First draw 2 lines at 45° L2R, move over an inch or so and draw 3 lines also at 45° L2R , these represent the 23. Now turn your paper through 90° and draw 3 lines crossing the first set at 90° and above, right, draw 4 lines. These lines represent the 34. Now turn the paper level and you will see that the intersection sets form three columns. In the left column there is one set of intersections with 6 intersections in it. In the right column there is one set of intersections with 12 intersections in it. In the centre column there are two sets of intersections with 8 and 9 intersections in them respectively, so the centre column totals 8+9=17 intersections. So we write the three columns sums as 6 - 17 - 12 respectively. Then we carry as necessary (here two ones) to get 7 - 8 - 2.

Now you can check the result on your calculator if you like, but 23 * 34 = 782 as we just saw. With practice you can do this all in your head!

My good friend Matthias's daughter, who is a bright young lass, top of her class in physics and maths,
after some thought asked *"What do you do if there's a zero in the numbers?
Show me ten times ten please"*

Well, for a zero we just draw a dotted line, and anything crossing a dotted line has zero intersections. Thus 10 * 10 = 100 looks like the sketch shown here on the left.

Now Matthias is mistrustful of me, because I have pulled his leg so often, something you
can only do with very good friends :-)
He takes everything I say (rightly) with a pinch of salt. So he said
*"Ah, Stu, but you chose the numbers.
Now I'll choose the numbers and we'll see if it still works"*. Fair do's.
He chose 132 * 421 (so he needed 5 columns as well). And as we see on the left, his column totals were 4 - 14 - 15 - 7 - 2.
Doing the carries, he got 55572 which is the correct answer :-)
Try it with your choice of numbers too. It'll work :-)

They then both asked simultaneously *"Does it always work? Can you prove it?"*

So here's a proof for all of you blogreaders, not just Matthias and his 2 daughters :-

Assume we are multiplying two-digit numbers AB * CD as we did with 23 * 34. These numbers really represent (10*A + B) * (10*C + D). Multiplying out the terms, we get 100*A*C ( that's the left column) + B * D (that's the right column) + [(10*A*D) + (10*C*B)] which is the centre column. The algebra remains the same even when I choose any number of columns. It just gets harder to draw the (crowded) pictures. Mind you, 99*99 makes for a crowded sketch too, which is why I chose smaller digits for these examples. But yes, folks, it always works. Just be careful drawing the diagrams so that the columns line up correctly :-)

Bet they didn't teach you that in school ! (Except in traditional classes in India :-).

Now go visit my blog please, or look at other interesting maths stuff :-)

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