If n > 6, and if n is a multiple of 6, which of the following is alway

Method 2: Methodical/Conceptual Approach

• If n is a multiple of 6, we can write n as:

o n = 6*k, where k > 1 (since n > 6)

• \(n – 6 = 6k – 6 = 6 * (k -1)\)

o n and n – 6 are co-prime to each other. Hence n – 6 cannot be a factor of n.
o Another way to look at it is by dividing n by n – 6

 \(\frac{n}{(n-6)} = \frac{6k}{{6(k-1)}} = \frac{k}{(k-1)}\)
• From, here we can clearly see that k and k – 1 are two consecutive integers and we know that they are always co-prime.
• Hence n – 6 is not a factor of n.

• \(n + 6\)

o Now, \(n + 6 > n\)

 So, n + 6 cannot be a factor of n. Factors of n must be equal to or lesser than n.

• \(\frac{n}{3} = \frac{6k}{3} = 2k\)

o Now, if we divide n by 2k, we get an integer.

 \(\frac{n}{2k} = \frac{6k}{2k} = 3\)
 Or we can also look at it this way: \(\frac{n}{(n/3)} = 3\).

• \(\frac{n}{2} + 3 = 3k + 3 = 3 * (k + 1)\)

o If we divided n by \(\frac{n}{2} + 3\), we get:

 \(\frac{6k}{(3*(k +1))} = \frac{2k}{(k+1)}\)
 We cannot be sure if 2k is fully divisible by k + 1 or not. So, we can discard this.

• \(\frac{n}{2} + 6 = 3k + 6 = 3 * (k + 2)\)

o If we divided n by \(\frac{n}{2} + 6\), we get:

 \(\frac{6k}{(3*(k +2))} = \frac{2k}{(k+2)}\)
 We cannot be sure if 2k is fully divisible by k + 2 or not. So, we can discard this.