Hi,

On Feb 13, 2008 1:11 AM, Dave <[EMAIL PROTECTED]> wrote:
>
> Sum log 1 + log 2 + log 3 + ... + log n, where log represents the
> base-10 logarithm function. The ceiling of the sum gives the number of
> digits.
>
> Dave

Yeh, that work fine. Only to explain a bit more to Hariharan:

**let N your number

I)- First, make cases by hand and convince yourself  that
the number of digits in a number is equals to exponent of the minor
power of 10(ten) which is
bigger then N.

II)- Remember of the logarithm property learned at high school:

- log(x.y) = log(x) + log(y)

III)- N! = N.(N-1).(N-2)....1 => log10(N!)= log10(N.(N-1).(N-2)....1)
= log10(N) + log10(N-1) + log(N-2)+...+1

Best regards,
Caio  Valentim

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