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- October 12th 2011, 08:20 AM #1

- October 12th 2011, 07:36 PM #2

- October 12th 2011, 07:42 PM #3

- October 12th 2011, 09:05 PM #4

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## Re: 90!

but...the answer given on the yahoo link isn't one of the choices....

here is a better answer (not my own, from this link: Wilmott Forums - problem with factorial)

how about this:

the problem essentially asks for 90!/10^21 (mod 100), where 21 comes from counting the power of 5 as already pointed out.

since 100=25*4 with 25 and 4 are coprime, and 90!/10^21 apparently is divisible by 4, so we just need to find 90!/10^21 (mod 25).

[if x=0 (mod 4) and x=a (mod 25), the Chinese remainder theorem gives x=-24*a (mod 100)]

now note (5n+1)(5n+4)=4 (mod 25) and (5n+2)(5n+3)=6 (mod 25), one gets (5n+1)(5n+2)(5n+3)(5n+4)=-1 (mod 25) [*]

cancel all the 5's first, one gets (-1)^18*{1*2*3*4*6*7*8*9*2*11*12*13*14*3*16*17*18}/2^21 or (-1)^22*2*3/(19*2^21) (mod 25),

where all the factors in the curly brackets are from 5,10,15,..., and [*] has been repeatedly used.

note 19*4=76=1 (mod 25) and 2*13=1 (mod 25), the expression above boils down to 3*(13^18) (mod 25). note 13^3=-3 (mod 25), so 3*(3^6)=3*(27^2)=3*(2^2)=12 (mod 25).

since 12 is divisible by 4, the answer is indeed 90!/10^21=12 (mod 100).

- October 12th 2011, 09:06 PM #5