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Updated on: 22 Mar 2025, 09:55
Originally posted by EgmatQuantExpert on 04 May 2015, 05:43.
Last edited by Bunuel on 22 Mar 2025, 09:55, edited 7 times in total.
Last edited by Bunuel on 22 Mar 2025, 09:55, edited 7 times in total.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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X is the largest prime number less than positive integer N. P is an integer such that P = X – 16. Also, Z = 1*2*...*\(\sqrt{P}\). If N is the first non-zero perfect square whose tens digit and units digit are same, How many different prime factors does Z have?
A. 4
B. 5
C. 6
D. 160
E. 320
Question 1 of the e-GMAT Primes Trio: 3 Questions on Number of factors and prime factors
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Regards,
The e-GMAT Quant Team
P.S.: Solutions with clarity of thought and elegance will get kudos! :-D
A. 4
B. 5
C. 6
D. 160
E. 320
Question 1 of the e-GMAT Primes Trio: 3 Questions on Number of factors and prime factors
Next Question
Regards,
The e-GMAT Quant Team
P.S.: Solutions with clarity of thought and elegance will get kudos! :-D
04 May 2015, 06:25
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If N is the first perfect square whose tens digit and units digit are same, then N=100.
X is the largest prime number less than positive integer N , then X is
97.
P is an integer such that P = X – 16.
So P=81
Z=(P^1/2)! = 9!
How many different prime factors does Z have?
Answer 4 (2,3,5,7)
X is the largest prime number less than positive integer N , then X is
97.
P is an integer such that P = X – 16.
So P=81
Z=(P^1/2)! = 9!
How many different prime factors does Z have?
Answer 4 (2,3,5,7)
05 May 2015, 23:18
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Here is the solution for this question. 
Given: We are given that \(X\) is the largest prime number less than positive integer \(N\) and that \(N\) is the first perfect square whose tens digit and units digit are same.
We are also given that \(Z = 1 * 2 * 3 * … * \sqrt{P}\), where \(P = X – 16\)
The question asks us to find the number of prime factors of \(Z\).
Approach: Since \(Z\) is of the form of a factorial, q! in this case. (where \(q = \sqrt{P}\)), the number of prime factors of \(Z\) will be simply the number of prime numbers less than or equal to \(\sqrt{P}\).
(For example, \(4!\) is simply \(1*2*3*4 = 1*2*3*2^2\). As you can see the only prime factors of \(4!\) are \(2\) and \(3\) which are essentially the prime numbers less than or equal to \(4\) itself.)
Once we find the value of \(\sqrt{P}\), the problem simply boils down to counting the number of prime numbers less than \(\sqrt{P}\).
To find \(\sqrt{P}\), we need the value of \(P\) and to find the value of \(P\), we need the value of \(X\). (Since \(P = X – 16\))
Since \(X\) is the largest prime number less than \(N\), to find the value of \(X\), we simply need the value of \(N\).
\(N\) has two constraints over it.
Using these constraints, let us try to determine the value of \(N\) and then work backwards to solve the problem.
Working Out: Since \(N\) is a non-zero perfect square, possible values of \(N\) are: \(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121\)
Notice that \(100\) is the first value that satisfies the condition that the tens digit and units digit of \(N\) must be same.
Therefore \(N = 100\).
Since \(X\) is the largest prime number less than \(100\), \(X\) must be \(97\).
This gives us \(P = 97 – 16 = 81\) which in turn gives us \(\sqrt{P} = 9\).
This means \(Z\) is simply \(1*2*…*9 = 9!\)
Therefore prime factors of \(Z\) are simply the prime numbers less than or equal to \(9\), which are \(2, 3, 5, and 7\).
Therefore \(Z\) has \(4\) different prime factors.
Correct Answer: Option A
Foot Note: If you are unclear about how the number of different prime factors of n! is same as number of prime numbers less than or equal to n itself, then you can refer to this post.
if-n-is-the-product-of-the-integers-from-1-to-8-inclusive-135542.html#p1522399
Here is another question that tests your conceptual understanding of primes and factors.
x-y-are-integers-find-the-number-of-even-factors-of-4x-197375.html
Regards,
Krishna
Given: We are given that \(X\) is the largest prime number less than positive integer \(N\) and that \(N\) is the first perfect square whose tens digit and units digit are same.
We are also given that \(Z = 1 * 2 * 3 * … * \sqrt{P}\), where \(P = X – 16\)
The question asks us to find the number of prime factors of \(Z\).
Approach: Since \(Z\) is of the form of a factorial, q! in this case. (where \(q = \sqrt{P}\)), the number of prime factors of \(Z\) will be simply the number of prime numbers less than or equal to \(\sqrt{P}\).
(For example, \(4!\) is simply \(1*2*3*4 = 1*2*3*2^2\). As you can see the only prime factors of \(4!\) are \(2\) and \(3\) which are essentially the prime numbers less than or equal to \(4\) itself.)
Once we find the value of \(\sqrt{P}\), the problem simply boils down to counting the number of prime numbers less than \(\sqrt{P}\).
To find \(\sqrt{P}\), we need the value of \(P\) and to find the value of \(P\), we need the value of \(X\). (Since \(P = X – 16\))
Since \(X\) is the largest prime number less than \(N\), to find the value of \(X\), we simply need the value of \(N\).
\(N\) has two constraints over it.
- a. \(N\) is a non-zero perfect square.
b. The tens digit of \(N\) = Units digit of \(N\)
Using these constraints, let us try to determine the value of \(N\) and then work backwards to solve the problem.
Working Out: Since \(N\) is a non-zero perfect square, possible values of \(N\) are: \(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121\)
Notice that \(100\) is the first value that satisfies the condition that the tens digit and units digit of \(N\) must be same.
Therefore \(N = 100\).
Since \(X\) is the largest prime number less than \(100\), \(X\) must be \(97\).
This gives us \(P = 97 – 16 = 81\) which in turn gives us \(\sqrt{P} = 9\).
This means \(Z\) is simply \(1*2*…*9 = 9!\)
Therefore prime factors of \(Z\) are simply the prime numbers less than or equal to \(9\), which are \(2, 3, 5, and 7\).
Therefore \(Z\) has \(4\) different prime factors.
Correct Answer: Option A
Foot Note: If you are unclear about how the number of different prime factors of n! is same as number of prime numbers less than or equal to n itself, then you can refer to this post.
if-n-is-the-product-of-the-integers-from-1-to-8-inclusive-135542.html#p1522399
Here is another question that tests your conceptual understanding of primes and factors.
x-y-are-integers-find-the-number-of-even-factors-of-4x-197375.html
Regards,
Krishna
