total number of factors of a natural number N is 45. what is the maximum number of prime numbers that can divide N without leaving a remainder
Answers
Step-by-step explanation:
What is the maximum value of natural number n for which 21^n divides 50 factorial(!) is?
Basically what we are looking for is that how many times we can divide 50! by 21 such that remainder is 0.
We know that 21 = 7*3, therefore, the problem now comes down to finding how many 7s are there and how many 3s are there in 50! .
We know that, in 50! , we have following factors of 7
7, 14, 21, 28, 35, 42, 49. Here the count is 7 but 7 appears twice in 49 and thus 7 appears 8 times (directly + indirectly) in 50! .
Similarly, 3 appears 22 (verification is being left as an exercise) times, directly + indirectly in 50! .
We know that 21n , is same as (7∗3)n .
Since 3 is appearing 22 times but since 7 is appearing 8 times in 50! , our n , can’t go beyond 8, when it comes to 21.
Thus, the maximum value of n is 8, such that 21n divides 50! , completely i. e. sans any remainder.
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