The Number n Is The Product Of The First 49 Natural Numbers GMAT Problem-Solving

This is a GMAT Problem solving question. Here, the data given in the questions has to be analyzed to answer the question. Several areas of mathematics can be involved in the process. The way options are given is very close to correct answer, and normal guessing can lead to mistakes. Students need to understand the question properly and use proper methods to approach the answer.

Given:
The product of the first 49 natural numbers yields the number n.
The digits \(n/(24)^p\) and \(n/(36)^p\) are digits

To determine: • The highest value that p + q can have

If \(n/(24)^p\) is given as an integer, then p is unambiguously the maximum power of 24 that can divide n.

•         In a similar way, we can assert that q is the largest power of 36 that can divide the number n if \(n/(36)^p\) is an integer.

Due to the fact that n is now defined as the sum of the first 49 natural numbers, n is equal to 49.
In other words, we're just looking for the highest power of 24 (which is p) and 36 (which is q) that can divide 49!

• Because the numbers 24 and 36 are composite numbers, we must represent them in terms of their prime factors and then identify each instance of those prime factors in order to get their maximum power.

The result of factoring the numbers 24 and 36 is: • 24 = 2^3 * 3^1

• 36 = 2^2 ∗ 3^2

We will first discover the specific instances of 2 and 3 in the number 49 as both 24 and 36 are just powers of 2 and 3.

• The number of 2s in 49! = 49/2 + 49/2^2 + 49/2^3 + 49/2^4 + 49/2^5 = 24 + 12 + 6 + 3+1 = 46.

• The amount of 3s in 49! = 49/3 + 49/3^2 + 49 / 3^3 = 16 + 5 + 1 = 22.

Taking into account that the number 24 is equal to 2^3 * 3^1

Number of 3^1s present = 22/1 = 22
Number of 2^3s present = 46/3 = 15

• As a result, the number of viable combinations for 2^3 * 3^1 = 15.

So, the highest power of 24 is present in 49! = max (p) = 15
In a similar way, taking into account the fact that the number 36 is equal to 2^2 * 3^2
Number of 2^2s present = 46/2 = 23
Number of 3^2s present = 22/2 = 11

• As a result, the number of viable combinations for 2^2 * 3^2= 11.

Consequently, we can state that 49 has the highest power of 36. = max (q) = 11
We may state that Max (p + q) = 15 + 11 = 26 because we have the maximum values of p and q, respectively.

As a result, choice D is the right response.

This is a GMAT Problem solving question. Here, the data given in the questions has to be analyzed to answer the question. Several areas of mathematics can be involved in the process. The way options are given is very close to correct answer, and normal guessing can lead to mistakes. Students need to understand the question properly and use proper methods to approach the answer.
Based on the given information, we can determine the maximum values of p and q, which represent the highest powers of 24 and 36, respectively, that can divide the number 49!.

The prime factorization of 24 is 2^3 * 3^1, and the prime factorization of 36 is 2^2 * 3^2.
To determine the highest power of 2 that can divide 49!, we calculate the number of 2s in 49! using the formula: 49/2 + 49/2^2 + 49/2^3 + 49/2^4 + 49/2^5 = 24 + 12 + 6 + 3 + 1 = 46.
Similarly, to determine the highest power of 3 that can divide 49!, we calculate the number of 3s in 49! using the formula: 49/3 + 49/3^2 + 49/3^3 = 16 + 5 + 1 = 22.
Since the maximum power of 2^3 (from 24) that can divide 49! is 46/3 = 15, and the maximum power of 3^2 (from 36) that can divide 49! is 22/2 = 11, the highest value that p + q can have is 15 + 11 = 26.

The right response is D.

This is a GMAT Problem solving question. Here, the data given in the questions has to be analyzed to answer the question. Several areas of mathematics can be involved in the process. The way options are given is very close to correct answer, and normal guessing can lead to mistakes. Students need to understand the question properly and use proper methods to approach the answer.

Both 24 and 36 have a prime factorization of 23 * 31, whereas 36 has a prime factorization of 22 * 32.
We count the amount of 2s in 49! to discover the largest power of 2 that can divide 49! 49/2 + 49/2 + 49/2 + 49/2 + 49/2 + 49/2 = 24 + 12 + 6 + 3 + 1 = 46 is the formula.
Similar to this, we count the amount of 3s in 49! using the following formula: 49/3 + 49/32 + 49/33 = 16 + 5 + 1 = 22. This yields the maximum power of 3 that can divide 49!.
The largest number that p + q can have is 15 + 11 = 26 because the highest power of 23 (from 24) that can divide 49! is 46/3 = 15, and the highest power of 32 (from 36) that can divide 49! is 22/2 = 11.

The appropriate answer is D.