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): $9^{k}$ does not end with 2.
Now for n=k+1, P(k+1): $9^{k+1}$= $9^{k}$×9¹,
which does not end with 2 since $9^{k}$ 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.

Answer 2

Answer:

Step-by-step explanation:

We know that 9^n can ends with 2 it must have 2 as a factor

But,9^n= (3×3)^n shows that only 3 is prime factor of 9^n

Also, we know that fundamental therom of arithmetic that each prime factorization is unique.

So, 2 is not a factor of 9^n

Hence,9^n can never ends with 2