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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.
So the answer is A.
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Re: For any integer m greater than 1, $m denotes the [#permalink]
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$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]
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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


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
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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


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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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
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Re: For any integer m greater than 1, $m denotes the product of [#permalink]
Expert Reply
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.

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 [#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


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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Re: For any integer m greater than 1, $m denotes the product of [#permalink]
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