If 9^n end with 2 it must have 2 in its prime factorization(That is it must be a multiple of 2). 9^n = 3^n * 3^n
We can see prime factorization on 9^n does not contain 2. Hence it can't end with 2.

17

Jitendar Singh answered this

to end with number 2, 9n show divisible by 2 and the prime factorisation of 9n should cannot the number 2 but the prime factorisation of 9n will be 1 and 3 thus cannot end with digit 2 for any natural number.

-10

Pk answered this

Because 9^n is not of the form 2^1+4n where n is a non negative integar.

-8

Nikhil answered this

HOPE ITS HELPFUL.

-9

Igdgos answered this

If odd no. like (3×3)^n is 3^2n
so 3 is a odd number and its power is even then remember it will not end with 2 .
if you dont trust we do 3^2=9,3^3=27,3^4=81,3^5=243 and so on ..
hope it must be helpfull to you

1

Pk answered this

expand (10-1)^n binomially the product will be 10*integar +9 or -1 depending on the n