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        please answer it as fast as you can and it is of mathematics
2 years ago

jagdish singh singh
173 Points
							$\hspace{-0.70 cm}exponent of prime no. p in n! is E_{p}(n!)=\bigg\lfloor \frac{n}{p} \bigg\rfloor +\bigg\lfloor \frac{n}{p^2} \bigg\rfloor +\bigg\lfloor \frac{n}{p^3} \bigg\rfloor +\cdots\\\\\\ Where \lfloor x \rfloor  represent greatest integer function.\\\\\\ So E_{17}(2050!) = \bigg\lfloor \frac{2015}{17}\bigg\rfloor+\bigg\lfloor \frac{2015}{17^2}\bigg\rfloor+\bigg\lfloor \frac{2015}{17^3}\bigg\rfloor+\bigg\lfloor \frac{2015}{17^4}\bigg\rfloor+\cdots\\\\\\ So we get E_{17}(2015) = 120+7+0+0+0\cdots =127$

2 years ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions