Napier's Bones Part 2

After seeing how students multiply multiple digit numbers by the lattice method, I was reminded of Napier’s Bones. John Napier (1550-1617) developed this tool for increasing the speed and accuracy of multiplications. His Napier’s Bones consisted of rectangular rods inside a board, or frame. On each rod, or bone, is written the multiplication of a single digit by 1, 2, 3, 4, 5, 6, 7, 8, and 9. Each number is written within a square divided by a diagonal line. Each tens digit is above the diagonal line and the ones digit is written below the diagonal line. The left side of the board is divided into squares marked with the digit 1 through 9. The squares on the side of the board are the same size as the squares on the bones. In fact, the fifth square on the left side of the board aligns with the fifth square in any bone. And that particular bone’s square has the value for five times the value of the bone. Since I have an interest in Baroque science, I decided to make my own set of bones.

This bone is for 9 and you can see it has written on it (starting from the top and working our way done) 9, 18, 27, 36, 45, 54, 63, 72, and 81. The best way to see how Napier ’s Bones are used is to work an example. So let’s multiply 25,806 by 79. You’ll need a sheet of paper and pencil to write the intermediate results.

First, load the bones for 25,806 into the board as shown below.

Now, we’ll first multiply 25,806 by 9 by reading off the sum of two digits in every diagonal formed by the numbers in the bones.

The product from multiplying 25,806 by 9 is read across the bones at the nine level of the board. Look on the left side of the board for the 9 and then start reading off numbers beginning on the right side and working your way to the left. First is the 4 all by itself in the lower right-hand corner. There is no other digit in its diagonal, so there is no other digit to add to 4, therefore just write a digit 4 on a sheet of paper.

Now move over to the next diagonal to the left, which contains 5 and 0 (0 is at its lower left of the 5). So add the 0 and the 5 to get 5. Write 5 to the left of the 4 you wrote first on the paper. You will have now written on your paper, 54

Now move over to the next diagonal to the left which contains the digits 0 and 2. Add these two digits together to get 2 and then write the digit 2 to the left of the 45 already written on the paper. You have 254 written on the paper now.

In the next diagonal as the digits 5 and 7. So add these two digits to get 12. Only write the 2 on the paper, the 1 (in the ten’s place) will be carried to the next diagonal. On your paper is now written 2254.

The next diagonal has the digits 4 and 8. Add those together and don’t forget to add the 1 carried from the previous diagonal. The result is 4 + 8 + 1, or 13. Again, only write the digit in the one’s place (a 2) on the paper (on the left side of the number you’ve written so far) and save the ten’s digit (a 1) so it can be carried to the next diagonal. The result on the paper so far is 32254.

The last diagonal is like the first diagonal in that there is only one digit. However, since the previous diagonal resulted in a carry, we’ll need to add that 1 to the 1 in this diagonal to get 2. Write 2 as the last digit on the paper. The result on the paper up to now is 232254. The number, 232,254 is the product of 25,805 X 9. Easy, wasn’t it?

In my next blog posting, we’ll add the product of 25,806 X 7. However, if you remember you multiplication, you know we’re going to write a 0 in the next line below the 232254 we’ve written so far and then add the digits for the product of 25,806 X 7.