yasar434 wrote:
can anyone tell me how to calculate the no.of factors for 20!. well this was a small number if i would have to answer a bigger number like no.of factors for 720! what will be the solution.
You won't need this for GMAT but still let's see if we can do it:
First we should make prime factorization of 20!. 20! will have all primes from 0 to 20, so we should find the powers of these primes in 20!.
Powers of 2 --> \(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\);
Powers of 3 --> \(\frac{20}{3}+\frac{20}{9}=6+2=8\);
Powers of 5 --> \(\frac{20}{5}=4\);
Powers of 7 --> \(\frac{20}{7}=2\);
Powers of 11 --> \(\frac{20}{11}=1\);
Powers of 13 --> \(\frac{20}{13}=1\);
Powers of 17 --> \(\frac{20}{17}=1\);
Powers of 19 --> \(\frac{20}{19}=1\).
So \(20!=2^{18}*3^8*5^4*7^2*11^1*13^1*17^1*19^1\).
Next: How to Find the Number of Factors of an Integer.First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\).
NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
So, the # of positive factors of \(20!=2^{18}*3^8*5^4*7^2*11^1*13^1*17^1*19^1\) will be \((18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1)=19*9*5*3*2*2*2*2=41040\).
The same way we can find for 720!, but we'll need much more time.
Hope it helps.
1) Can you explain the logic behind / the steps to get to the formula (p+1)(q+1)(r+1)? Thank you!
2) Can you explain the logic behind the fact that there will be at least as many factors of 2 as there will be of 5 in a factorial?