# Unusual Maths

Here is a novel way of doing multiplication, based around intersecting lines. We have not seen this one before, but it certainly seems to work.

Ian N. Dennis kindly sent us the following comment...

Your article on “unusual maths” is a visual representation of Jakow Trachtenberg’s system of “speed maths”. Basically, instead of performing the multiplications first (123 x 300, 123 x 20, 123 x 1) and then adding up the results (36900 + 2460 + 123 = 39483), Trachtenberg’s system has you multiplying and adding at the same time (100 x 300) + ((100 x 20) + (20 x 300)) + ((3 x 300) + (20 x 20) + (100 x 1)) + ((20 x 3) + (20 x 1)) + (3 x 1) = 30000 + (2000 + 6000) + (900 + 400 + 100) + (60 + 20) + 3 = 30000 + 8000 + 1400 + 80 + 3 = 39483.

Actually, in practice it’s a little easier, because you write the results as you go, so having multiplied the 2 leftmost numbers together (1 x 3) you write the answer 3, then you take the pairs 12 and 32 and multiply the outside numbers together (1 x 2), multiply the inside numbers together (2 x 3), and add the two results (2 + 6 = 8) and write that down next to the 3 … 38. Then you take all three numbers … 123 and 321, multiply the outer pair (1 x 1), the middle pair (2 x 2) and the inner pair (3 x 3), add the 3 results together (1 + 4 + 9 = 14), so the 1 gets “carried” to the 8 and the result becomes 394. Then you take the two rightmost pairs (23 and 21) and multiply the outside numbers together (2 x 1), multiply the inside numbers together (3 x 2), and add the two results (2 + 6 = 8) and add that to the result … 3948. Finally, you multiply the two rightmost numbers together (3 x 1) and the final result becomes 39483.

Interestingly, despite the seeming awkwardness of this method, remedial students often find it easier than using more traditional methods. I read a translation of Trachtenberg’s book many years ago and still use many of his techniques, and can often beat a pocket calculator.

Ian N. Dennis

Sr. Application Developer

Aerojet

Sacramento, CA

## Bookmark with: