IMO E is the answer!!!
Statement 1
If n*(n+2) is an odd number, then both n and n+2 are odd which implies n+1 is an even number
If n+1 is an even multiple of 2, \(12\) is a factor of \(n×(n+1)×(n+2)\)
But if n+1 is an odd multiple of 2, then \(12\) is not a factor of \(n×(n+1)×(n+2)\)
Insufficient
Statement 2
If n+1 is even, both n and n+2 are odd numbers
If n+1 is an even multiple of 2, \(12\) is a factor of \(n×(n+1)×(n+2)\)
But if n+1 is an odd multiple of 2, then \(12\) is not a factor of \(n×(n+1)×(n+2)\)
Combining both equations
Both statements implying that n+1 is even, but none tells us that n+1 is even multiple of 2
Insufficient!!
Dillesh4096 wrote:
Asad wrote:
If \(n\) is a positive integer, is \(12\) a factor of \(n×(n+1)×(n+2)\)?
1) \(n×(n+2)\) is odd
2) \(n+1\) is even
n×(n+1)×(n+2)
--> 2 possibilities:
1st case: odd*even*odd --> {1*2*3 or 3*4*5 or 5*6*7} = ALWAYS divisible by 6 (Need not be by 12)
2nd case: even*odd*even --> {2*3*4 or 4*5*6 or 6*7*8} = ALWAYS divisible by 24
1) n×(n+2) is odd --> 1st case = Sufficient (A definite NO)
--> n is odd
--> 1st case -
Sufficient (A definite NO)
2) (n+1) is even
--> n is odd
--> 1st case -
Sufficient (A definite NO)
IMO Option D
Pls Hit Kudos if you like the solution