digits, modulus

p = 0 (2), and p^3 - p^2 = 0 (mod 10), find p + 3 (mod 10).


(p-1)p^2 = 0 (mod 10)
or p(p-1) = 0 (mod 10)

or 10 | p(p-1)
since 2 | p, (2,p-1) = 1
Therefore, 5 | (p-1)

p = 1 (mod 5) and p = 0 (mod 2)

Find evens among {6,11,16,21, ..}
{6,16,26}

p = 6 (mod 10)

p+3 = 9 (mod 10)