Some interesting maths calculation and results to made formula.

1) n! + 1 is not divisible by any number between 2 and n

n = 5 ==> 5! + 1 = 120 + 1 = 121 (121 is not divisible by between 2 and 5)

2) n(n + 1)(2n + 1) is always divisible by 6.

n = 2 ==> 2*(3)*(5) = 30 / 6 = 5

3) 3^{2n} leaves remainder = 1 when divided by 8

n = 3 ==> 3

4) n^{3} + (n + 1 )^{3} + (n + 2 )^{3} is always divisible by 9

n = 4 ==> 4

5) 10^{2n + 1} + 1 is always divisible by 11

n = 2 ==> 10

6) n(n^{2} - 1) is always divisible by 6

n = 5 ==> 5(5

7) n^{2} + n is always even

n = 5 ==> 5

8 ) 2^{3n} - 1 is always divisible by 7

n = 5 ==> 2

9) 15^{2n-1} + 1 is always divisible by 16

n = 2 ==> 15

10) 3^{4n} - 4^{3n} is always divisible by 17

n = 2 ==> 3

11) n^{3} + 2n is always divisible by 3

n = 4 ==> 4

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