Alg10.12

Show that $676\,$ divides $3^{3n}-26n-1\,$ , where $n\,$ is an integer.

$3^{3n}-26n-1=(27)^n-26n-1=(1+26)^n-26n-1\,$

$(27)^n-26n-1=(1+26)^n-26n-1=[1^n+n\!C_1 26+n\!C_2 (26)^2+......+n\!C_r (26)^r+.....+n\!C_n (26)^n]-26n-1\,$

$=n\!C_2 (26)^2+n\!C_3 (26)^3+....+n\!C_r (26)^r+.....+n\!C_n (26)^n\,$

$(26)^2[n\!C_2+n\!c_3 (26)+n\!C_4 (26)^2+.....+n\!C_r (26)^{r-2}+....+n\!C_n (26)^{n-2}]\,$

$676\,$ [integer]

Therefore $676\,$ divides the given expression.