82 81 80 Rewrite these as single numbers: 64 8 1 You don’t need any powers of 8 larger than your original number (in this case, 98). Since 83 = 512, and 512 is larger than 98, we can leave it off the chart.
98÷ 64 8 1= 1 ← This is the first digit of your octal number.
98 34÷ 64 8 1= 1
98 34÷ ÷ 64 8 1= = 1 4
98 34 2÷ ÷ ÷ 64 8 1= = = 1 4 2 The final answer: 98 base 10 = 142 base 8. You can write this as 9810 = 1428
2 x 80 = 2 x 1 =2 4 x 81 = 4 x 8 = 32 1 x 82 = 1 x 64 = 64 2 + 32 + 64 = 98, the number we started with.
Highlight this area: 327 7 7÷ ÷ ÷ 64 8 1= = = 5 0 7 The answer is 507. (Hint: it’s fine to have 0 as the answer to a division problem. )
This method is faster than the successive division method. Most people find it more difficult to understand why it works, and may want to start with the easier method above.
This method is faster than the successive division method. Most people find it more difficult to understand why it works, and may want to start with the easier method above.
In our example: 670 ÷ 8 = 83.
In our example: 670 ÷ 8 = 83 remainder 6. Our octal number so far is ???6. If your calculator has a “modulus” or “mod” button, you can find this value by entering “670 mod 8. "
In our example: The answer to our last division problem was 83. 83 ÷ 8 = 10 remainder 3. Our octal number so far is ??36.
In our example: The answer to our last division problem was 10. 10 ÷ 8 = 1 remainder 2. Our octal number so far is ?236.
In our example: The answer to our last division problem was 1. 1 ÷ 8 = 0 remainder 1. Our final answer is the octal number 1236. We can write this as 12368 to show that it is an octal number.
You start with a pile of 670 units. The first division problem divides these into groups, with 8 units in each group. Anything left over, the remainder, doesn’t fit into the octal 8s place. It must be in the 1s place instead. Now you take your pile of groups, and divide them into sections with 8 groups each. Each section now has 8 groups with 8 units each, or 64 units total. The remainder doesn’t fit into these, so it can’t fit into the octal 64s place. It must be in the 8s place. This continues until you discover the entire number.
