Question

Class 11 maths
JEE Question

If 100! is divided by (24)^k (where k belongs to n ), then find the maximum value of k.​

Answer 1

$\huge\star{\mathfrak{\underline{\underline{\red{Answer}}}}}$

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$\large{\tt{\boxed{\blue{Permutation \:And \:Combination}}}}$

Exponent of 2 in 100! is represented by $\bold {E}_{2}$ 100!=

$\implies$ $\frac{100}{2} + \frac{100}{{2}^{2}} + \frac{100}{{2}^{3}} + ...........+\frac{100}{{2}^{6}}$

=50 +25 +12 + 6 + 3 +1

=97

$\implies$ Exponent of 2³ in 100! is 32

Exponent of 3 in 100! is represented by $\bold {E}_{3}$100!=

$\implies$$\frac{100}{3} + \frac{100}{{3}^{2}} + \frac{100}{{3}^{3}} + \frac{100}{{3}^{4}}$

=33+11+3+1

=48

$\implies$ Exponent of (2³ x 3) in 100! is min{48,32} =32

$\implies$ Exponent of (24) in 100! is =32

$\large\implies$ Maximum value of k is $\large{\boxed{\boxed{32}}}$

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