Bunuel
If 100!/30^n is an integer, which of the following is the greatest possible value of n?
A. 28
B. 24
C. 20
D. 15
E. 10
Here's the explanation.
Find the prime factors of \(30^n\)
30 = 2 x 3 x 5.
Therefore \(30^n = 2^n\) x \(3^n\) x \(5^n\)
Formula to find powers of prime numbers present in \(n!\) is \(= \frac{n}{p} + \frac{n}{p^2} + \frac{n}{p^3} .... + \frac{n}{p^k}\) ----- (p is the prime factor. k must be such that \(p^k \geq{n}\) )
Therefore we can find powers of prime factors (2, 3 and 5) in 100! by using the above formula.
Number of 2 in 100!
\(\frac{100}{2} + \frac{100}{4} + \frac{100}{8} + \frac{100}{16} + \frac{100}{32} + \frac{100}{64} = 50 + 25 + 12 + 6 + 3 + 1 = 97\)
Number of 3 in 100!
\(\frac{100}{3} + \frac{100}{9} + \frac{100}{27} + \frac{100}{81} = 33 + 11 + 3 + 1 = 48\)
Number of 5 in 100!
\(\frac{100}{5} + \frac{100}{25} = 20 + 4 = 24\)
Greatest value of power of n! would be the power of the prime factor which has the lowest power. In this case power of 5 which is 24.
Hence \(30^n = 30^{24}\). Greatest value of
n = 24. Answer B.
Hope its clear now.