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Exponent of a prime number p  in a number N  is given as  a  if
       N = p^a * M,  where  M is the product of all remaining factors.

If p, q, r are prime numbers or relatively prime, then
      N=p^a * q^b * r^c * ...
Then exponent of (pq) in N is the smaller of the two numbers a and b.

Exponent of a number p in N!  is given by the expression :

E_p(N!) = \frac{N}{p}+\frac{N}{p^2}+\frac{N}{p^3}+... +\frac{N}{p^s}\\\\ where,\ s\ is\ maximum\ integer\ such\ that\ p^s<=N\\

This is so because,  there are N/p multiples of p in N like p, 2p, 3p etc., then for each p^2, there is one extra p.  For p^3 , there is one more extra  p  as compared to p^2.   Hence we get the above formula.
12 = 2^2 * 3.  So we find exponents of 2 and 3 in 100 !.    Then we find exponent of 12.

E_2(100!)=\frac{100}{2}+\frac{100}{4}+\frac{100}{8}+\frac{100}{16}+\frac{100}{32}+\frac{100}{64} = 97\\\\E_3(100!)=\frac{100}{3}+\frac{100}{9}+\frac{100}{27}+\frac{100}{81}=48\\\\100!=2^{97}*3^{48}*M,\ \ M=product\ of\ other\ factors.\\\\ .\ \ \ =2*4^{48}*3^48*M=2*12^{48}*M\\&#10;

Hence exponent of 12 in 100! is 48.

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