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Updated on: 18 Feb 2019, 01:51
Originally posted by EgmatQuantExpert on 01 Jun 2018, 11:57.
Last edited by EgmatQuantExpert on 18 Feb 2019, 01:51, edited 5 times in total.
Last edited by EgmatQuantExpert on 18 Feb 2019, 01:51, edited 5 times in total.
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D
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Difficulty:
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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?
To access all the questions: Question of the Week: Consolidated List
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 the questions: Question of the Week: Consolidated List
01 Jun 2018, 12:20
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EgmatQuantExpert
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.
05 Jun 2018, 11:59
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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
- • \(24 = 2^3 * 3^1\)
• \(36 = 2^2 * 3^2\)
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 \(49! = \frac{49}{2} + \frac{49}{2^2} + \frac{49}{2^3} + \frac{49}{2^4} + \frac{49}{2^5} = 24 + 12 + 6 + 3 + 1 = 46\)
• The number of 3s present in \(49! = \frac{49}{3} + \frac{49}{3^2} + \frac{49}{3^3} = 16 + 5 + 1 = 22\)
Considering the number 24, as it is equal to \(2^3 * 3^1\)
- • Number of \(2^3\)s present = \(\frac{46}{3}\) = 15
• Number of \(3^1\)s present = \(\frac{22}{1}\) = 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 \(2^2 * 3^2\)
- • Number of \(2^2\)s present = \(\frac{46}{2}\) = 23
• Number of \(3^2\)s present = \(\frac{22}{2}\) = 11
• Hence, number of combinations possible for \(2^2 * 3^2 = 11\)
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
