Re: Re: Re: Print[Plot] vs Print[text,Plot]? (*now Do and Table*)

*To*: mathgroup at smc.vnet.net*Subject*: [mg88210] Re: [mg88113] Re: [mg88090] Re: Print[Plot] vs Print[text,Plot]? (*now Do and Table*)*From*: DrMajorBob <drmajorbob at att.net>*Date*: Mon, 28 Apr 2008 04:40:18 -0400 (EDT)*References*: <200804240957.FAA28500@smc.vnet.net> <4714471.1209200633216.JavaMail.root@m08> <op.t97t0dee2c6ksp@bobbys-imac>*Reply-to*: drmajorbob at longhorns.com

It occurs to me the proof that (n + 2)/6 is integer could be far simpler. That is, n is even, so n+2 is even. n is also a power of 10, so summing the digits of n+2 gives 3, hence n+2 is divisible by 3. (That's just the "threes" version of "casting out nines".) Bobby On Sat, 26 Apr 2008 06:25:15 -0500, DrMajorBob <drmajorbob at att.net> wrote: > Your curious result was > > Clear[n, result] > result = Sum[i (i + 1)/2, {i, 1, 10^7}] > > 166666716666670000000 > > In general, the sum is > > Clear[k, result, n] > result = Sum[i (i + 1)/2, {i, 1, n}] > > 1/6 n (1 + n) (2 + n) > > If n is the kth power of 10, the factor n causes "result" to have k > trailing zeroes, and the factor (n+1) causes the repetition you noticed, > but it also requires (n+2)/6 to be an integer, which occurs when > Mod[10^k+2,6]==0. > > But, modulo 6, we see that 10 is congruent to 4, and 4 is idempotent: > > Mod[10, 6] > > 4 > > Mod[4^2, 6] > > 4 > > Union@Table[Mod[10^k, 6], {k, 100}] > > {4} > > As a consequence, the pattern you noticed will always appear when n is a > power of 10. > > Table[Mod[10^k, 6], {k, 100}] > > {4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, > 4, \ > 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 , > 4, 4, \ > 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 , > 4, 4, \ > 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4} > > Oops! I almost forgot the final requirement, that (n+2)/6 < n. But > that's pretty obvious. > > Bobby > > On Fri, 25 Apr 2008 04:28:10 -0500, Syd Geraghty <sydgeraghty at mac.com> > wrote: > >> Craig, >> >> Just a (tongue in cheek) reminder that Mathematica is a wonderful all >> purpose environment in which to do Mathematics. >> >> 221 years ago the 10 year old Gauss would have solved the problem by >> coding:- >> >> n = 10000000; >> >> Timing[total = n (n + 1)/2] >> >> {0.000035, 50000005000000} >> >> Which is a whole lot faster! >> >> I realize that this is not very helpful in deciding usage of Table vs >> Do. >> >> But the precocious Gauss also might have tried >> >> Timing[total = Table[i (i + 1)/2, {i, 10000000}]] >> >> {20.46425400000001, {1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, >> 91, <<9999974>>, >> 49999885000066, 49999895000055, 49999905000045, 49999915000036, >> 49999925000028, 49999935000021, 49999945000015, 49999955000010, >> 49999965000006, 49999975000003, 49999985000001, 49999995000000, >> 50000005000000}} >> >> or better yet >> >> Timing[Total[Table[i (i + 1)/2, {i, 10000000}]]] >> >> {31.7433, 166666716666670000000} >> >> Just to show off ... :) >> >> Cheers ... Syd >> >> PS Where would we be now if Stephen Wolfram had provided Gauss with a= >> free Mathematica License back then? >> >> BTW What a surprising result (1666667)(1666667)(0000000) >> >> I wonder if anyone has computed this result before and explained its >> repetition of the first 7 digits? >> >> >> Syd Geraghty B.Sc, M.Sc. >> >> sydgeraghty at mac.com >> >> My System >> >> Mathematica 6.0.2.1 for Mac OS X x86 (64 - bit) (March 13, 2008) >> MacOS X V 10.5.2 >> MacBook Pro 2.33 Ghz Intel Core 2 Duo 2GB RAM >> >> >> >> >> >> >> On Apr 24, 2008, at 2:57 AM, W_Craig Carter wrote: >> >>> This is problem with a known simple result, but it will serve: let's= >>> find the sum of the first 10000000 integers. >>> >>> (*let's use a do loop*) >>> Timing[ >>> icount = 0; >>> Do[icount = icount + i, {i, 1, 10000000, 1}]; >>> icount >>> ] >>> >>> (*this returns {10.2826, 50000005000000} on my machine.*) >>> (*10.28 being a measure of how long the computation took to run*) >>> >>> (*lets try a table*) >>> Timing[ >>> Total[Table[i, {i, 1, 10000000, 1}]] >>> ] >>> >>> (*This returns {3.25516, 50000005000000} on my machine*) >> >> >> > > > -- = DrMajorBob at longhorns.com

**References**:**Re: Print[Plot] vs Print[text,Plot]? (*now Do and Table*)***From:*"W_Craig Carter" <ccarter@mit.edu>