tag:blogger.com,1999:blog-8070362.post115730501486090437..comments2022-03-18T11:24:32.869+05:30Comments on Death Ends Fun: Simple but satisfyingDilip D'Souzahttp://www.blogger.com/profile/08221707482541503243noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-8070362.post-1157520349373075062006-09-06T10:55:00.000+05:302006-09-06T10:55:00.000+05:30Dilip,I just meant that mathematics is full of thi...Dilip,<BR/>I just meant that mathematics is full of this kind of gymnastics and there may be other interesting ones like Simpson's Paradox which are 'useful' and challenging.<BR/>Swarupgaddeswaruphttps://www.blogger.com/profile/16509075029154476375noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1157507145168080312006-09-06T07:15:00.000+05:302006-09-06T07:15:00.000+05:30This follows as a consequence of Euclid's algorith...This follows as a consequence of Euclid's algorithm (the one used to find the GCD -- greatest common divisor of two integers). Any multiple of the gcd(m,n) can be written as am + bn, where a,b are integers. So the smallest in both of the integer cases will be the gcd of the sets of numbers that you have given. <BR/><BR/>The gcd of 231 and 3250 is 1. So 1 is the smallest positive number of the form 231a + 3250b.<BR/><BR/>The gcd of 231 and 2750 is 11. So the smallest positive number of the form 231a + 2750b is 11.<BR/><BR/>Since sqrt(2) is irrational, one can show (somewhat tediously) that we can produce arbitrarily small positive numbers (but not zero) of the form sqrt(2)a + 231b.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1157500580858263972006-09-06T05:26:00.000+05:302006-09-06T05:26:00.000+05:30dilip,a and b can be negative, if necessary. thus,...dilip,<BR/>a and b can be negative, if necessary. thus, the smallest non-zero positive number of the type 3a + 10b is 1, obtained, for example, when a = -3, b = 1.<BR/><BR/>the "extension" problem came up in a discussion yesterday. thought it was a nice coincidence that you put up a prime-number problem this morning.<BR/><BR/>b.b.https://www.blogger.com/profile/11007863932313868604noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1157470034523649542006-09-05T20:57:00.000+05:302006-09-05T20:57:00.000+05:30Thanks Suresh. Of course, n could also be 5k for s...Thanks Suresh. Of course, n could also be 5k for some k (which much more trivially makes M divisible by 5). <BR/><BR/>Thanks Swarup too -- there's no point really, it's just that I like figuring out little problems like this from time to time just to keep the grey cells churning!<BR/><BR/>b, thanks! It's good to be back. I'm a little puzzled by your problem, am I missing something? Because surely the smallest number of the form 231*a + 3250*b, assuming you mean positive numbers, is when you have a = b = 1, which gives us 3481. (Similarly for the others). <BR/><BR/>Did you mean something else?Dilip D'Souzahttps://www.blogger.com/profile/08221707482541503243noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1157465170480651692006-09-05T19:36:00.000+05:302006-09-05T19:36:00.000+05:30dilip, welcome back. a delight to read you, as alw...dilip, <BR/>welcome back. a delight to read you, as always.<BR/><BR/>here's one for you. consider non-zero numbers of the form 231*a + 3250*b, where a and b are integers. What is the smallest such number? How about numbers of the form 231*a + 2750*b?<BR/><BR/>Extension: What about sqrt(2)*a + 231*b? How small can you make it?b.https://www.blogger.com/profile/11007863932313868604noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1157309466809135562006-09-04T00:21:00.000+05:302006-09-04T00:21:00.000+05:30Never liked number theory; but even I can do it. F...Never liked number theory; but even I can do it. First n is even and divisible by 3 since one of three successive numbers is divisible by 3 and the other two are not. Since we are squaring and since n^2+16 is also divisible by 4, 144 divides the given number. Now n must be of the form 5k+2 or 5k+3 if it is not divisible by 5. So n^2+16 is divisible by 5. But, what is the point about this question?gaddeswaruphttps://www.blogger.com/profile/16509075029154476375noreply@blogger.comtag:blogger.com,1999:blog-8070362.post-1157309040447290632006-09-04T00:14:00.000+05:302006-09-04T00:14:00.000+05:30not the most elegant, but here goes.Let M = n^2(n^...not the most elegant, but here goes.<BR/><BR/>Let M = n^2(n^2 + 16).<BR/><BR/>1. if n > 2 then n must be even (n-1/n+1 are both prime, and therefore are odd). <BR/><BR/>2. if n > 5, then n must be divisible by 3 (of the triple n-1, n, n+1, exactly one must divide 3, and the other two are prime).<BR/><BR/>3. If n > 6, n must be either 5k+2 or 5k-2 for some k (same reason as in (2))<BR/><BR/>therefore, n^2 is divisible by 4, n^2+16 is divisible by 4 (from 1.) and therefore 16 divides M.<BR/><BR/>n^2 is divisible by 9, and so 9 divides M. (from 2.)<BR/><BR/>(5k+2)^2 = 5( some terms) + 4, and so n^2+16 is 5(some terms) + 20 which is divisible by 5 (same holds for 5k-2)<BR/><BR/>so M is divisible by 5. (from 3.)<BR/><BR/>therefore M is divisible by 16 * 9 * 5 = 720.<BR/><BR/>QED.Suresh Venkatasubramanianhttps://www.blogger.com/profile/15898357513326041822noreply@blogger.com