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X is the largest prime number less than positive integer N. P is an in

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X is the largest prime number less than positive integer N. P is an in [#permalink]

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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

Here is a fresh question from the e-GMAT bakery! Go ahead and give it a shot! :)

This is Question 1 of the e-GMAT Primes Trio: 3 Questions on Number of factors and prime factors

Regards,
The e-GMAT Quant Team

P.S.: Solutions with clarity of thought and elegance will get kudos! :-D

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[Reveal] Spoiler: OA

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Last edited by EgmatQuantExpert on 13 Dec 2016, 07:13, edited 5 times in total.

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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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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)
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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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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.

    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
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X is the largest prime number less than positive integer N. If N is the first non-zero perfect square whose tens digit and units digit are same.

N ~ (25,36...81,100) = 100 Unit/Ten digit same

N= 100 , X largest prime less than 100 = 97 ; X = 97

P is an integer such that P = X – 16.
P = 97 - 16 = 81

Also, Z = 1*2*…*\(\sqrt{P}\). , How many different prime factors does Z have?

Z = 1*2*…*\(\sqrt{81}\).

Z = 1*2*3*4*5*6*7*8*9 = prime { 2,3,5,7} ~ 4

Ans : A
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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 07 May 2015, 10:52
One Doubt ... The question says that N is the first non zero perfect square ... 100 has zero in them .. shoudnt N be 144???

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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 17 May 2015, 22:58
Udai wrote:
One Doubt ... The question says that N is the first non zero perfect square... 100 has zero in them .. shoudnt N be 144???


Dear Udai

By non-zero perfect square, we mean a perfect square whose magnitude is not equal to zero. Yes, 100 does have 2 zeroes in it, but is the magnitude of 100 equal to zero? Not at all! Therefore, 100 doesn't violate the 'non-zero perfect square' bit.

Hope this helped! :)

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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 18 May 2015, 04:39
EgmatQuantExpert wrote:
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

Here is a fresh question from the e-GMAT bakery! Go ahead and give it a shot! :)

This is Question 1 of the e-GMAT Primes Trio: 3 Questions on Number of factors and prime factors

Regards,
The e-GMAT Quant Team

P.S.: Solutions with clarity of thought and elegance will get kudos! :-D


Would you say that this is a 650 level question? I got confused when reading the prompt on what I was suppose to do, but once I figured out what it was asking the problem only took me 1:45.
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Gmat prep 6 720 (48Q 41V)

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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 27 Mar 2016, 01:55
I was able to derive all information from the question prompt except that Z = 1*2*…*√P denotes the factorial of √P. Hence my question is:
Is Z = 1*2*…*√P a common way of denoting the factorial of √P? I have seen the factorial be denoted √P!, but I have never seen this denotation before.

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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 22 Aug 2017, 04:11
Pretty straight forward question.
However took around 3 minutes to solve.
But, I love the way e-gmat has built their course around such concepts.

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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 03 Sep 2017, 05:27
@e-gmat, abhimahna, and Bunuel

If it was not given that N is a non-zero integer, then could the value of N be 0?
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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 03 Sep 2017, 09:12
Shiv2016 wrote:
@e-gmat, abhimahna, and Bunuel

If it was not given that N is a non-zero integer, then could the value of N be 0?


Hi Shiv2016 ,

We are already given that N is a positive integer, which means N > 0

Hence, even if we were not given N is a first non zero perfect square, we cannot take N = 0

Does that make sense?
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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 03 Sep 2017, 09:26
abhimahna wrote:
Shiv2016 wrote:
@e-gmat, abhimahna, and Bunuel

If it was not given that N is a non-zero integer, then could the value of N be 0?


Hi Shiv2016 ,

We are already given that N is a positive integer, which means N > 0

Hence, even if we were not given N is a first non zero perfect square, we cannot take N = 0

Does that make sense?



If that information was also not given, then N could be anything - even 0?

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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 03 Sep 2017, 09:28
Shiv2016 wrote:
If that information was also not given, then N could be anything - even 0?


Yes, if we are not given anything about N, it could be +ve number, -ve number or zero. But yes, this question cannot be solved in that case.
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Re: X is the largest prime number less than positive integer N. P is an in [#permalink]

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New post 03 Sep 2017, 10:14
It's clear now. Thank you for your reply.
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Re: X is the largest prime number less than positive integer N. P is an in   [#permalink] 03 Sep 2017, 10:14
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