The greatest integer r such that 30^{r} divid | vedclass

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(B) To find the greatest integer $r$ such that $30^{r}$ divides $30!$,we first note the prime factorization of $30 = 2 	imes 3 	imes 5$.Since $30^{r

Vedclass · Accepted Answer

$7$

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(B) To find the greatest integer $r$ such that $30^{r}$ divides $30!$,we first note the prime factorization of $30 = 2 	imes 3 	imes 5$.Since $30^{r} = 2^{r} 	imes 3^{r} 	imes 5^{r}$,$r$ must be the minimum of the exponents of $2, 3,$ and $5$ in the prime factorization of $30!$.Using Legendre's Formula,the exponent of a prime $p$ in $n!$ is given by $E_{p}(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^{k}} floor$.For $p=5$: $E_{5}(30!) = \lfloor \frac{30}{5} floor + \lfloor \frac{30}{25} floor = 6 + 1 = 7$.For $p=3$: $E_{3}(30!) = \lfloor \frac{30}{3} floor + \lfloor \frac{30}{9} floor + \lfloor \frac{30}{27} floor = 10 + 3 + 1 = 14$.For $p=2$: $E_{2}(30!) = \lfloor \frac{30}{2} floor + \lfloor \frac{30}{4} floor + \lfloor \frac{30}{8} floor + \lfloor \frac{30}{16} floor = 15 + 7 + 3 + 1 = 26$.The greatest integer $r$ is $\min(26, 14, 7) = 7$.

The greatest integer $r$ such that $30^{r}$ divides $30!$ is

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