KCC's Quiz AQQ300 about Divisibility by 6

if we expand n power 5 - n then it will be n(n power 4 - 1)
which is (n square - 1)(n square + 1) => (n square + 1) (n + 1)(n - 1) => (n-1) *  n * (n +1)* (n square + 1)
except 0 and 1 (n-1) *  n * (n +1) will be three consecutive number so the product of three number alwasy divisble by 6 because divisibility rule of 2 and 3 Since at least one of the numbers in the product is even, it is divisible by 2, and one of the numbers is divisible by 3, the product must also be divisible by 3. Therefore, the product is divisible by 6
for 0 and 1 value is 0 so divisible by 0 also

if we expand n power 5 - n then it will be n(n power 4 - 1)
which is (n square - 1)(n square + 1) => (n square + 1) (n + 1)(n - 1) => (n-1) *  n * (n +1)* (n square + 1)
except 0 and 1 (n-1) *  n * (n +1) will be three consecutive number so the product of three number alwasy divisble by 6 because divisibility rule of 2 and 3 Since at least one of the numbers in the product is even, it is divisible by 2, and one of the numbers is divisible by 3, the product must also be divisible by 3. Therefore, the product is divisible by 6
for 0 and 1 value is 0 so divisible by 0 also

Bingo Sona22 ! You find it!