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P(m)

by venmic » Sat Oct 01, 2011 10:17 pm
what is the best way to answer the below question


For any integer m greater than 1, P(m) denotes the product of all the integers from 1 to m, inclusive. How many prime numbers are there between P(9) + 2 and P(9) + 8, inclusive?

None

(A) None

(B) One

(C) Two

(D) Three

(E) Four
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by knight247 » Sat Oct 01, 2011 10:31 pm
What the problem says is that P(m)=m!
Now the factorial of any number greater than 1 is ALWAYS even.

P(9)=9! and we need to find how many primes are there betn 9!+2 and 9!+8
Lets list all the numbers


9!+2
9!+3
9!+4
9!+5
9!+6
9!+7
9!+8

In the above list, the highlighted ones are all even as they are the sum of two even numbers. And we know that Even ± Even = Even

So, we have left
9!+3
9!+5
9!+7

Now, remember that when two multiples of a certain number are added/subtracted the resultant number is also a multiple of the same number.

9! is a multiple of 3 so when added with 3 the sum is going to be a multiple of 3.
9! is a multiple of seven as 9!=9*8*7.... And when added to 7 the sum is going to be a multiple of 7
9! is a multiple of 5 as 9!=9*8*7*6*5.... And when added to 5 the sum is going to be a multiple of 5

Seeing that all of the seven numbers are composite multiples of some number, none of them are prime.
Hence A
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by venmic » Sun Oct 02, 2011 6:09 am
Awesome thanks knight great explanation .. is this like a 700+ problem
knight247 wrote:What the problem says is that P(m)=m!
Now the factorial of any number greater than 1 is ALWAYS even.

P(9)=9! and we need to find how many primes are there betn 9!+2 and 9!+8
Lets list all the numbers


9!+2
9!+3
9!+4
9!+5
9!+6
9!+7
9!+8

In the above list, the highlighted ones are all even as they are the sum of two even numbers. And we know that Even ± Even = Even

So, we have left
9!+3
9!+5
9!+7

Now, remember that when two multiples of a certain number are added/subtracted the resultant number is also a multiple of the same number.

9! is a multiple of 3 so when added with 3 the sum is going to be a multiple of 3.
9! is a multiple of seven as 9!=9*8*7.... And when added to 7 the sum is going to be a multiple of 7
9! is a multiple of 5 as 9!=9*8*7*6*5.... And when added to 5 the sum is going to be a multiple of 5

Seeing that all of the seven numbers are composite multiples of some number, none of them are prime.
Hence A
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by gmatboost » Mon Oct 03, 2011 10:36 am
This is either an exact copy or an almost exact copy of an OG question. Make sure that you read all the OG questions and explanations (where needed). You should be comfortable with the reasoning used in every single one.
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