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divakarbio7
Responding to a pm:
Yes, you do not have to find the actual value of 30!.
All you have to do is find the highest power of 3 in 30!.
Now, the process is simple: Highest power of 3 in 30! will be given by:
30/3 = 10
10/3 = 3
3/3 = 1
10 + 3 + 1 = 14
06 Jan 2021, 12:57
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divakarbio7
Powers of a prime number in n!
\(=\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+\frac{n}{p^4}+\frac{n}{p^5}+..............\frac{n}{p^k}\)
So, \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}\)
\(=10+3+1\)
\(=14\)
The answer is \(C\)
There is another way, I don't know whether it has any flaws but it works:
30/3=10
10/3=3
3/3=1
10+3+1=14
The answer is C.
Whatever understanding the basic formula is very important. : https://gmatclub.com/forum/everything-a ... 85592.html
18 Jun 2021, 04:32
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Solution:
Given p is the product of numbers from 1 to 30
=> p=30!
So we have to find the highest power of 3 in p
We need to use the method of successive division where we divide continuously, consider the quotients an add them up
In this case these successive quotients on dividing p by 3 would be 10, 3, 1 and zero
The sum of all these quotients would be 10 +3 + 1 + 0= 14
Hence the highest power of three present in p is 14 (option c)
Happy studying
Devmitra Sen
GMAT SME
Given p is the product of numbers from 1 to 30
=> p=30!
So we have to find the highest power of 3 in p
We need to use the method of successive division where we divide continuously, consider the quotients an add them up
In this case these successive quotients on dividing p by 3 would be 10, 3, 1 and zero
The sum of all these quotients would be 10 +3 + 1 + 0= 14
Hence the highest power of three present in p is 14 (option c)
Happy studying
Devmitra Sen
GMAT SME
