5 + 6 = 11. We write 11 below and not write 1 carry 1. Do the same for the other numbers. 7 + 3 = 10. 6 + 4 = 10.
Here's what we do with the 2 digit numbers. We find multiples of 8 ( 0, 8, 16, 24, 32, 40...) closest to the sum. For the first digit , we have 11. The closest would be 8 so we subtract 8 to 11 to get 3. We subtracted it by 8 once so we add 1 to the next digit. 10 + 1 will be 11. Find the multiple closest to 11 which is 8. Perform the subtraction to get 3 and carry 1 to the next digit once again. Repeat the process until all digits are below 8.
Here's what we do with the 2 digit numbers. We find multiples of 8 ( 0, 8, 16, 24, 32, 40...) closest to the sum. For the first digit , we have 11. The closest would be 8 so we subtract 8 to 11 to get 3. We subtracted it by 8 once so we add 1 to the next digit. 10 + 1 will be 11. Find the multiple closest to 11 which is 8. Perform the subtraction to get 3 and carry 1 to the next digit once again. Repeat the process until all digits are below 8.
The final answer is 1 3 3 3 (Octal).
If we were to do this normally in decimal number system ( 0 - 9) , every time we reach a value of 10 or above, we always carry a 1 and input the remainder. So in octal number system ( 0 - 7) we carry a 1 every time we reach a value of 8 or above. This concept is applied to ANY number system that you can think of.
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