What is the max value of k for which 105! / 12 ^ k is an integer
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Answer:
50
Step-by-step explanation:
Highest power of 3 in 105! =
[105/3^1]+[105/3^2]+[105/3^3]+[105/3^4]
= 35+11+3+1 => 50
where [ ] this bracket signifies greatest integer function.
As highest power of 2 will be greater in 105! We don't need to calculate this for all the prime factors of 12 to get the maximum value of k for 105!/12^k to be an integer.
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