In the prime factorization of 37! = 2^{alpha_ | vedclass

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Question

(B) The exponent of a prime $p$ in the prime factorization of $n!$ is given by Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k

Vedclass · Accepted Answer

$17 : 8$

Vedclass · Answer

(B) The exponent of a prime $p$ in the prime factorization of $n!$ is given by Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$. For $n = 37$ and $p = 3$: $\alpha_3 = \lfloor \frac{37}{3} \rfloor + \lfloor \frac{37}{9} \rfloor + \lfloor \frac{37}{27} \rfloor = 12 + 4 + 1 = 17$. For $n = 37$ and $p = 5$: $\alpha_5 = \lfloor \frac{37}{5} \rfloor + \lfloor \frac{37}{25} \rfloor = 7 + 1 = 8$. Thus,the ratio $\alpha_3 : \alpha_5 = 17 : 8$.

In the prime factorization of $37! = 2^{\alpha_2} \cdot 3^{\alpha_3} \cdot 5^{\alpha_5} \cdots 37^{\alpha_{37}}$,the ratio $\alpha_3 : \alpha_5$ is:

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