Thanks Geva! Can you elaborate on the last point a little more [b]"there's no common factor (other than 1) you can extract from this both 5! and 7, so the sum is not divisible by any integer other than 1 and itself. "[/b]
I still don't quite see the logic there yet.
[quote="Geva@MasterGMAT"][quote="yellowho"]For any integer p greater than 1, &p& denotes the product of
all the integers from 1 to p, inclusive. How many prime
numbers are there between &5& and &5& +7, inclusive?
How do you do this without multiplying out?
The only way I can think of is factor out 5!+1,5!+2, 5!+3,etc..... all the way to 5!+6 and 5!+7. Then you have to multiply and test.[/quote]
You dont need to actually factor each number: all you care about is whether the sum of 5!+integer is divisible by another integer or not: If it's divisible by another number, it's not prime.
The principle tested here is that result of adding multiples of the same number. Multiple of x + multiple of x must also result with a multiple of x, since you can extract x as a common factor.
For example, 5!+2 is divisible by 2, since 5! is also a multiple of 2: 5! is equal to 5*4*3*2*1, so it is possible to extract 2 as a common factor:
5!+2 = 2(5*4*3*1 +1)
So by the same rationale, 5!+any integer up to 5 will not be prime, since 5! itself is also divisible by the integers up to 5, resulting with a sum of two multiples of x.
The same could be said about 6: 5!+6 is divisible by 3 and 2, since both 5! and 6 are multiples of 3 and 2. So 5!+6 is not prime.
The only prime here is 5!+7 - there's no common factor (other than 1) you can extract from this both 5! and 7, so the sum is not divisible by any integer other than 1 and itself.
Answer is 1 - there's only 1 prime from the 7 integers described.[/quote]
