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# If F is the prime factorization of N!, how many factors in F have an e

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If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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09 Apr 2015, 06:33
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If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 32 ≤ N ≤ 40

(2) 27 ≤ N ≤ 35

Kudos for a correct solution.

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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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09 Apr 2015, 08:42
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Bunuel wrote:
If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 25 ≤ N ≤ 35

Kudos for a correct solution.

1) N can be 30 and have 17, 19, 23, 29 with exponent 1 or can be 31 and have 17, 19, 23, 29, 31 with exponent 1 (4 and 5 prime factors) Insufficient
2) N can be 30 and have 17, 19, 23, 29 with exponent 1 or can be 31 and have 17, 19, 23, 29, 31 with exponent 1 (4 and 5 prime factors) Insufficient

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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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09 Apr 2015, 09:07
1
Bunuel wrote:
If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 25 ≤ N ≤ 35

Kudos for a correct solution.

For statement A for 40!, the factors with exponent 1 would be 23, 29, 31, 37 etc. If N=40, there could be upto 5 factors with exponent 1. (as 34 = 17*2 and 38=19*2)
and if N=30, there will be upto 4 factors with exponent 1 (13, 17, 19, 23 and 29). Insufficient

For statement B for 35!, the factors with exponent 1 would be 19, 23, 29, 31 etc. If N=35, there could be upto 4 factors with exponent 1. (as 34 = 17*2)
and if N=25, there will be upto 5 factors with exponent 1 (13, 17, 19, 23 and 29). Insufficient

Combining A & B, 25 ≤ N ≤ 40; again, N can assume any value, say 30 or 40, for which the no. of factors with exponents 1 would vary. Insufficient

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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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Updated on: 09 Apr 2015, 23:24
Bunuel wrote:
If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 25 ≤ N ≤ 35

Kudos for a correct solution.

(1) 30 ≤ N ≤ 40
If N=30 There are say X prime factors with exponent 1.
for N=31 , we have X+1 prime factors with exponent 1.
so we cannot find the total number of primer factor with exponent 1 till we do not know value of N .
Insufficient.

(2) 25 ≤ N ≤ 35
same issue with this statement
if N=25 , in N!, we will have Y prime factors with exponent 1
N=29, Y+1
N=31, Y+2
Insufficient.

Together:
$${30}\le{N}\leq{35}$$
If N=30 There are X prime factors with exponent 1.
but If N=31 There are X+1 prime factors with exponent 1.

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Originally posted by Lucky2783 on 09 Apr 2015, 23:07.
Last edited by Lucky2783 on 09 Apr 2015, 23:24, edited 1 time in total.
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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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09 Apr 2015, 23:17
King407 wrote:
Bunuel wrote:
If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 25 ≤ N ≤ 35

Kudos for a correct solution.

For statement A for 40!, the factors with exponent 1 would be 23, 29, 31, 37 etc. If N=40, there could be upto 5 factors with exponent 1. (as 34 = 17*2 and 38=19*2)
and if N=30, there will be upto 4 factors with exponent 1 (13, 17, 19, 23 and 29). Insufficient

For statement B for 35!, the factors with exponent 1 would be 19, 23, 29, 31 etc. If N=35, there could be upto 4 factors with exponent 1. (as 34 = 17*2)
and if N=25, there will be upto 5 factors with exponent 1 (13, 17, 19, 23 and 29). Insufficient

Combining A & B, 25 ≤ N ≤ 40; again, N can assume any value, say 30 or 40, for which the no. of factors with exponents 1 would vary. Insufficient

This is not important for this question but it can be a pitfall in another question:
combining A & B gives us $${30}\le{N}\leq{35}$$
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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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13 Apr 2015, 04:43
Bunuel wrote:
If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 25 ≤ N ≤ 35

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

This is a tricky one.

Let’s think about, say, 40! This number, 40!, is the product of all the integers from one to 40. Let’s think about its prime factorization. It would have at least one factor of 2 for every even number from 2 to 40, and a second factor for every multiple of 4, and a third factor for every multiple of 8, etc.; a lot of factors of two. Think about the factors, say, of 7: there are five multiples of 7 from 7 to 35, so in the prime factorization of 40!, the factor 7 would have an exponent of 5. Which factors would have exponents of 1? Well, the prime numbers that are less than N, but have no other multiples less than N. For example, in 40!, the factor 37 would have an exponent of 1 since it appears once and no other multiple of it is less than 40.

Statement #1: 32 ≤ N ≤ 40

As we move through different N’s in this region, we cross the prime number 37, which will have an exponent of 1 if it appears. Some N’s include this prime number and some don’t, so the number of factors with an exponent of 1 is different for different values of N. This statement, alone and by itself, is not sufficient.

