KCC's Quiz AQQ300 about Divisibility by 6

Factors of 6 are 2 and 3, so if a number is divisible by both 2 and 3, it's divisible by 6.

First, tackle the divisible by 2 part.

n^5 - n = n(n^4 - 1)

For odd values of n, the square is also odd, square again and it's still odd.  Any odd number -1 is even, so the n^4-1 term is always divisible by 2 for odd n's.

For even values of n, the n outside  n(n^4-1) is always even so it is divisible by 2.

Now the divisible by 3 part.

All numbers can be represented by n = 3x, 3x+1, or 3x-1 for some value of x.

A) If it's a 3X number, then the n outside is divisble by 3.

B) When n = 3x+1 or 3x-1, you can show that n^4 - 1 is always divisible by 3:

  (3x+1) ^2 = 9x^2 + 6x + 1

Square that again for n^4 and you get 81x^4 + 108x^3 + 54x^2 + 12x + 1.

So n^4 - 1 = 3(27x^4 + 36x^3 + 18x^2 + 4x) which is always divisible by 3 due to the factor of 3 outside.

(3x-1)^4 power gives a very similar result, where all factors are multiples of 3 except the final +1 which is cancelled with the -1.

This shows that n(n^4 - 1) is always divisible by both 2 and 3, so divisible by 6.

Factors of 6 are 2 and 3, so if a number is divisible by both 2 and 3, it's divisible by 6.

First, tackle the divisible by 2 part.

n^5 - n = n(n^4 - 1)

For odd values of n, the square is also odd, square again and it's still odd.  Any odd number -1 is even, so the n^4-1 term is always divisible by 2 for odd n's.

For even values of n, the n outside  n(n^4-1) is always even so it is divisible by 2.

Now the divisible by 3 part.

All numbers can be represented by n = 3x, 3x+1, or 3x-1 for some value of x.

A) If it's a 3X number, then the n outside is divisble by 3.

B) When n = 3x+1 or 3x-1, you can show that n^4 - 1 is always divisible by 3:

  (3x+1) ^2 = 9x^2 + 6x + 1

Square that again for n^4 and you get 81x^4 + 108x^3 + 54x^2 + 12x + 1.

So n^4 - 1 = 3(27x^4 + 36x^3 + 18x^2 + 4x) which is always divisible by 3 due to the factor of 3 outside.

(3x-1)^4 power gives a very similar result, where all factors are multiples of 3 except the final +1 which is cancelled with the -1.

This shows that n(n^4 - 1) is always divisible by both 2 and 3, so divisible by 6.

Very clear and convincing demonstration you have made BMackin595 ! Congratulations!