A) \[3\]
B) \[4\]
C) \[5\]
D) \[6\]
Correct Answer: C
Solution :
[c] Number of zeroes at the end of n! is equal to the exponent of 5 in n! \[\therefore \] Number of zeroes \[=x=\sum\limits_{i=1}^{\infty }{\left[ \frac{n}{{{5}^{i}}} \right]}\] \[\Rightarrow \,\,\,x<\sum\limits_{i=1}^{\infty }{\frac{n}{{{5}^{i}}}}\] \[(\because \,\,\,[x]\le x)\] \[\Rightarrow \,\,\,\,x<n\frac{\frac{1}{5}}{1-\frac{1}{5}}\] \[(\because \,\,\,\,[x]\le x)\] \[\Rightarrow \,\,\,\,x<\frac{n}{4}\] \[\Rightarrow \,\,\,\,\,n>80\] (Given\[x=20\]) If \[n=80,\] \[x=\left[ \frac{80}{5} \right]+\left[ \frac{80}{25} \right]=19\] If \[n=85,\] \[x=\left[ \frac{85}{5} \right]+\left[ \frac{85}{25} \right]=20\] If \[n=90,\] \[x=\left[ \frac{90}{5} \right]+\left[ \frac{90}{25} \right]=21\] \[\Rightarrow \,\,\,n\in \{85,86,87,88,89\}\]
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