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# For any integer m greater than 1, $m denotes the product of SORT BY: Tags: Show Tags Hide Tags Manager Joined: 28 May 2009 Posts: 138 Own Kudos [?]: 807 [25] Given Kudos: 91 Location: United States Concentration: Strategy, General Management GMAT Date: 03-22-2013 GPA: 3.57 WE:Information Technology (Consulting) Most Helpful Reply Math Expert Joined: 02 Sep 2009 Posts: 93447 Own Kudos [?]: 626273 [13] Given Kudos: 81954 General Discussion Intern Joined: 18 Feb 2013 Posts: 22 Own Kudos [?]: 77 [3] Given Kudos: 14 GMAT 1: 710 Q49 V38 Senior Manager Joined: 26 Jun 2017 Posts: 319 Own Kudos [?]: 327 [2] Given Kudos: 334 Location: Russian Federation Concentration: General Management, Strategy WE:Information Technology (Other) Re: For any integer m greater than 1,$m denotes the [#permalink]
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gmatexam439 wrote:
For any integer m greater than 1, $m denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between$7 + 2 and $7 + 10, inclusive? (A) None (B) One (C) Two (D) Three (E) Four$7 = 7*6*5*4*3*2*1=5040.
So we need to check all numbers between 5042 and 5050.
All even numbers are divisable by 2.
So what about 5043, 5045, 5047, 5049.
5045 is divisable by 5.
5043 and 5049 are divisable by 3, because the sum of all digits is divisable by 3.
And the last one - 5047. We can easily find that it is divisable by 7.
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Re: For any integer m greater than 1, $m denotes the [#permalink] 1 Kudos$m is m!, so between 7! + 2 and 7! + 10 how many prime number are there?

All of prime numbers except 2 are odd, 7! is even so only 7! + 3, 5, 7, or 9 can make an odd number.

Prime number is divisible by 1 and itself, examine the contenders:
- 7! divisible by 3, so adding 3 or 9 will result in a number divisible by 3.
- 7! is divisible by 5 and 7 also, so adding 5 or 7 will result in a number divisible by 5 or 7.

In sum, A- None.
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Re: For any integer m greater than 1, $m denotes the [#permalink] 1 Bookmarks Expert Reply gmatexam439 wrote: For any integer m greater than 1,$m denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between $7 + 2 and$7 + 10, inclusive?

(A) None
(B) One
(C) Two
(D) Three
(E) Four

hi..

$7 is nothing but 7!.. LOGIC why none is prime so we are looking at PRIMES between 7!+2 and 7!+10.. 7!= 1*2*3*4*5*6*7, so 7! is multiple of 1,2,3,4,5,6,7,8(2*5),9(3*6),10(2*5) so when you add any of these number starting from 2 till 10 to 7! the RESULTING SUM will be a MULTIPLE of that number 2,3,..or 10.. so NONE is PRIME A Target Test Prep Representative Joined: 14 Oct 2015 Status:Founder & CEO Affiliations: Target Test Prep Posts: 18895 Own Kudos [?]: 22311 [0] Given Kudos: 285 Location: United States (CA) Re: For any integer m greater than 1,$m denotes the [#permalink]
gmatexam439 wrote:
For any integer m greater than 1, $m denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between$7 + 2 and $7 + 10, inclusive? (A) None (B) One (C) Two (D) Three (E) Four We see that$m is the conventional notation of m!. Thus, the problem asks for the number of prime numbers between 7! + 2 and 7! + 10, inclusive. Let’s analyze each of these numbers.

7! + 2: Since 2 divides into 7! and 2, 7! + 2 has 2 as a factor, and thus it’s not a prime.

7! + 3: Since 3 divides into 7! and 3, 7! + 3 has 3 as a factor, and thus it’s not a prime.

7! + 4: Since 4 divides into 7! and 4, 7! + 4 has 4 as a factor, and thus it’s not a prime.

7! + 5: Since 5 divides into 7! and 5, 7! + 5 has 5 as a factor, and thus it’s not a prime.

7! + 6: Since 6 divides into 7! and 6, 7! + 6 has 6 as a factor, and thus it’s not a prime.

7! + 7: Since 7 divides into 7! and 7, 7! + 7 has 7 as a factor, and thus it’s not a prime.

7! + 8: Since 8 divides into 7! (notice that 7! has factors 2 and 4) and 8, 7! + 8 has 8 as a factor, and thus it’s not a prime.

7! + 9: Since 9 divides into 7! (notice that 7! has factors 3 and 6) and 9, 7! + 9 has 9 as a factor, and thus it’s not a prime.

7! + 10: Since 10 divides into 7! (notice that 7! has factors 2 and 5) and 10, 7! + 10 has 10 as a factor, and thus it’s not a prime.

Thus, none of the integers between 7! + 2 and 7! + 10, inclusive, are prime.

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For any integer m greater than 1, $m denotes the product of [#permalink] Bunuel wrote: megafan wrote: For any integer m greater than 1,$m denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between $7 + 2 and$7 + 10, inclusive?

