Of course it is. Working from right to left, take 4x4=16, put the 6 under the last column and remember the 1 to carried over. Next multiply 4 by 3=12, put the 2 to the left of the 6 but there was a carry so 2 becomes 3. Next, multiply 4 by 2=8, put the 9 left of the 3, (don’t forget there was a carry (1)). Next multiply 4 by 1=4. Put the 4 to the left of the 9. So now the first row shows 4936.
Do the same for the next row, but before that as you are working in the order of tens (10) you have to put a zero under the 6 first of all.
Then 6x4=24 put the 4 under the 3 and carry 2. Now 6x3=18 plus the carry is 20 put the zero under the 9 and carry is again 2. Now 6x2=12 plus carry of 2=14 put the 4 under the 4 and carry is 1. Lastly 6x1=6 plus carry=7. Put the 7.

So the answer is after adding up:
4,936+74,040= 78,976.

Obviously you will have to know your tables. Fortunately some schools still teach them.
Try it.

But now there is a wonderful method called Napier. Multiplying a two-digit number with another two-digit number would require a 2 by 2 by grid if you wanted to multiply 26 x 15 or a 3 by 2 grid if you wanted to use 658 x 46.

So a 2x2 grid is simply 4 boxes - 2 under another 2.
The second step is to draw a diagonal in each box from top-right to bottom-left. The diagonal line separates the tens and the units. Always write the tens above the diagonal line in each box.
Now by multiplying 26x15 you would start with the tens values so 2x1, you would put a zero in the top left box upper diagonal and a 2 in the same box bottom diagonal. You have to put a zero because there are no tens values to carry by multiplying 2x1. Now multiply 6x1 to fill the right top box. That would be 6 so a 0 in the top right box upper diagonal ( no tens value) and a 6 in the bottom diagonal.
Now do the same way to fill the bottom boxes. That would be 2x5 and 6x5. After you have completed the grid, add the columns along the diagonals, starting at the bottom-right diagonal. You might have to carry over a value if over ten, from one diagonal to the next.
So you start with 0 then 0+3+6 =9 and lastly 1+2+0=3, and the last diagonal is a zero, so can be dismissed. Therefore the answer is 390.
Interesting method but in my opinion slow and easy to make mistakes if you get the sequence wrong. But to be fair it works. For me multiplication is knowing your tables.
Not forgetting 'carries' and working from right to left as shown in the first example.

## No comments:

## Post a Comment