If the integer represented by 100! has K cons | vedclass

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Question

(A) The number of zeroes at the end of $n!$ is determined by the exponent of the highest power of $5$ that divides $n!$,as there are always more facto

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) The number of zeroes at the end of $n!$ is determined by the exponent of the highest power of $5$ that divides $n!$,as there are always more factors of $2$ than $5$ in the prime factorization of $n!$. We use Legendre's Formula to find the exponent of $5$ in $100!$: $K = \lfloor \frac{100}{5} \rfloor + \lfloor \frac{100}{5^2} \rfloor$ $K = \lfloor 20 \rfloor + \lfloor 4 \rfloor$ $K = 20 + 4 = 24$. Thus,there are $24$ consecutive zeroes at the end of $100!$.

If the integer represented by $100!$ has $K$ consecutive zeroes at the end,then $K=$

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