For any integer greater than 1, Ln denotes the product of all integers from 1 to n, inclusive. How many prime numbers are there between L6 + 2 and L6 + 6, inclusive? (Note, the "L" is used as a symbol in the original question.) (A) None (B) One (C) Two (D) Three (E) Four I get the long way of getting 6! + 2 and 6! + 6, and then determining primes between them. The book has a shorter way that I'm hoping someone can figure out and explain.

Is the answer A? Multiple of x + multiple of x will be divisible by x. 3(2) + 3 = divisible by 3. So 6! = multiple of 2, 3, 4, 5, 6; and +2 --> +6 are all within those multiples. So any of those sum should result with the number being divisible, and thus there isn't any prime between the them? Hopefully I am not wrong on this, or I might have to go on a drinking binge.

uwhusky wrote:Is the answer A? Multiple of x + multiple of x will be divisible by x. 3(2) + 3 = divisible by 3. So 6! = multiple of 2, 3, 4, 5, 6; and +2 --> +6 are all within those multiples. So any of those sum should result with the number being divisible, and thus there isn't any prime between the them? Hopefully I am not wrong on this, or I might have to go on a drinking binge. You dont have to. I think you are right!!

If I am wrong, I guess we're both wrong...and I have to buy drinks =).

L6=(1.2.3.4.5.6) L6+2=2+(1.2.3.4.5.6)=2[1+(1.3.4.5.6)]-->Multiple of 2.Not prime L6+3=3+(1.2.3.4.5.6)=3[1+(1.2.4.5.6)]-->Multiple of 3.Not prime L6+4=4+(1.2.3.4.5.6)=4[1+(1.2.3.5.6)]-->Multiple of 4.Not prime L6+5=5+(1.2.3.4.5.6)=5[1+(1.2.3.4.6)]-->Multiple of 5.Not prime L6+6=6+(1.2.3.4.5.6)=6[1+(1.2.3.4.5)]-->Multiple of 6.Not prime No prime between L6+2 and L6+6 Note that we don't repeat all the steps.Just from step 1 itself we can deduce the rem steps. Pick A

uwhusky - you're right, answer is (A). I wasn't familiar with Multiple of x + multiple of x will be divisible by x. . Is that a named rule I should know? selengo - I take it you're factoring out a "2" in the first step you show? L6+2 = 2 + (1 x 2 x 3 x 4 x 5 x 6) = 2 + 2(1 x 3 x 4 x 5 x 6) = 2[1 + 1(1 x 3 x 4 x 5 x 6)] = 2(1 + 3 x 4 x 5 x 6)