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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


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