Question

Show that 9 ^n cannot end with digit 2 for any n = N​

Answer 1

*Solution:-

Let, P(n) denotes the event that 9^n can not end with the digit 2

When, n=1, P(1): 9^1 = 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 × 9^1 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): 9^n does not end with 2.

Answer 2

Answer:

LET US TAKE EXAMPLE TO UNDERSTAND IT

9¹=9

9²=81

9³=729

9⁴= 6561

AS WE OBSERVE THAT WHEN N IS ODD THEN IT END WITH DIGIT 9 & WHEN N IS EVEN THEN IT END WITH DIGIT 1.

SO WE CAN SAY THAT 9^n never ends with digit zero .

I HOPE IT HELPS U then PLZ MARK ME AS BRAINLIEST.