Statement #2: 27 ≤ N ≤ 35

As we move through different N’s in this region, we cross two prime numbers, 29 and 31, each of which will have an exponent of 1 if it appears. Some N’s include neither, some include 29 and not 31, and some include both, so the number of factors with an exponent of 1 is different for different values of N. This statement, alone and by itself, is not sufficient.

Combined: 32 ≤ N ≤ 35

Now, there are no prime values in the range specified. But, here’s a tricky thing. If N = 32 or 33, then either 32! or 33! contains exactly one factor of the prime numbers {17, 23, 29, 31}: four prime factors with an exponent of one. BUT, if N = 34 or 35, there are now two factors of 17 (one from 17 and one from 34), either 34! or 35! contains exactly one factor of the prime numbers {23, 29, 31}: three prime factors with an exponent of one. Even in this narrow range, different choices lead to different answers for the prompt question. Even together, the statements are not sufficient.

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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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18 Jun 2016, 13:15
Lucky2783 wrote:
Bunuel wrote:
If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 25 ≤ N ≤ 35

Kudos for a correct solution.

(1) 30 ≤ N ≤ 40
If N=30 There are say X prime factors with exponent 1.
for N=31 , we have X+1 prime factors with exponent 1.
so we cannot find the total number of primer factor with exponent 1 till we do not know value of N .
Insufficient.

(2) 25 ≤ N ≤ 35
same issue with this statement
if N=25 , in N!, we will have Y prime factors with exponent 1
N=29, Y+1
N=31, Y+2
Insufficient.

Together:
$${30}\le{N}\leq{35}$$
If N=30 There are X prime factors with exponent 1.
but If N=31 There are X+1 prime factors with exponent 1.

This is where I started. I figured if I could find a difference in prime factors in 30<=N<=35, I could go straight to answer E. Did I over look something?
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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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12 Jul 2016, 07:13
Bunuel wrote:
Bunuel wrote:
If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 25 ≤ N ≤ 35

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

This is a tricky one.

Let’s think about, say, 40! This number, 40!, is the product of all the integers from one to 40. Let’s think about its prime factorization. It would have at least one factor of 2 for every even number from 2 to 40, and a second factor for every multiple of 4, and a third factor for every multiple of 8, etc.; a lot of factors of two. Think about the factors, say, of 7: there are five multiples of 7 from 7 to 35, so in the prime factorization of 40!, the factor 7 would have an exponent of 5. Which factors would have exponents of 1? Well, the prime numbers that are less than N, but have no other multiples less than N. For example, in 40!, the factor 37 would have an exponent of 1 since it appears once and no other multiple of it is less than 40.

Statement #1: 32 ≤ N ≤ 40

As we move through different N’s in this region, we cross the prime number 37, which will have an exponent of 1 if it appears. Some N’s include this prime number and some don’t, so the number of factors with an exponent of 1 is different for different values of N. This statement, alone and by itself, is not sufficient.

Statement #2: 27 ≤ N ≤ 35

As we move through different N’s in this region, we cross two prime numbers, 29 and 31, each of which will have an exponent of 1 if it appears. Some N’s include neither, some include 29 and not 31, and some include both, so the number of factors with an exponent of 1 is different for different values of N. This statement, alone and by itself, is not sufficient.

Combined: 32 ≤ N ≤ 35

Now, there are no prime values in the range specified. But, here’s a tricky thing. If N = 32 or 33, then either 32! or 33! contains exactly one factor of the prime numbers {17, 23, 29, 31}: four prime factors with an exponent of one. BUT, if N = 34 or 35, there are now two factors of 17 (one from 17 and one from 34), either 34! or 35! contains exactly one factor of the prime numbers {23, 29, 31}: three prime factors with an exponent of one. Even in this narrow range, different choices lead to different answers for the prompt question. Even together, the statements are not sufficient.

Why 19 is missing from the set of exactly one factor of the prime numbers i.e. {17, 23, 29, 31} ?
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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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21 Aug 2016, 01:40
1
Dear Bunuel and Magoosh
I think this question has a language issue
here when you say that how many factors have power of one => " EVERY FACTOR OF THE NUMBER WILL HAVE A POWER OF ONE"
e.g => N=24 say
that N=2^3*3 => It has 8^1 ; 3^1,
or it can have 24^1
Hence i marked it E.
I know you will say it says Prime factorization.
But even if it says prime factorization then too the facyors would be the same .

GMAT like?
I doubt that..!

P.S => in both cases E is the answer

Regards
Stone Cold
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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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21 Aug 2016, 01:55
stonecold wrote:
Dear Bunuel and Magoosh
I think this question has a language issue
here when you say that how many factors have power of one => " EVERY FACTOR OF THE NUMBER WILL HAVE A POWER OF ONE"
e.g => N=24 say
that N=2^3*3 => It has 8^1 ; 3^1,
or it can have 24^1
Hence i marked it E.
I know you will say it says Prime factorization.
But even if it says prime factorization then too the facyors would be the same .

