How to find the first 2 nonzero terms in Mod100 of n! without a prefix?

[heartyface]

hello
so I know how to get the 0s in n!, but how do I find the first 2 nonzero terms?
Is there a shortcut?
thanks :)

 

[Deveno]

do you mean the 2 right-most non-zero digits?

if so, that is not mod 100, unless you only have 2 "trailing 0's".

it WILL be mod 10^k for some k, but n! grows very quickly, and so unless n is between 10 and 14 (inclusive), k is not 2.

 

[heartyface]

hehe... that's why I said '2' first *nonzero* terms to n!

 

[Deveno]

hehe... that's why I said '2' first *nonzero* terms to n!

i understood what i thought you meant. but, realize the convention in english, is to parse left-to-right, so the first 2 digits of 5687 are 5 and 6, not 7 and 8.

i don't think there is a general formula for all n, usually, finding the right-most non-zero digits of n! requires a fair amount of work (finding the number of 0's at the end is not so bad, you can count occurences of factors of 5 (since we get 5 even numbers for every 2 multiples of 5, the factors of 5 will be the limiting factor, here)).

do you have a specific n! in mind?

 

[heartyface]

<random number generator> hmm, let's go with 109!
and yeah my english sux :P also, that's interesting! hmm, so if I actually did mean the 'first' 2 digits of n!, that seems... way more work(?!) hehe, let me try that out

 

[willem2]

To find the last 2 non-zero digits before all the zeros at the end.

factor n!, which is easy, because it has [n/p] + [n/p^2] + [n/p^3] + ... factors of p.

(where [x] is the largest integer, smaller or equal to x)

remove all factors of 5, and an equal number of factors of 2, now multiply what is left together modulo 100, so you never have to multiply numbers with more than 2 digits.

 

[heartyface]

To find the last 2 non-zero digits before all the zeros at the end.

factor n!, which is easy, because it has [n/p] + [n/p^2] + [n/p^3] + ... factors of p.

(where [x] is the largest integer, smaller or equal to x)

remove all factors of 5, and an equal number of factors of 2,
.

yuh... i knew that:)

now multiply what is left together modulo 100, so you never have to multiply numbers with more than 2 digits

it's easy to say that.. but how do you 'multiply the numbers' when there are some 100s of them.. then simply modulo100 it :(

 

[chiro]

yuh... i knew that:)
it's easy to say that.. but how do you 'multiply the numbers' when there are some 100s of them.. then simply modulo100 it :(

There are quite a lot of results in number theory that help you do these kinds of problems and also problems that are useful for doing this for really really large numbers (this kind of thing is common in cryptographic applications).

 

[heartyface]

ummm so like from this problem http://www.artofproblemsolving.com/Wiki/index.php/2010_AMC_10A_Problems/Problem_24
I don't understand it from the M=(1*2*3*4)(6*7*8*9)...*(86*87*88*89)......blabla
yup :( enlightenments?

 

[heartyface]