For a positive integer n, find number of zeroes at right end of n!

Hi tanmaykothari2710

Statement 2 : -
Number of zeroes at right end of n! = 4
SUFFICIENT

Please correct OA.

Please re check the solution .
Thanks

The second option states that n-1 is divisible by 19. So clearly n-1 can be 19 or its multiple. And other multiples are above 30 so they are not applicable. So, n will be 20 which is not a prime number. So how is the answer C??

________________________
Its (n-1)!

Hello tanmaykothari2710 ... I am not sure how you say N can be greater than 23 in statement 2. if N is greater than 23 then greatest primer factor dividing it would be 23, which contradicts the statement 2. And thus number of zeros would be same for N =(20,21,22,23).
Please correct me if I am wrong.

Number of zeroes in a factorial would depend on numbers of 2 or 5, and as number of 5s will be lesser, 5s will give us our answer.

Now what about number of 5s -- They will increase only when a multiple of 5 is added, that is number of 5s will remain same from 20! to 24!, and will increase in 25!.

Statement 1: n is a prime number lying between 18 and 30
So n can be 19, 23 or 29..
19! will have 3* 5s so 3 zeroes in the end.
23! will have 4* 5s so 4 zeroes in the end.
29! will have 6* 5s so 6 zeroes in the end.
Insuff

Statement 2: Greatest prime number dividing (n-1)! is 19
Now greatest prime number dividing (n-1)! is 19, so (n-1)! will not be divisible by the next prime number after 19, that is 23.
so \(19\leq{n-1}<{23}............20\leq{n}<{24}......n=20, 21, 22, 23\)
As we have seen above that the number of 5s will remain same from values of n as 20 to n as 24.
Hence, the number of Zeroes will always be 4, irrespective of value of n as 20, 21, 22 or 23.
Suff

B

OA given as C is wrong

For a positive integer n, find number of zeroes at right end of n! ( \(n! = n*(n-1)*(n-2)......*1\) )

Statement 1: n is a prime number lying between 18 and 30
n can be 19,23, and 29
But note that 19! and 20! itself differ by one '0'. So, even if we consider n = 19 we can say that the number of zeroes differ when we take the factorial of next number i.e. 20!, whether n is prime or not doesn't matter.

INSUFFICIENT.

Statement 2: Greatest prime number dividing (n-1)! is 19
So,
n - 1 = 19 \(\implies\) n = 20 BUT
when n - 1 = 20, n = 21 given condition is still satisfied and so for n = 22 and 23. However, not for n = 24.
Hence 20 ≤ n ≤ 23.

As there are no addl. 5's are multiplied with 2's to result in extra zeroes, the statement is SUFFICIENT.

Answer B.

chetan2u,bunuel,Kinshook
would you help me I have a doubt statement 2 talks about the greatest factor dividing (n-1)! and not n!
so (n-1)! can be 19,20,21,22,23,24,25,26,27and n! can be 20,21,22,23,24,25,26,27,28
this sum does not make sense to me

But the moment you take n-1 as 23, the greatest prime number dividing (n-1)! will be 23, so the statement II becomes false.
Thus, n-1 cannot be 23 and above.
n-1 can be 19,20,21,22, and n can, therefore, be 20,21,22,23

why are you focusing on (n-1) when the question is asking find number of zeroes at right end of n!
and not (n-1)! I can take n as 23 and (n-1)! factorial greatest prime factor will still be 19 because (n-1) will be 22!

Chetan's explanation is correct. Statement 2 tells us that 19 is the largest prime divisor of a certain factorial. That factorial certainly must be at least 19!, because 19 is not a divisor at all of any factorial from 1! through 18!. But the factorial cannot be 23! or larger, because 23!, and every larger factorial, is divisible by 23, a prime larger than 19. So if 19 is the largest prime divisor of a certain factorial, that factorial can only be 19!, 20!, 21! or 22!. Statement 2 tells us that factorial is (n-1)!, so n-1 must be 19, 20, 21 or 22, and n itself must therefore be 20, 21, 22 or 23. The question asks for the number of 'trailing zeros' in n!, and 20!, 21!, 22! and 23! all have four zeros at the end, so Statement 2 is sufficient.

I was so engrossed in doubt that I was not able to think it clearly you are right Ianstewart and chetan's

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