A. 11

B. 15

C. 20

D. 26

E. 30

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**A**

**B**

**C**

**D**

**E**

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

**To access all questions:** Consolidated list of Questions of the Week

A. 11

B. 15

C. 20

D. 26

E. 30

- e-GMAT
- GMAT Instructor
**Posts:**645**Joined:**26 Jul 2010**Location:**US**Thanked**: 527 times**Followed by:**226 members

- The number n is the product of the first 49 natural numbers

The numbers $$\frac{n}{24^p}$$ and $$\frac{n}{36^q}$$ are integers

- The maximum possible value of p + q

- When it is given that $$\frac{n}{24^p}$$ is an integer, it necessarily means p is the highest power of 24 that can divide the number n

In a similar way we can say that, if $$\frac{n}{36^q}$$ is an integer, then q is the highest power of 36 that can divide the number n

So, effectively we are trying to find out the highest power of 24 (which is p) and 36 (which is q) respectively which can divide 49!

- Now, as the numbers 24 and 36 are composite numbers, to find the highest power of them, we need to express them in terms of the prime factors and then figure out the individual instances of those prime factors.

- 24 = 2^3 x 3^1

36 = 2^2 x 3^2

- The number of 2s present in 49! = 49/2 + 49/22 + 49/23 + 49/24 + 49/25 = 24 + 12 + 6 + 3 + 1 = 46

The number of 3s present in 49! = 49/3 + 49/32 + 49/33 = 16 + 5 + 1 = 22

- Number of 23s present = 46/3 = 15

Number of 31s present = 22/1 = 22

Hence, number of combinations possible for 2^3 âˆ— 3^1 =15

In a similar way, considering the number 36, as it is equal to 2^2 x 3^2

- Number of 22s present = 46/2 = 23

Number of 32s present = 22/2 = 11

Hence, number of combinations possible for 2^2 x 3^2 =11

As, we have the maximum values of p and q respectively, we can say

- Max (p + q) = 15 + 11 = 26

Answer: D