(A) None
(B) One
(C) Two
(D) Three
(E) Four

$is basically a factorial of a number. So, we are asked to find the number of primes between 7!+2 and 7!+10, inclusive. From each number 7!+k where $$2\leq{k}\leq{10}$$ we can factor out k, thus there are no primes in the given range. For example: 7!+2=2(3*4*5*6*7+1) --> a multiple of 2, thus not a prime; 7!+3=3(2*4*5*6*7+1) --> a multiple of 3, thus not a prime; ... 7!+10=10(3*4*6*7+1) --> a multiple of 10, thus not a prime. Answer: A. Hope it's clear. Hi Bunuel , I am trying to better understand this concept. So with this thought process, would any number 7!+k where k is a postive integer lead to a prime number? Example k is 29, well 29(2*3*4*5*6*7+1) is not divisible by 29. it leads to 5069 which is prime. Do we know that any number k that is within 7! (7,6,5...) will be a factor of k itself? And then we just have to decide for k=8,9, and 10? k=8 and 10 are even so obviously not prime. And it just so happens k=9 leads to 5049 which is divisible by 9. I guess my question is how can we be sure that 7!+k where k is between 2 and 10 is prime? I see that 7!+7 is divisible by 7, but why is 7!+13 not divisible by 13? Thanks Math Expert Joined: 02 Sep 2009 Posts: 93447 Own Kudos [?]: 626273 [0] Given Kudos: 81954 Re: For any integer m greater than 1,$m denotes the product of [#permalink]
msurls wrote:
Bunuel wrote:
megafan wrote:
For any integer m greater than 1, $m denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between$7 + 2 and $7 + 10, inclusive? (A) None (B) One (C) Two (D) Three (E) Four$ is basically a factorial of a number.

So, we are asked to find the number of primes between 7!+2 and 7!+10, inclusive.

From each number 7!+k where $$2\leq{k}\leq{10}$$ we can factor out k, thus there are no primes in the given range.

For example:
7!+2=2(3*4*5*6*7+1) --> a multiple of 2, thus not a prime;
7!+3=3(2*4*5*6*7+1) --> a multiple of 3, thus not a prime;
...
7!+10=10(3*4*6*7+1) --> a multiple of 10, thus not a prime.

Hope it's clear.

Hi Bunuel , I am trying to better understand this concept. So with this thought process, would any number 7!+k where k is a postive integer lead to a prime number?

Example k is 29, well 29(2*3*4*5*6*7+1) is not divisible by 29. it leads to 5069 which is prime.

Do we know that any number k that is within 7! (7,6,5...) will be a factor of k itself? And then we just have to decide for k=8,9, and 10? k=8 and 10 are even so obviously not prime. And it just so happens k=9 leads to 5049 which is divisible by 9.

I guess my question is how can we be sure that 7!+k where k is between 2 and 10 is prime? I see that 7!+7 is divisible by 7, but why is 7!+13 not divisible by 13?

Thanks

7! + k will NOT be a prime if 7! and k have a common factor other than 1. In this case we would be able to factor out that factor out of 7! + k. For example, 7! + 7 is not a prime because 7! and 7 have 7 as a common factor so we can factor out 7 out of 7! + 7: 7! + 7 = 7(6! + 1).

If 7! and k does NOT have a common factor other than 1, then 7! + k might or might not be a prime. For example, 7! + 11 = 5051, which IS a prime but 7! + 13 = 5053, which is NOT a prime.
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Re: For any integer m greater than 1, $m denotes the product of [#permalink] 2 Kudos Expert Reply megafan wrote: For any integer m greater than 1,$m denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between $7 + 2 and$7 + 10, inclusive?

(A) None
(B) One
(C) Two
(D) Three
(E) Four

We see that $m is the conventional notation of m!. Thus, the problem asks for the number of prime numbers between 7! + 2 and 7! + 10, inclusive. Let’s analyze each of these numbers. 7! + 2: Since 2 divides into 7! and 2, 7! + 2 has 2 as a factor, and thus it’s not a prime. 7! + 3: Since 3 divides into 7! and 3, 7! + 3 has 3 as a factor, and thus it’s not a prime. 7! + 4: Since 4 divides into 7! and 4, 7! + 4 has 4 as a factor, and thus it’s not a prime. 7! + 5: Since 5 divides into 7! and 5, 7! + 5 has 5 as a factor, and thus it’s not a prime. 7! + 6: Since 6 divides into 7! and 6, 7! + 6 has 6 as a factor, and thus it’s not a prime. 7! + 7: Since 7 divides into 7! and 7, 7! + 7 has 7 as a factor, and thus it’s not a prime. 7! + 8: Since 8 divides into 7! (notice that 7! has factors 2 and 4) and 8, 7! + 8 has 8 as a factor, and thus it’s not a prime. 7! + 9: Since 9 divides into 7! (notice that 7! has factors 3 and 6) and 9, 7! + 9 has 9 as a factor, and thus it’s not a prime. 7! + 10: Since 10 divides into 7! (notice that 7! has factors 2 and 5) and 10, 7! + 10 has 10 as a factor, and thus it’s not a prime. Thus, none of the integers between 7! + 2 and 7! + 10, inclusive, are prime. Answer: A Non-Human User Joined: 09 Sep 2013 Posts: 33159 Own Kudos [?]: 829 [0] Given Kudos: 0 Re: For any integer m greater than 1,$m denotes the product of [#permalink]
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Re: For any integer m greater than 1, \$m denotes the product of [#permalink]
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