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?
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Thanks! man
That's a relief
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Yes, that's correct. There is an example about power of 900 in 50! at: everything-about-factorials-on-the-gmat-85592-20.html



You'll need only to know how to determine the number of trailing zeros and the power of primes in n! (everything-about-factorials-on-the-gmat-85592.html), the above example is out of the scope of GMAT.
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nice explanation.
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Hi Anan,
Not to worry much as this particular concept has been explained very well was Bunuel in math book as well as in one of the topics related to factorial.

You can find them here at everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html.

If still concept remains unclear, please ask.

The key thing to see is that 40! / 80^n = Int.

After factoring and finding the respective powers for 5 and 2 ... manipulate the original equation in the question stem to 40! = 80^n * int.

So, we know that 40! will equal 80^n * some int. value (in this case, we'll mark that int. as "p" -- for simplicity's sake)

Since 80^n = (5^9 * 2^38) --> We now have 40! = (5^9 * 2^38) * p

Now, we need to match the higher power down to the lower power. We need to deduce 2^38 to the 9th power, so that it'll match up with 5^9.
Knowing that when we factorized 80^n ... we were left with 5^n and 2^4n ...

Using 2^4 --> we want to reduce the 38 down by dividing 4 into 38. Well, that obviously won't work. So, we take out 2^2:

40! = (5^9 * 2^36) * 2^2 * p

Then, since we have 2^4 as a factor of 80, we plug that in for 2^36. We divide 4 into 36, and left with this ---> 40! = (5^9 * (2^4)^9) * 2^2 * int

2^4 becomes 8: 40! = (5^9 * 16^9) * 2^2 * int -----> 40! = (5 * 16)^9 * 2^2 * int ----> 40! = (80)^9 * 2^2 * int

So: n = 9

Not sure if this is helpful to anyway, but this is how I think about these types of problems:

What are the components of 80? --> factor 80 --> (2^4)*(5).

So we are asking how many "lots" of four 2's and one 5 are there in 40? Well, there are a ton of 2's but 5's are the limiting factor. To phrase it a different way, I imagine a house that requires four 2's and one 5 for each house. Given that 5 is the limiting factor we can search for that.

40 --> one 5
35 --> one 5
30 --> one 5
25 --> two 5's
20 --> one 5
15 --> one 5
10 --> one 5
5 --> one 5

Total 5's = 9, therefore the total number of houses you can build is 9, therefore you can have nine 80's.

hi

here 5 is a limiting factor...so we can only work on 5, 5^9 ....

thanks