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3
12 00:00

Difficulty:   55% (hard)

Question Stats: 63% (01:47) correct 37% (02:09) wrong based on 421 sessions

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8
2
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.

Hope it's clear.
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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, $m denotes the product of [#permalink] ### Show Tags megafan wrote: For any integer m greater than 1,$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..

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For any integer m greater than 1, $m denotes the product of [#permalink] ### Show Tags Bunuel wrote: megafan wrote: For any integer m greater than 1,$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 _________________ Would I rather be feared or loved? Easy. Both. I want people to be afraid of how much they love me. How to sort questions by Topic, Difficulty, and Source: https://gmatclub.com/forum/search.php?view=search_tags "Genuine learning is impossible without curiosity." - Naval Math Expert V Joined: 02 Sep 2009 Posts: 64318 Re: For any integer m greater than 1,$m denotes the product of  [#permalink]

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

We see that $m is the conventional notation of m!. Thus, the problem asks for the number of prime numbers between 7! + 2 and 7! + 10, inclusive. Let’s analyze each of these numbers. 7! + 2: Since 2 divides into 7! and 2, 7! + 2 has 2 as a factor, and thus it’s not a prime. 7! + 3: Since 3 divides into 7! and 3, 7! + 3 has 3 as a factor, and thus it’s not a prime. 7! + 4: Since 4 divides into 7! and 4, 7! + 4 has 4 as a factor, and thus it’s not a prime. 7! + 5: Since 5 divides into 7! and 5, 7! + 5 has 5 as a factor, and thus it’s not a prime. 7! + 6: Since 6 divides into 7! and 6, 7! + 6 has 6 as a factor, and thus it’s not a prime. 7! + 7: Since 7 divides into 7! and 7, 7! + 7 has 7 as a factor, and thus it’s not a prime. 7! + 8: Since 8 divides into 7! (notice that 7! has factors 2 and 4) and 8, 7! + 8 has 8 as a factor, and thus it’s not a prime. 7! + 9: Since 9 divides into 7! (notice that 7! has factors 3 and 6) and 9, 7! + 9 has 9 as a factor, and thus it’s not a prime. 7! + 10: Since 10 divides into 7! (notice that 7! has factors 2 and 5) and 10, 7! + 10 has 10 as a factor, and thus it’s not a prime. Thus, none of the integers between 7! + 2 and 7! + 10, inclusive, are prime. Answer: A _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Senior Manager  P Joined: 05 Feb 2018 Posts: 440 Re: For any integer m greater than 1,$m denotes the product of  [#permalink]

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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 $7 + 10, inclusive? (A) None (B) One (C) Two (D) Three (E) Four Question asks if there's a prime number between 7!+2 and 7!+10 7! = 7*6*5*4*3*2*1 We can check to see if from +2 to +10 there are any extra factors we can't factor out from 7! 2 to 7 seems self-evident... there's always a factor of each in 7! 8 is 2^3 and we have that from 2 and 4=(2^2) 9 is 3^2 and we have that from 3 and 6=(2*3) 10 is 5*2 and we have that as well. So there's no primes in this range, E. Re: For any integer m greater than 1,$m denotes the product of   [#permalink] 20 Jan 2019, 18:35

# For any integer m greater than 1, \$m denotes the product of   