What is the max value of k for which 105! / 12 ^ k is an integer
Answers
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
As heighest 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
Given : 105! / 12 ^ k is an integer
To find : Maximum value of k
Solution:
12 = 2 * 2 * 3 = 2² * 3
Factor of 3 in 105!
[ 105 / 3] + [105 / 3²] + [105 / 3³] + [105 / 3⁴]
= 35 + 11 + 3 + 1
= 50
Factor of 2 in 105!
[ 105 / 2] + [105 / 2²] + [105 / 2³] + [105 / 2⁴] + [105 / 2⁵] + [105 / 2⁶]
= 52 + 26 + 13 + 6 + 3 + 1
= 101
> 2(50)
Hence both are 50
Maximum possible value of k = 50
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