A) 48
B) 49
C) 96
D) None of these
Correct Answer: A
Solution :
We have, \[12={{2}^{2}}\times 3=2\times 2\times 3\] Now, \[a={{E}_{2}}(100!)=\left[ \frac{100}{2} \right]+\left[ \frac{100}{{{2}^{2}}} \right]\] \[+\left[ \frac{100}{{{2}^{3}}} \right]+\left[ \frac{100}{{{2}^{4}}} \right]+\left[ \frac{100}{{{2}^{5}}} \right]+\left[ \frac{100}{{{2}^{6}}} \right]\] \[=50+25+12+6+3+1=97\] \[b={{E}_{3}}(100!)=\left[ \frac{100}{3} \right]+\left[ \frac{100}{{{3}^{2}}} \right]\] \[+\left[ \frac{100}{{{3}^{3}}} \right]+\left[ \frac{100}{{{3}^{4}}} \right]\] \[=33+11+3+1=48\] \[\therefore \] \[{{E}_{12}}(100!)=\min (a,b)\] Hence, \[{{E}_{12}}(100!)=48\]
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