The exponent of 6 in 72! is | vedclass

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(A) To find the exponent of $6$ in $72!$,we need to find the exponents of its prime factors $2$ and $3$ in the prime factorization of $72!$.Using Lege

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$34$

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(A) To find the exponent of $6$ in $72!$,we need to find the exponents of its prime factors $2$ and $3$ in the prime factorization of $72!$.Using Legendre's formula,the exponent of a prime $p$ in $n!$ is given by $E_p(n!) = \sum_{k=1}^{\infty} \left[ \frac{n}{p^k} ight]$.For $p=2$: $E_2(72!) = \left[ \frac{72}{2} ight] + \left[ \frac{72}{4} ight] + \left[ \frac{72}{8} ight] + \left[ \frac{72}{16} ight] + \left[ \frac{72}{32} ight] + \left[ \frac{72}{64} ight] = 36 + 18 + 9 + 4 + 2 + 1 = 70$.For $p=3$: $E_3(72!) = \left[ \frac{72}{3} ight] + \left[ \frac{72}{9} ight] + \left[ \frac{72}{27} ight] = 24 + 8 + 2 = 34$.Since $6 = 2 	imes 3$,the exponent of $6$ in $72!$ is $\min(E_2(72!), E_3(72!)) = \min(70, 34) = 34$.

The exponent of $6$ in $72!$ is

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