AtifS wrote:
This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book.
When finding the power of non-prime number in \(n!\) we first do prime-factorization of the non-prime number and then find the powers of each prime number in \(n!\) one by one using the following formula
\(\frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}}\) such that \(p^x <n\), where \(p\) is the prime number.
Let's suppose, we want to find the powers of \(80\) in \(40!\).
Prime factorization of \(80=2^4 * 5^1\).
Now first find the power of \(2\) in \(40!\);
\(\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38\)powers of \(2\) in \(40!\) --> \(2^{38}\)
Now find the powers of \(5\) in \(40!\);
\(\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9\)
And \(40!=80^x*q=(2^4 * 5^1)^x*q\), where \(q\) is the quotient and \(x\) is any power of \(80\), now from above calculation
\(40!=(2^{38}*5^9)*q=(2^4*5^1)^9*2^2*q=(80)^9*4q\), So we have \(80\) in the power of \(9\) in \(40!\).
Now, the main reason for why did I do all of the above is that whether I am doing it right or not? I took values (\(80\) & \(40!\)) randomly and tried to apply the rule/method and I was confused whether I am (or should it be
am I?
) right or not. Would like expert opinions. -->Bunuel? or Shrouded1? or Gurpreetsingh?
We can also determine this by factoring the power required and then we can just find the power of the highest prime and dividing the result so obtained by the number of power of the highest prime so obtained by factoring the required number.
Eg., In the above example, by factoring 80 = 2^4 * 5^1
Now, find the power of highest prime, i.e., 5 in 40!
{(40/5) + (40/5^2)}
{8 + (40/25)}
{8 + 1}
{9}
Now, for 80, we need four 2's and one 5.
Hence, the power of 80 in 40! = 9/1 (1 in denominator comes from number of 5's, which is the highest prime after factoring 80, required).
Answer = 9