tag:blogger.com,1999:blog-264226589944705290.post964455542558024676..comments2022-08-07T01:05:01.413-07:00Comments on God Plays Dice: Divisibility by 7Michael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-264226589944705290.post-5096732352356893692010-03-16T23:49:40.943-07:002010-03-16T23:49:40.943-07:00Do You interesting of [b]Viagra 100mg dosage[/b]? ...Do You interesting of [b]Viagra 100mg dosage[/b]? 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This is just because, for example, 10 mod 7 = 3, so multiply the ten's digit by three. I'm also considering, for example 1000 mod 7 = -1, rather than 1000 mod 7 = 6, because it's easier to do the addition problem that way.<BR/><BR/>You can do this even faster if you take everything mod seven, so for example to test <BR/><BR/>4728402<BR/><BR/>you can do<BR/><BR/>2 + 0 + 1 - 1 - 6 - 0 + 4 = 0<BR/><BR/>which is a multiple of seven. As you go from digit to digit, you keep a running tally of your current total, which is always somewhere from minus 7 to 7 (because if it goes outside you just take mod 7). You can do an arbitrarily long number in your head easily, as long as you keep track of the<BR/><BR/>1<BR/>3<BR/>2<BR/>-1<BR/>-3<BR/>-2<BR/><BR/>pattern.Markkimarkkonnenhttps://www.blogger.com/profile/11122368349695914223noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-2254075436392489652009-02-03T16:18:00.000-08:002009-02-03T16:18:00.000-08:00Well, strictly speaking for divisibility by 7, tak...Well, strictly speaking for divisibility by 7, taking the mod7 remainder of a 3 digit number in base 10 isn't too difficult. Not sure what you mean.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-55628468728487048882009-02-03T16:14:00.000-08:002009-02-03T16:14:00.000-08:00David: yes, but that assumes you can reduce things...David: yes, but that assumes you can reduce things mod 7, which one might argue is a bit harder than my rule.<BR/><BR/>Anonymous: this requires dealing with some pretty big numbers and arbitrary constants. Could you remember it?<BR/><BR/>John: well, yeah. But I didn't prove anything new today.Michael Lugohttps://www.blogger.com/profile/01950197848369071260noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-50954280104835233962009-02-03T16:10:00.000-08:002009-02-03T16:10:00.000-08:00Subtract 21 times the last digit from the number. ...Subtract 21 times the last digit from the number. Clearly this is divisible by seven. Further, the new final digit is zero, so we can divide by 10. This also leaves divisibility by 7 unchanged.<BR/><BR/>All you're doing is a slightly slicker way of computing the same result.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-67175505259339995132009-02-03T16:01:00.000-08:002009-02-03T16:01:00.000-08:00Since (10mod7)=3, (10^2mod7)=2, and (10^3mod7)=-1,...Since (10mod7)=3, (10^2mod7)=2, and (10^3mod7)=-1,...(10^6mod7)=1...(10^9mod7)=-1... We can check divisibility by 7 by the following: <BR/><BR/>Is 108425723 divisible by 7?<BR/><BR/>Well, this number mod 7 could be written as 108425723 = 723 + 425*-1 + 108*1 which is just 2 + -5 + 3 (mod 7) which is 0. <BR/><BR/>Therefore, I know 5329809714 isn't divisible by 7 since r7(714-809+329-5) = r7(0-4+0-5) = 5.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-22187499804222942012009-02-03T13:25:00.000-08:002009-02-03T13:25:00.000-08:00Wouldn't it be easier to repeatedly replace th...Wouldn't it be easier to repeatedly replace the top two digits by their value mod 7?<BR/><BR/>4728402 => (47 mod 7)28402 => 528402<BR/>528402 => (52 mod 7)8402 => 38402<BR/>38402 => (38 mod 7)402 => 3402<BR/>3402 => (34 mod 7)02 => 602<BR/>602 => (60 mod 7)2 => 42<BR/>42 => (42 mod 7) => 0<BR/><BR/>It's clearer from this, I think, that there's nothing special about 7. You can make up an alternative left-to-right rule that works more similarlly to the right-to-left one (remove the top digit and add its triple to the next digit) but it's much slower.<BR/><BR/>But then I also tend to add pairs of multi-digit numbers left-to-right instead of right-to-left so maybe I'm doing this backwards as well.Anonymousnoreply@blogger.com