Find the maximum value of n such that 50! is perfectly divisible by $2520^n$ .
2520 = $2^3 \times 3^2 \times 5 \times 7$
Here 7 is the Highest prime So find the number of 7s in 50! only.
Number of 7s in 50! =$ \left[\dfrac{50}{7}\right] +\left[ \dfrac{50}{7^2}\right]$ = 7+1 = 8
For n(max) = 8, 50! is perfectly divisible by $2520^8$
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