SOLUTION: Q)it is desired to extract the maximum power of 3 from 24! where n!= n.(n-1). (n-2)....321. What will be the exponent of 3? (a)8 (b) 9 (c)11

Question 471386: Q)it is desired to extract the maximum power of 3 from 24! where n!= n.(n-1). (n-2)....321. What will be the exponent of 3?
(a)8 (b) 9 (c)11 (d) 10

Found 2 solutions by Alan3354, richard1234:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
If you mean from 24!, it's 10.

Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
An algorithm that always works is to repeatedly divide 24 by 3, to count the number of 3's in the factorization. For example,

24/3 = 8

8/3 = 2 (exclude remainders)

2/3 = 0

There are 8+2+0 = 10 factors of 3.
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