tag:blogger.com,1999:blog-4147841314301981805.post9080601423151459732..comments2014-01-31T07:06:18.285-07:00Comments on Heisencoder: Is 91 Prime?Matthew V Ballhttp://www.blogger.com/profile/04512577643269096659noreply@blogger.comBlogger31125tag:blogger.com,1999:blog-4147841314301981805.post-67672834656022989172011-10-25T16:28:01.621-06:002011-10-25T16:28:01.621-06:00Wonderful. Thank you very much.Wonderful. Thank you very much.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-80664443098874647412008-09-22T17:30:00.000-06:002008-09-22T17:30:00.000-06:00pretty and nice issue.. thank you.pretty and nice issue.. thank you.Transpalethttp://www.transpaletevi.com/noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-25490729870883801782008-08-25T22:46:00.000-06:002008-08-25T22:46:00.000-06:00This comment has been removed by a blog administrator.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-38139285577031318192008-05-05T04:22:00.000-06:002008-05-05T04:22:00.000-06:00Much of this is based off of the Euclidean Algorit...Much of this is based off of the Euclidean Algorithm. (which actually isn't an algorithm, but is still called that) Everyone is coming up with their own tricks when this has already been around for hundreds of years...Sinokehttps://www.blogger.com/profile/12328550208778908773noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-77963743286792063312008-03-14T23:05:00.000-06:002008-03-14T23:05:00.000-06:00There's an interesting trick I found for (working ...There's an interesting trick I found for (working toward) factoring numbers of 4 digits (or more) and it's based on 91 not being prime.<BR/><BR/>91*11 = 1001<BR/>7*11*13 = 1001<BR/><BR/>Thus, if you have any 4 digit number abcd (where a,b,c,d are single digits), you can simplify it sigificantly by subtracting either a00a or d00d, whichever is less, leaving bc(d-a) or (a-d)bc0. In the case of (a-d)bc0, you can divide by 10 to get the 3 digit number (a-d)bc. <BR/><BR/>If a=d, you get a 2 digit number to work with (well bc0, but you can divide by 10). <BR/><BR/>In any case, you have turned a 4 digit number into at least a 3 digit number and can quickly test that 3 digit number for factors of 7,11 and 13.<BR/><BR/>Also, you don't have to actually add the digits to check for divisibility by 3, just throw out 3s (and multiples of 3). For example, is 3240184320741 divisible by 3? throw out 3,24,18,432,741 leaving no more digits. Not all are this easy, but you get the idea.Jordan Hendersonhttps://www.blogger.com/profile/06586821482992546930noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-20165552742331952972008-03-14T11:07:00.000-06:002008-03-14T11:07:00.000-06:00Anonymous #1: Yes, of course if (a-b)=1 then (a+b)...Anonymous #1: Yes, of course if (a-b)=1 then (a+b)(a-b) might be prime. Good catch; I wasn't thinking of the degenerate case.<BR/><BR/>Anonymous #2: It was obviously a typo. Should have been<BR/><BR/>91 = 10^2 - 3^2drdranghttp://www.leancrew.com/all-thisnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-16095423695755035322008-03-13T11:20:00.000-06:002008-03-13T11:20:00.000-06:00Checking divisibility by 11 is easier by adding an...Checking divisibility by 11 is easier by adding and subtracting alternate digits. For 12345, you compute 1-2+3-4+5. If the result (3 in this case) is zero or a multiple of 11, the original number is divisible by 11.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-53674467599941359932008-03-13T07:44:00.000-06:002008-03-13T07:44:00.000-06:00Hello,Divide by 91 was the topic of my first video...Hello,<BR/><BR/>Divide by 91 was the topic of my first video. I teach this to my MathCounts team. The goal is to beat the calculator. I call it Club 91. Just thought I would share this.<BR/><BR/>Kevin<BR/><BR/>www.siyensya.comKevinwww.siyensya.comnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-35208147511763292642008-03-12T23:55:00.000-06:002008-03-12T23:55:00.000-06:00* Although the divisible by seven (7) test is not ...* Although the divisible by seven (7) test is not well known (especially compared to the popular divisible by three (3) test), there is an easy way to test if a natural number is evenly divisible by seven (7). See also Divisibility rule.<BR/><BR/> 1. Remove the last digit,<BR/> 2. Double it, and<BR/> 3. Subtract it from the remaining digits.