Question

Show that 21^n, n €N can never ends with 0,2,4,6 and 8​

Answer 1

Answer:

Step-by-step explanation:

$21 ^n = (3 \times 7)^n$

21 has no 2 in it's prime factorization. So it can never end in 0, 2 , 4 , 6 , 8.

 OR

Odd number raise to any number is always odd, Which does not include 2, 4, 6 , 8.

So $21^n$ does not end with 0, 2, 4 , 6 , 8.