Glarmorgan Posted November 26, 2009 Share Posted November 26, 2009 That works with 3 too...obviously as it's the square root of 9. Actually no, I'm wrong. With 3 it's if the sum of the digits add up to something that is divisible by 3, then the wholenumber is divisible by 3. So, 477 = 4+7+7 = 18 which is divisible by 3, so 477 is also divisible by 3. It's also used with 6 - the sum of digits is divisible by 3 AND 2 - the whole number is divisible by 6 (not surprising, considering the amazing fact that 6=3*2...). Link to comment Share on other sites More sharing options...
leviramsey Posted November 27, 2009 VT Supporter Share Posted November 27, 2009 That works with 3 too...obviously as it's the square root of 9. Divisibility tests for the non-zero digits: 1: all whole numbers are divisible by 1 2: if the last digit is a multiple of 2 (i.e. even) 3: if the digital root is 3, 6, or 9 (the digital root being the sum of the digits in the number; if the sum is greater than nine than re-sum the digits... thus the digital root of 578 is 5+7+8=20 => 2+0=2) 4: if the last two digits are a multiple of 4... this further simplifies to, if the tens digit is even, then the last digit is 0, 4 or 8; if odd, then the last digit must be 2 or 6 5: if the last digit is 5 or 0 6: if it's divisible by 2 and divisible by 3 (i.e. even and with a digital root of 3, 6, or 9) 7: there is a test but it's too complex to explain in a small space... 8: if the last three digits are a multiple of 8... there's a further simplification that's too complex to explain in a small space... 9: if the digital root is 9 In general if you have a composite number with no repeated factors, the divisibility test for that number is application of the tests for each factor. Link to comment Share on other sites More sharing options...
granville Posted November 27, 2009 Share Posted November 27, 2009 That works with 3 too...obviously as it's the square root of 9. Divisibility tests for the non-zero digits: 1: all whole numbers are divisible by 1 2: if the last digit is a multiple of 2 (i.e. even) 3: if the digital root is 3, 6, or 9 (the digital root being the sum of the digits in the number; if the sum is greater than nine than re-sum the digits... thus the digital root of 578 is 5+7+8=20 => 2+0=2) 4: if the last two digits are a multiple of 4... this further simplifies to, if the tens digit is even, then the last digit is 0, 4 or 8; if odd, then the last digit must be 2 or 6 5: if the last digit is 5 or 0 6: if it's divisible by 2 and divisible by 3 (i.e. even and with a digital root of 3, 6, or 9) 7: there is a test but it's too complex to explain in a small space... 8: if the last three digits are a multiple of 8... there's a further simplification that's too complex to explain in a small space... 9: if the digital root is 9 In general if you have a composite number with no repeated factors, the divisibility test for that number is application of the tests for each factor. I was thinking about posting something along these lines but couldn't be ****. Well done to you, even though you didn't bother with some of them. Link to comment Share on other sites More sharing options...
leviramsey Posted November 27, 2009 VT Supporter Share Posted November 27, 2009 The only digit I left out was zero, for obvious reasons... Link to comment Share on other sites More sharing options...
granville Posted November 27, 2009 Share Posted November 27, 2009 The only digit I left out was zero, for obvious reasons... I was talking about 7 . If anyone else actually gives a shit the 7 one goes like this (I think). Take off the last digit, double it and subtract that from the remaining number. If the result is divisible by 7 then so was the original number. You can repeat it as many times as necessary. Eg. 7945. Take off the 5 and double it (10), then take that from 794, (784). Still can't tell? Take off the 4, double it and subtract from 78 (78-8=70). Still can't tell? Get back in your cage small heathen. Link to comment Share on other sites More sharing options...
leviramsey Posted November 27, 2009 VT Supporter Share Posted November 27, 2009 It's actually... compute the reversed alternating sum of the ordered 3-tuples of digits and then subtract double the last digit. If the result is divisible by seven then the original number is divisible by 7. Link to comment Share on other sites More sharing options...
mjmooney Posted November 27, 2009 VT Supporter Share Posted November 27, 2009 Maths.... :cry: Link to comment Share on other sites More sharing options...
Stevo985 Posted November 27, 2009 VT Supporter Share Posted November 27, 2009 I'm convinced Levi is just a computer that has somehow learned to post on a forum. And he runs Opera obviously Link to comment Share on other sites More sharing options...
Phumfeinz Posted November 27, 2009 Share Posted November 27, 2009 He could be an Opera client that has become self-aware. Link to comment Share on other sites More sharing options...
leviramsey Posted November 27, 2009 VT Supporter Share Posted November 27, 2009 I will admit to being a singular individual... Link to comment Share on other sites More sharing options...
ED Posted November 27, 2009 Share Posted November 27, 2009 He could be an Opera client that has become self-aware. Nail on the head imo Link to comment Share on other sites More sharing options...
Qwpzxjor1 Posted November 27, 2009 Share Posted November 27, 2009 Little known trivia: maths is crap, boring and too complicated for 4pm on a Friday, let's have some stats about how many people get their cocks stuck in hoover nozzles per year or something. Link to comment Share on other sites More sharing options...
Gringo Posted December 2, 2009 Share Posted December 2, 2009 The children of Tony Blair have all applied for and received Irish passports. Link to comment Share on other sites More sharing options...
TheSufferingVilla Posted December 2, 2009 Share Posted December 2, 2009 The children of Tony Blair have all applied for and received Irish passports. I had no idea they were good at football. :? Link to comment Share on other sites More sharing options...
The_Rev Posted December 2, 2009 Share Posted December 2, 2009 You dont even have to be good at football to get an Irish passport. Link to comment Share on other sites More sharing options...
Dom_Wren Posted December 2, 2009 Share Posted December 2, 2009 The children of Tony Blair have all applied for and received Irish passports. Why is that? I was thinking about applying for an Irish passport so i could enter the green card lottery, but found out tha in roder to enter on an Irish passport you had to have been born in the emerald isle. Hairy muff i suppose. Link to comment Share on other sites More sharing options...
BOF Posted December 2, 2009 Moderator Share Posted December 2, 2009 Could it be to do with being in areas of conflict like Iraq, Afghanistan, Chad etc? Having an Irish passport would increase your chances in any sticky situations as we tend to be peacekeepers. That was my first thought anyhoo (not that I even know what the junior Blair's do for a living). Link to comment Share on other sites More sharing options...
mjmooney Posted December 2, 2009 VT Supporter Share Posted December 2, 2009 Well they're Catholics, so they may as well go the whole hog. Link to comment Share on other sites More sharing options...
mjmooney Posted December 3, 2009 VT Supporter Share Posted December 3, 2009 "Big Country" is an anagram of "Bi word removed orgy". Link to comment Share on other sites More sharing options...
Stevo985 Posted December 3, 2009 VT Supporter Share Posted December 3, 2009 Word filter doesn't help there Link to comment Share on other sites More sharing options...
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