<BR/> 4. If the result is negative and there are 2 or more digits, drop the negative sign.<BR/> 5. Repeat until you end up with a result that is a multiple of seven (7). (i.e. -7, 0, or +7)<BR/><BR/> For example, the number 1358 is evenly divisible by seven, since:<BR/><BR/> 135 - (8*2) = 119<BR/> 11 - (9*2) = -7<BR/><BR/> Using Number Theory the proof is rather simple, once the number n is rewritten in the form:<BR/><BR/> n = 10a + b<BR/><BR/> Where:<BR/><BR/> a is the remaining digits, and<BR/> b is the last digit.<BR/><BR/> Then:<BR/><BR/> 10a + b = 0 (mod 7)<BR/> 5 * (10a + b) = 0 (mod 7)<BR/> 49a + a + 5b = 0 (mod 7)<BR/> a + 5b - 7b = 0 (mod 7)<BR/> a - 2b = 0 (mod 7)Michaelhttps://www.blogger.com/profile/16918754955275589020noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-18117724517701375092008-03-12T20:04:00.000-06:002008-03-12T20:04:00.000-06:00Fantastic. The beauty of mathematics!Fantastic. The beauty of mathematics!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-83153037150948505402008-03-12T19:17:00.000-06:002008-03-12T19:17:00.000-06:00More tricks (4,8,9,25):4:If the last 2 digits of a...<B>More tricks (4,8,9,25):</B><BR/><BR/><B>4</B>:<BR/><BR/>If the last 2 digits of a number are divisible by 4, then the number is divisible by 4.<BR/><BR/>Example: 148 -- 48 is divisible by 4<BR/><BR/><B>8</B>:<BR/><BR/>If the last 3 digits of a number are divisible by 8, then the number is divisible by 8.<BR/><BR/>Example: 9168 -- 168 is divisible by 8<BR/><BR/><B>9</B>:<BR/><BR/>If the digits add up to a number divisible by 9, then the number is divisible by 9.<BR/><BR/>Example: 78885 -- 7 + 8 + 8 + 8 + 5 = 36, which is divisible by 9<BR/><BR/><B>25</B>:<BR/><BR/>If the last 2 digits are 00, 25, 50, or 75.. it is divisible by 25.<BR/><BR/>Example: 450 -- last 2 digits are 50<BR/><BR/><B>Combinations</B>:<BR/><BR/>You can get easy checks for 6, 12, 14, 15, 18, 21 by combining other simpler checksJasonhttps://www.blogger.com/profile/06700502509871390544noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-35021214762009959142008-03-12T17:36:00.000-06:002008-03-12T17:36:00.000-06:00This is like the scientific clock post all over!I ...This is like the scientific clock post all over!<BR/><BR/>I believe the answer is 42? Isn't that what Douglas Adams taught us?<BR/><BR/>Here... kill your time with this one. Much easier.<BR/><BR/>Top 10 Awesome Websites That Sell Cool Products You Probably Have Never Visited But Need To.<BR/><BR/>http://www.comember.net/blogs/firepixel/fire-pixel.comhttp://fire-pixel.comnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-89509659501091279692008-03-12T16:55:00.000-06:002008-03-12T16:55:00.000-06:00@ drdrang91 = 100 - 9 = 10^2 - 3^391 = 100 - 9 = 1...@ drdrang<BR/><BR/>91 = 100 - 9 = 10^2 - 3^3<BR/>91 = 100 - 9 = 100 - 27 ?<BR/><BR/>hmm, sup wit dat?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-33482203914299188522008-03-12T15:57:00.000-06:002008-03-12T15:57:00.000-06:00Check out the 37 factoring trick, it's really impr...Check out the 37 factoring trick, it's really impressive<BR/><BR/>http://www.flickr.com/groups/multiplesof37/discuss/72157594196046188/<BR/><BR/>and it works!Alvyhttp://www.microsiervos.comnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-48815808633554447592008-03-12T14:33:00.000-06:002008-03-12T14:33:00.000-06:00If you want to get some practice in, try my game P...If you want to get some practice in, try my game <A HREF="http://www.1729.com/math/integers/PrimeShooter.html" REL="nofollow">PrimeShooter</A>. <BR/>When 91 (or 57) comes at you, try shooting it with the "p" (for prime) bullet. It will ignore your bullet and keep coming at you.<BR/>Actually, I did consider adding an advanced "Grothendieck" mode to the game, but this remains on my to-do list.Philip Dorrellhttp://www.1729.com/noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-9432051330572303402008-03-12T13:40:00.000-06:002008-03-12T13:40:00.000-06:00Graham: the real question is "Is 57 prime?" :-)Graham: the real question is "Is <A HREF="http://en.wikipedia.org/wiki/57_%28number%29#In_mathematics" REL="nofollow">57</A> prime?" :-)Mileshttps://www.blogger.com/profile/07136909835648629963noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-35788184582281998912008-03-12T13:37:00.000-06:002008-03-12T13:37:00.000-06:00@drdrang: The difference between two squares will ...