Question

provethat 8power n last digit cannot be zero when n is some integer​

Answer 1

Step-by-step explanation:

First we have to look for a  pattern in the last digit for 8^n each time , and prove that 0 does not occur in that pattern .

We know 8^1 has last digit 8 .

8² is 64 , so it has last digit 4 .

8³ is 512 , so it has last digit 2 .

8^4 is 4096, so it has last digit 6

8^5 is 32768 , so it has last digit 8 .

Basically what am I doing over here to get the last digit in each step ? . In 8^1 I get the last digit as 8 , and from next step onwards each time I am multiplying 8 with the previous last digit , and take the last digit of the product . (For example in 8² we had last digit 4, to get the last digit of 8³ we just multiplied 4 with 8 to get 32, so we get last digit 2.)

So we actually did not need to do the actual multiplication to get the last digit of any 8^n .

So we find a pattern of last digits as 8,4,2,6,8,4,2,6,...... and so on (since at 8^5 we got last digit 8 , so the same pattern will continue again).

Since there is no 0 occuring in the pattern , the last digit of 8^n can never be 0 , where n is some integer .

Answer 2

Answer:

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