Question

8. Show that 9^n can not end with digit 0 for any natural number​

Answer 1

Answer:

Let us suppose that 9n can end with digit o for any natural number n.

>it is divisible by 10

>it is divisible by 2 and 5 -------1)

prime factors of 9n=3n ×3n--------2)

by unique factors theorem this is unique

from 1 and 2 prime factors of 9n don't have 2 and 5 as factors

it isn't divisible by 2 and 5

it isn't divisible by 10

thus our contradiction was wrong