@drdrang: The difference between two squares will not always be a composite number. The difference between any two adjacent squares n^2 and (n+1)^2 is always an odd number 2n+1. Every odd number is the difference between two squares, including odd prime numbers. For example, the prime 23 is the difference between 12^2 (144) and 11^2 (121). In this case (a-b) turns out to be 1.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-2293211902028946532008-03-12T12:45:00.000-06:002008-03-12T12:45:00.000-06:00Hi Plex (and JM),This is how the divisibility by 7...Hi Plex (and JM),<BR/>This is how the divisibility by 7 works: write your number n as n = 10a + b for b an integer from 0 to 9. Since -2 and 7 have no factors in common, n is divisible by 7 if and only if -2n = -20a -2b is. Since 21a is divisible by 7, you can add this to it, so this is also equivalent to a - 2b being divisible by 7.<BR/>(mathematical explanation: -2 is the inverse of 10 modulo 7)<BR/><BR/>DoetoeAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-27409817072933675202008-03-12T10:38:00.000-06:002008-03-12T10:38:00.000-06:00The divisible-by-7 technique is clever -- I wasn't...The divisible-by-7 technique is clever -- I wasn't aware of it before -- but personally I find it easier to just do it the straightforward way, from left to right, in my head:<BR/><BR/>Is 8638 divisible by 7? Subtract 7000<BR/>Is 1638 divisible by 7? Subtract 1400<BR/>Is 238 divisible by 7? Subtract 210<BR/>Is 28 divisible by 7? It sure is.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-17548014680160330782008-03-12T10:31:00.000-06:002008-03-12T10:31:00.000-06:00Also, if you have a number abcdef, calculate ijk =...Also, if you have a number abcdef, calculate ijk = abc-def. <BR/>If ijk is divisible by 7,11, or 13, then abcdef is too - based on the fact that 1001=7*11*13.<BR/>You can actually group a number into chunks of size 3, alternately add and subtract, and the same rule holds: if your end result is divisible by 7,11,13 then the original number is too.gnaffhttps://www.blogger.com/profile/17742070409534417521noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-66660267692364355602008-03-12T10:23:00.000-06:002008-03-12T10:23:00.000-06:0091 is not prime. 91 divides by both 7 and 13.You c...91 is not prime. 91 divides by both 7 and 13.<BR/><BR/>You can test for primes pretty easily in Excel. :^)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-81832429123595886602008-03-12T10:17:00.000-06:002008-03-12T10:17:00.000-06:00Nice article, but 91 is a triangle number so it is...Nice article, but 91 is a triangle number so it isn't prime. It's also a hexagonal number (the difference between two cubes) AND the sum of two cubes.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-12566338429930676122008-03-12T10:13:00.000-06:002008-03-12T10:13:00.000-06:00I'm stumped on the trick for 7. I actually figure...I'm stumped on the trick for 7. I actually figured out on my own very similar tricks for 3, 9, and 11, but I don't understand why the trick for 7 always works. Is there a simple explanation?plexhttps://www.blogger.com/profile/17251720520390777580noreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-58302510048988498812008-03-12T10:09:00.000-06:002008-03-12T10:09:00.000-06:00sum of digits != 345 - (3 * 100) + 3 - (4 * 10) + ...sum of digits != 345 - (3 * 100) + 3 - (4 * 10) + 4<BR/><BR/>sum of digits = 345 - (3 * 100) + 3 - (4 * 10) + 4 + 5monkeymannoreply@blogger.comtag:blogger.com,1999:blog-4147841314301981805.post-86469567081927563292008-03-12T09:53:00.000-06:002008-03-12T09:53:00.000-06:00Nice little article. You missed one big tip -- no...Nice little article. You missed one big tip -- not sure if it is a "trick" or not, but it's worth noting that you never need to evaluate a number for prime factors that are larger than the number's square root.<BR/><BR/>In your '91' example, the square root of 81 is 9, the square root of 100 is 10, so the square root of 91 is somewhere between there. The largest prime number less than 10 is 7 ... so for 91, you do not even need to consider 11, 13, etc.<BR/><BR/>Another example, 151. 13 squared is 169, so the square root of 151 must be less than that, so the largest prime that needs to be considered is 11.Anonymousnoreply@blogger.com