as before, but I haven't heard any more about it. Bubba73 ^{You talkin' to me?} 21:31, 2 June 2025 (UTC)[reply]

Women in Red World Contest

Hi. We're into the last five days of the Women in Red World Contest. There's a new bonus prize of $200 worth of books of your choice to win for creating the most new women biographies between 0:00 on the 26th and 23:59 on 30th November. If you've been contributing to the contest, thank you for your support, we've produced over 2000 articles. If you haven't contributed yet, we would appreciate you taking the time to add entries to our articles achievements list by the end of the month. Thank you, and if participating, good luck with the finale!

Right to use picture?

Hi

I like to use your picture of the Sherman necktie in our publication "All Scale Rails" We would give you photo credit, but will need to have you sign our release form. Let me know if we could and where I could email the release form?

Our email is:

[email protected]

Thank You

Kevin Domrois

allscalerails.com

Pascal Rounding

"nearest" integer could be misconstrued when dealing with positive and negative numbers. I believe the term "Rounding half away from zero" is necessary and sufficient as described in the Wikipedia definition:

https://en.wikipedia.org/wiki/Rounding

The formal definition in ISO/IEC 7185:1990(E) is:

6.6.6.3 Transfer functions

round(x)

From the expression x that shall be of real-type, this function shall return a result of integer-type. If x is positive or zero, round(x) shall be equivalent to trunc(x+0.5); otherwise, round(x) shall be equivalent to trunc(x-0.5). It shall be an error if such a value does not exist.

Examples: round(3.5) yields 4 round(-3.5) yields -4 Chris Burrows (talk) 23:51, 22 April 2025 (UTC)[reply]

Thanks. The problem is with negative numbers. Bubba73 ^{You talkin' to me?} 01:57, 23 April 2025 (UTC)[reply]

Magic Squares

Hi Bubba73,

I know you all don't want to put this on the Magic Squares Wiki Page because it is original research. I still think it is sort of gray area that falls in the routine math exception, but, whatever... All I did was solve them using standard routine algebra, and it is very easy to doublecheck whether the results are correct or not. In any case, I would still like someone interested in Magic Squares to at least have a look. I notice that we have similar backgrounds (math & physics). John

Method of algebra

Magic squares can be generally solved by assigning variables to each cell and doing the algebra.

1x1 Magic Square, with Magic Constant=1M

This square is trivally magic, since it is a single number. It can be a "normal" magic square by setting M=1, so that the magic constant = (1+1)/2*(1/1) = 1.

M

2x2 Magic Square, with Magic Constant=2M

This square is not really considered magic, since all entries are the same number.

M	M
M	M

3x3 Magic Square, with Magic Constant=3M

This square is the first non-trivial magic square. It can be a "normal" magic square by setting M=5, so that the magic constant = (1+9)/2*(9/3) = 15, and setting the other variables appropriately, ex. a=4, b=9.

a	b	3M-a-b
4M-2a-b	M	-2M+2a+b
-M+a+b	2M-b	2M-a

4x4 Magic Square, with Magic Constant=4M

This particular algebraic solution is also panmagic and most perfect. It can be a "normal" magic square by setting M=8.5, so that the magic constant = (1+16)/2*(16/4) = 34, and setting the other variables appropriately, ex. a=4, b=14, e=5, g=8.

a	b	a+e-g	4M-2a-b-e+g
e	4M-a-b-e	g	a+b-g
2M-a-e+g	-2M+2a+b+e-g	2M-a	2M-b
2M-g	2M-a-b+g	2M-e	-2M+a+b+e

5x5 Magic Square, with Magic Constant=5M

This particular algebraic solution is also panmagic. It can be a "normal" magic square by setting M=13, so that the magic constant = (1+25)/2*(25/5) = 65, and setting the other variables appropriately, ex. a=20,b=8,c=21,d=14,g=4.

a	b	c	d	5M-a-b-c-d
3M-a-b	g	3M-b-d	M+b-c+d-g	-2M+a+b+c
4M-a-c-g	-M+a+b+d-g	M	3M-a-b-d+g	-2M+a+c+g
M-d+g	M-b+c-d+g	-M+b+d	2M-g	2M-c+d-g
-3M+a+b+c+d	5M-a-b-c-g	2M-c	-M+a+c-d+g	2M-a

6x6 Magic Square, with Magic Constant=6M

This particular algebraic solution is also panmagic, which means that it cannot be "normal" (having consequtive number from 1..36) A "normal" 6x6 magic square has a magic constant = (1+36)/2*(36/6) = 111.

a	b	-a-b+3M	2M-s	2M-t	-M+s+t
g	h	-g-h+3M	j	2M-z	-j+M+z
-a-g+3M	-b-h+3M	a+b+g+h-3M	-j+M+s	-M+t+z	j+3M-s-t-z
s	t	3M-s-t	-a+2M	-b+2M	a+b-M
-j+2M	z	j+M-z	-g+2M	-h+2M	g+h-M
j+M-s	3M-t-z	-j-M+s+t+z	a+g-M	b+h-M	-a-b-g-h+5M

12.188.175.154 (talk) 14:47, 19 May 2025 (UTC)[reply]