If 2^{n} divides 16! and 2^{n+1} does not div | vedclass

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

Vedclass · Accepted Answer

$15$

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(B) The exponent of a prime $p$ in the prime factorization of $m!$ is given by Legendre's Formula: $E_p(m!) = \sum_{k=1}^{\infty} \left[ \frac{m}{p^k} \right]$.Here,$m = 16$ and $p = 2$.$n = \left[ \frac{16}{2} \right] + \left[ \frac{16}{4} \right] + \left[ \frac{16}{8} \right] + \left[ \frac{16}{16} \right]$.$n = 8 + 4 + 2 + 1 = 15$.Since $2^{15}$ divides $16!$ and $2^{16}$ does not divide $16!$,the value of $n$ is $15$.

If $2^{n}$ divides $16!$ and $2^{n+1}$ does not divide $16!$,then $n=$

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