Thanks -- wouldn't the number of trailing zeroes simply be the number
of times 5 is a factor?
In other words (using my function) to find the trailing zeroes for
1,000,000,000,000!
f=: 2&{@:((0&{ , (0 , 2&{) + [: <. 1&{ % 0&{)^:_)
f 5 1000000000000x 0
249999999997
Or am I wrong?
gc
On Apr 13, 2008, at 12:45 AM, Arie Groeneveld wrote:
Last year (24 aug) I started a thread about counting the number of
trailing zero's of a factorial. Modifying some of these verbs will
give you:
fF=:[:+/[:<.]%[^>:@i.@<[EMAIL PROTECTED] NB. modified version of R Miller
or
mF=: +/@}.@(<[EMAIL PROTECTED]:a:)
2 mF 10
8
2 3 5 fF"0 [ 100000
99994 49995 24999
2 3 5 mF 100000
99994 49995 24999
Exponents of all prime factors of !20:
(i.&.(_1&p:) mF ]) 20
18 8 4 2 1 1 1 1
__ q: !20
2 3 5 7 11 13 17 19
18 8 4 2 1 1 1 1
=@@i
Geoff Canyon schreef:
I've created a function to figure out how many times a large
factorial will have a given number as a factor.
So for example, !10 has 2 as a factor 8 times:
2 1
4 2
6 1
8 3
10 1
Here's the function I defined to calculate this:
f=: 2&{@:((0&{ , (0 , 2&{) + [: <. 1&{ % 0&{)^:_)
It takes three arguments: the factor, the factorial, and the
starting number for the factor-count, which should always be 0.
It divides the factorial number by the factor repeatedly, keeping
track of the sum of the quotients, and at the end outputs just the
sum.
Is this a reasonable job of J programming? Is there a better way to
have coded this?
regards,
Geoff
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