e-GMAT Question of the Week #1

Solution
Given:

• 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

To find:

• The maximum possible value of p + q

Approach and Working:

• 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

Now, n is defined as the product of the first 49 natural numbers – means n = 49!

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.

If we factorise the numbers 24 and 36, we get

• 24 = 2^3 x 3^1
  36 = 2^2 x 3^2

As the numbers 24 and 36 both consist of powers of 2 and 3 only, first we will find out the instances of 2 and 3 individually in 49!

• 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

Considering the number 24, as it is equal to 2^3 x 3^1

• Number of 23s present = 46/3 = 15
  Number of 31s present = 22/1 = 22
  Hence, number of combinations possible for 2^3 ∗ 3^1 =15

Therefore, we can say highest power of 24 present in 49! = max (p) = 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

Therefore, we can say highest power of 36 present in 49! = max (q) = 11
As, we have the maximum values of p and q respectively, we can say

• Max (p + q) = 15 + 11 = 26

Hence, the correct answer is option D.
Answer: D