GMAT like?
I doubt that..!

P.S => in both cases E is the answer

Regards
Stone Cold

I think you are missing something here.

If the question step says prime factorization, then that doesn't mean we need to write same factors with an exponent of one.

For example, 24 = 2^3 *3 -- So, it has only One Prime Factor that has an exponent of one. It is NEVER 4.

Also, if it were not a prime factorization, then we could have

24 = 2^3 *3 -- 1 factor with exponent of one.

24 = 4 * 2 * 3 -- 3 factors with exponent of one.

24 = 2 * 3 -- 2 factors with exponent of one.

Hence, the question is correct as mentioned.

Am I am missing anything?
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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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21 Aug 2016, 02:13
abhimahna wrote:
stonecold wrote:
Dear Bunuel and Magoosh
I think this question has a language issue
here when you say that how many factors have power of one => " EVERY FACTOR OF THE NUMBER WILL HAVE A POWER OF ONE"
e.g => N=24 say
that N=2^3*3 => It has 8^1 ; 3^1,
or it can have 24^1
Hence i marked it E.
I know you will say it says Prime factorization.
But even if it says prime factorization then too the facyors would be the same .

GMAT like?
I doubt that..!

P.S => in both cases E is the answer

Regards
Stone Cold

I think you are missing something here.

If the question step says prime factorization, then that doesn't mean we need to write same factors with an exponent of one.

For example, 24 = 2^3 *3 -- So, it has only One Prime Factor that has an exponent of one. It is NEVER 4.

Also, if it were not a prime factorization, then we could have

24 = 2^3 *3 -- 1 factor with exponent of one.

24 = 4 * 2 * 3 -- 3 factors with exponent of one.

24 = 2 * 3 -- 2 factors with exponent of one.

Hence, the question is correct as mentioned.

Am I am missing anything?

Disagree ( i might b wrong too)
Its says FACTORS have the power of one not the prime factors
E.g => 24=2^3*3=2^2*2*3

Hmm now just as i write this i am getting a pretty weird feeling.

Anyways Try this one => is-x-divisible-by-61266.html?fl=similar
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Re: If F is the prime factorization of N!, how many factors in F have an e  [#permalink]

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21 Aug 2016, 09:12
1
stonecold wrote:
abhimahna wrote:
stonecold wrote:
Dear Bunuel and Magoosh
I think this question has a language issue
here when you say that how many factors have power of one => " EVERY FACTOR OF THE NUMBER WILL HAVE A POWER OF ONE"
e.g => N=24 say
that N=2^3*3 => It has 8^1 ; 3^1,
or it can have 24^1
Hence i marked it E.
I know you will say it says Prime factorization.
But even if it says prime factorization then too the facyors would be the same .

GMAT like?
I doubt that..!

P.S => in both cases E is the answer

Regards
Stone Cold

I think you are missing something here.

If the question step says prime factorization, then that doesn't mean we need to write same factors with an exponent of one.

For example, 24 = 2^3 *3 -- So, it has only One Prime Factor that has an exponent of one. It is NEVER 4.

Also, if it were not a prime factorization, then we could have

24 = 2^3 *3 -- 1 factor with exponent of one.

24 = 4 * 2 * 3 -- 3 factors with exponent of one.

24 = 2 * 3 -- 2 factors with exponent of one.

Hence, the question is correct as mentioned.

Am I am missing anything?

Disagree ( i might b wrong too)
Its says FACTORS have the power of one not the prime factors
E.g => 24=2^3*3=2^2*2*3

Hmm now just as i write this i am getting a pretty weird feeling.

Anyways Try this one => is-x-divisible-by-61266.html?fl=similar

if the questions starts with "If F is the prime factorization of N!", then how can you assume its not going to have prime factorization only. The question has itself made that clear.

And thanks for sharing that question. Its an easy one.
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Re: If F is the prime factorization of N!, how many factors in F have an  [#permalink]

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01 Feb 2017, 18:22
1
2
daboo343 wrote:
If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 32 ≤ N ≤ 40

(2) 27 ≤ N ≤ 35

To do this Q, we have to first understand what it means..
Prime number and exponent as 1 MEANS prime number that are coming only ones

we have to look for prime numbers from N/2 to N as PRIME numbers from 0 to N/2 will be ATLEAST 2 times.

Let's see the statements

(1) 32 ≤ N ≤ 40

So all prime numbers above 32/2 or 16 have to be checked
Check for 17..
32! will have 17^1 but 34! Will have 17^2..one of 17 and second of 34..
Insufficient

(2) 27 ≤ N ≤ 35
Same case of 32! And 34! Can be taken..
Insufficient

Combined
Case of 32! And 34! Still exists ..
Insufficient

E
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