If the greatest integer k for which 3^k is a factor of n! is 8, what

\(\frac{n}{3} + \frac{n}{3^2} + \frac{n}{3^3} … = 8\)

\(n! = 18!, 19!, 20!\)

\([\frac{18}{3}] + [\frac{18}{3^2}] = [\frac{19}{3}] + [\frac{19}{3^2}] = [\frac{20}{3}] + [\frac{20}{3^2}] = 8\)

From \(21!\) we have additional factor of \(3\) hence \(3^9\).

\(18!\) has \(5^3\)

\(19!\) Still \(5^3\)

\(20!\) One more factor of 5 – \(5^4\)

Max power of \(5\) in \(n!\) is \(4\)

Answer C

Hello everybody,

Is someone out there who can explain me the first step that was made? Why can you divide n by 3 and 3^2 to figure out what the maximum value of n might be? In general, I really don't understand the applied approach.

Kind regards

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Hi Bonachhilfe,

Please refer the Factorial section on the following page:
https://gmatclub.com/forum/math-number-theory-88376.html

Thanks.

I understand that n can be 18,19,20. However,the question never suggested that we have to consider the largest possible value of n. It says the largest possible value of k. So I substituted the value of n = 18 and not 20 while solving.

This is what the question says: what is the largest possible value of p so that 5^p is a factor of n! ?

We need the largest possible value of p. We will get the largest possible value of p when n takes the largest possible value it can take. n can be 18, 19 or 20.
If n is 18, p is 3
If n is 19, p is 3.
If n is 20, then p is 4.
So 4 is the largest possible value of p.

C

18!, 19! and 20! has 8 x 3's
So then 18! & 19! has 3 x 5's, while 20! has 4 x 5's
So 4 x 5's

Greatest integer k for which 3^k is a factor of n!, is a fancy way of asking how many 3's are present in n!.
Since it is given that k is 8, this means that the following numbers are present in the expansion of n!:
3,6,9,12,15,18
Each of the above number gives the following number of 3's
1,1,2,1,1,2 [3*1, 3*2, 3*3, 3*4, 3*5, 3*3*2], which sum to a total of '8'(1+1+2+1+1+2).
In the above case, we have considered n as 18, but even if we consider n as 20, the number of 3's would still be the same. It is only when we take n as 21 that another extra 3 gets added(21=3*7).
Also, as the question asks for the maximum value of p for 5^p to be a factor of n!(another fancy way of asking the number of 5's in n!), we can consider n as 20.
Numbers containing a 5 in the expansion of 20! would be 5,10,15,20(5*1,5*2,5*3,5*4) and each number is giving a single value of 5.
Therefore, the maximum value of p would be 1+1+1+1=4

Hope this helps.

I solved this question in the following way:

Step 1: find the largest possible value of n! by calculating how many "3" are presented in n!
Because greatest integer k for which 3^k is a factor of n! is 8. => 3,6,9,12,15,18 are presented in n!
Since we want to find the largest possible value of n! => we can increase the value of n as long as we dont add any more "3" =>n! = 20!

Step 2: find the largest p by calculating how may "5" are presented in 20!
We have 5,10,15,20 are presented in 20! and give us four "5" => p = 4

Hence, the answer is C.
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Kindly press +1kudos if the explanation is clear!
Thank you!

I see many complicated solutions but here is my under 1 min easy one -

When we have to find the power of a prime number for a factorial Eg: power of 2 in 10!, we do 10!/2 = 5, 10!/2^2 = 2, 10!/2^3 = 1 so total gets to 2^8

So in this case max n! can be 8x3 = 24!
when we'll find max power of 5 in 24! it would be 4 and 24!/5^2 is not possible. Hence p = 4.

Hi kanikaa9

The method you used luckily gave you a correct answer for this question, however I wouldn't recommend to go by your method as it has a couple of flaws. Let's suppose we had been asked to find largest possible value of p so that \(2^p\) is a factor of n!, your method will give an incorrect answer. That is because when you consider n as 24, the greatest integer k for which \(3^k\) is a factor of n! becomes 10, whereas it is explicitly stated in question that it is 8. Hope this helps.