In article <18103614183559@[EMAIL PROTECTED]
>, Marcel Hendrix <mhx@[EMAIL PROTECTED]
>
wrote:
>William James <w_a_x_man@[EMAIL PROTECTED]
> wrote Re: Euler problem #48
>
>>> \ FORTH> euler48
>>> \ The last ten digits of the series 1^1 + 2^2 + 3^3 + ... + 1000^1000
are xxxxxxxxxx
>>> \ 0.011 seconds elapsed. ok
>
>> In PARI/gp. 0 ms. It yields more than 10 digits.
>
>> sum(i=1,1000,lift( Mod(i,10^10)^i ) )
>
>You mean, it calculates the wrong answer, faster?
>
>Sum = Mod( sum(i=1,1000,lift( Mod(i,10^10)^i ) ), 10^10 )
Methink that for i<1001 Mod(i,10^10) always equals i.
What do I miss here?
>
>Fixes that, surely.
>
>-marcel
>
>
--
--
Albert van der Horst, UTRECHT,THE NETHERLANDS
Economic growth -- like all pyramid schemes -- ultimately falters.
albert@[EMAIL PROTECTED]
&=n http://home.hccnet.nl/a.w.m.van.der.horst
|
