Question

show that 9^n cannot end with the digit 2 for any natural number

Answer 1

Let, P(n) denotes the event that 9ⁿ  can not end with the digit 2.
When, n=1, P(1): 9¹=9 does not end with 2.
Let us assume that the statement is true for n=k;
Then, P(k): does not end with 2.
Now for n=k+1, P(k+1): = ×9¹,
which does not end with 2 since and 9 both does not end with 2.
Therefore, assuming P(n) is true for n=k we have that P(n) is true for n=k+1.
Thus, by the principle of mathematical induction, P(n): 9ⁿ does not end with 2.