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For any integer m greater than 1, $m denotes the product of
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04 Mar 2013, 11:31
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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
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Re: For any integer m greater than 1, $m denotes the product of
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04 Mar 2013, 11:40
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 were \(2\leq{k}\leq{10}\) we can factor out k, thus there are no pries 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.
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Re: For any integer m greater than 1, $m denotes the product of
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04 Mar 2013, 15:59
Nice way to solve the problem. If you factor out the k, it should be 7!+3=3(2*4*5*6*7+1) ... 7!+10=10(3*4*6*7+1) Quote: For example: 7!+2=2(3*4*5*6*7+1) > a multiple of 2, thus not a prime; 7!+2=3(2*4*5*6*7+1) > a multiple of 3, thus not a prime; ... 7!+10=2(3*4*6*7+1) > a multiple of 10, thus not a prime.



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Re: For any integer m greater than 1, $m denotes the product of
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08 May 2017, 11:31
7!+2=2(3*4*5*6*7+1)=not prime; 7!+2=3(2*4*5*6*7+1)=not prime; ......................................... 7!+10=10(3*4*6*7+1) = not prime.



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Re: For any integer m greater than 1, $m denotes the product of
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23 Oct 2017, 22:57
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 megafan its a really nice question and a real 700 level question which tests one's aptitude of quant. I just tried to calculate the value of 7! and then tried to find number which was surely a wrong approach. Bunuel approach and solution is awesome, which added one more trick to my quant bucket. 7!+2 = 2(7*6*...*3) 7!+3 = 3(7*6..*4*2) 7!+4 = 4(7*6*5*3*2) .. .. 7!+10 = 10(7*6*4*3) So, there are no prime numbers between $7 + 2 and $7 + 10, inclusive.. Answer A
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For any integer m greater than 1, $m denotes the product of
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11 May 2018, 05:16
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 w here \(2\leq{k}\leq{10}\) we can factor out k, thus there are no pri mes 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
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Re: For any integer m greater than 1, $m denotes the product of
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11 May 2018, 09:50
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 w here \(2\leq{k}\leq{10}\) we can factor out k, thus there are no pri mes 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 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
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14 May 2018, 16:41
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
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