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Today is Groundhog Day, the American permutation of Candlemas. [livejournal.com profile] fanw had a fine journal entry explaining that we celebrate one day late, but I'm cool with the difference. I think of it as rounding up from half a baker's dozen.

By the way, I was uncomfortable with the existence of the number seven when I was a kid. There are no tricks for figuring out a number's divisibility by seven. Seven was everywhere, such as seven days of the week and lots of sevens in Catholic school. I could only tell whether a number was divisible by seven by doing the long division and looking for permutations of seven's remainder cycle of 142857. In other words:
 1/7 is 0.142857...,
 2/7 is 0.285714...,
 3/7 is 0.428571... 
...and you're seeing the loop by now.
Candlemas is midwinter's day. If you get good weather on the pivot of winter, you're seeing the halftime show of a full-length winter season. In contrast, crummy weather (at least an overcast day, such as the one Boston is having) means the winter season is finishing. I hear Jefferson County of Pennsylvania got the same notice. Huzzah!

Oh, and my buddy Sean made an excellent appearance on television by being willfully absurdist. Yay! Kibo and I were so proud of both Sean and Peter outside the courthouse. Can't wait for the beat-the-rap party!

Okay, back to work for me. I had one of those moments where I got a pain in my chest and it was hurting and I was scared and then I thought "heartburn?" and I sipped some water and bam! All gone. It'll be a decent day.

-rawkin' onward, Dante

Date: 2007-02-02 06:10 pm (UTC)
From: [identity profile] dobrovolets.livejournal.com
There are no tricks for figuring out a number's divisibility by seven.

Yes there are. Start with the right-most digit. Multiply it by two. Subtract that from the remaining digits. Keep doing that with the resulting numbers until you get something whose absolute value is less than or equal to seven. If the result is 7, 0 or -7, then the number you started with is divisible by seven. It's a bit laborious, but a bit faster than brute force useful if you're in a math competition where speed counts.

Date: 2007-02-02 07:03 pm (UTC)
From: [identity profile] pseydtonne.livejournal.com
Oh man, you may be the Kwisatch Haderach of divisibility! I've wanted that answer for DECADES! I'm just having a hard time wrapping my head around the steps. Would you mind doing a demo using a number without a remainder and one with?

Date: 2007-02-02 07:30 pm (UTC)
From: [identity profile] dobrovolets.livejournal.com
The trick to doing it fast is knowing that every end result on the way down is going to be divisible by seven. So once you hit a result that you know for certain either is or is not divisible, you stop.

Example 1: 195,620
0x2=0. 19,562-0=19,562.
2x2=4. 1,956-4=1,952.
2x2=4. 195-4=191.
1x2=2. 19-2=17 (and now it should be obvious, but one more step:)
7x2=14. 1-14=-13. Not divisible by seven.

Example 2: 195,622.
2x2=4. 19,562-4=19,558.
8x2=16. 1,955-16=1,939.
9x2=18. 193-18=175.
5x2=10. 17-10=7. It is divisible by seven.

This is the kind of silliness one learns trying to become middle school math champion of the state of Florida.

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