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)

wdgolen -

Yes, that's a rule you will need to remember. Selango showed in his equations on the proof of such rule.