If n is a positive integer and n! is a multiple of 7^6 but not 7^9, th

Question

If n is a positive integer and  n!  is a multiple of  7^6  but not  7^9 , then how many value can n take ?

A. 21
B. 14
C. 13
D. 12
E. 7

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ArunSharma12 · Answer

when n! is a multiple of  7^6 , minimum value of n is 42.
when n! is a multiple of  7^9 , minimum value of n is 56.
number of values of n = 56 - 42 = 14
Ans: B

Kinshook · Answer

Asked: If n is a positive integer and  n!  is a multiple of  7^6  but not  7^9 , then how many value can n take ?

Maximum power of 7 in 56! = 8 + 1 = 9
Maximum power of 7 in 42! = 6
41<n<56

n = {42,43,44,45,46,47,48,49,50,51,52,53,54,55} : 14 values on n are feasible

IMO B

GaurSaini · Answer

Each multiple of 7 will add one 7 to the factorial, but 49 will add two 7 to the factorial.

i.e. 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8

Now n! has to have 6 powers of 7, but not 9. And 55! will have same number of 7 as 49!.

So n can take values from 42 to 55, both inclusive.

55-42+1 = 14

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If n is a positive integer and n! is a multiple of 7^6 but not 7^9, th

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