EgmatQuantExpert wrote:
e-GMAT Question of the Week #1 The number n is the product of the first 49 natural numbers. What is the maximum possible value of p + q such that both \(\frac{n}{{24}^p}\) and \(\frac{n}{{36}^q}\) are integers?
A. 11
B. 15
C. 20
D. 26
E. 30
The number n is the product of 49 natural numbers(49!)
n will contain \([\frac{49}{2^1}]+[\frac{49}{2^2}]+[\frac{49}{2^3}]+[\frac{49}{2^4}]+[\frac{49}{2^5}] = 24+12+6+3+1 = 46\) 2's.
n will contain \([\frac{49}{3^1}]+[\frac{49}{3^2}]+[\frac{49}{3^3}] = 16+5+1 = 22\) 3's.
The reason we find out the details of the prime number 2 and 3 in 49! is
because \(24 = 2^3 * 3\) and \(36 = 2^2 * 3^2\) (when prime-factorized)
n which contains the product of \(2^{46} * 3^{22}\) and we know that \(\frac{n}{{24}^p}\) is an integer | Max value(p) = 15.
n which contains the product of \(2^{46} * 3^{22}\) and we know that \(\frac{n}{{36}^q}\) is an integer | Max value(q) = 11.
Therefore, the maximum possible value for the sum of p and q is 15+11 =
26(Option D)P.S The exponent of any prime number p in n! = \([\frac{n}{p^1}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]\)+.. where [ ] is the greatest integer function.
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