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About LinkBacksThe number obtained from the last two nonzero digits of 90! is equal to n. What
is n?
(A) 12 (B) 32 (C) 48 (D) 52 (E) 68
I ended up with 92.
My hard work went to waste.
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).
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Originally Posted by alexmahone 