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show that 9n can't end with 2 for any integer n .

let us assume that 9n ends with a digit 2 for any integer n.so it should have 2 in its prime factorisation. but it has 3 in its factors .so our assumption is wrong it can 't end with a digit 2.

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 or.........for a rational number the form is p/q where q should be the factor of 2 or 5 or both but over here 9 is not divisible by 2 or 5 so we can say that 9n can 't end with 2 for any integer n...........

OR

as the divisibility rule of 2 is that it should be divisible by 2 only but 9 dosen 't has the 2 in it..........

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