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 )
Fixes that, surely.
-marcel
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