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

1 Answer

3 votes

Answer:

Step-by-step explanation:


21 ^n = (3 * 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.

User ThaDon
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