In another article (The "3M" Game) I showed you how that by the use of a special table of values (the Triangular Numbers), you could multiply two numbers by just adding and subtracting numbers appropriately chosen from that table. If you haven't read that one yet, you can go there by clicking here. Don't forget to come back though!

     This time we will do much the same thing, that is, multiply two numbers without using the regular multiplication algorithm. Of course, as mentioned, we will need a table to do this work. This time it is a simple table of the square numbers. There has been much mention in my website about square numbers, so I will assume you know what that means.

     Below there will be a table of squares of the numbers from 1 to 100 for your convenience in understanding this lesson. Many textbooks, especially older books, often carry tables such as this one; also they have tables of square roots, cubes, and cube roots, plus tables necessary for trigonometry. That's because in the B.C. years -- Before Calculators -- we needed some help along this line to speed things up.

The "Squares" Algorithm Explained

     This algorithm takes an unexpected turn for us. We must consider two cases for our factors. Case I is when both factors are odd numbers or both are even numbers. Case II is when we have one of each, an odd and an even. Let's take Case I first.

     By way of explanation, I'll use the odd numbers 123 & 27.

  1. Add the numbers: 123 + 27 = 150.

  2. Take half that sum: ½ × 150 = 75

  3. Subtract the numbers: 123 - 27 = 96.

  4. Take half that difference: ½ × 96 = 48.

  5. Look up the squares in the table of 75 and 48:

    752 = 5625 and 482 = 2304

  6. Subtract the two squares found in Step 5.

    5625 - 2304 = 3321

     That's your product for 123 × 27! (Really.)

     For Case II, let's use these numbers: 38 & 145.

     We will do a little algebra "trick" here before proceeding to do the algorithm described above. We mentally convert 38 to the form of (37 + 1). Now using the distributive property, we have

(37 + 1) × 145 = 37 × 145 + 145

     By this means we have converted our problem back to the product of two odd numbers, plus another number. Something new from something old!

     Of course, we could have done it differently if we had rewritten the 145 as (144 + 1). This would give us:

38 × (144 + 1) = 38 × 144 + 38

     This time we merely have to perform the algorithm with two even numbers, then add another number to their product.


     Now I can hear you saying, "Gee, that was a lot of work just to multiply two numbers!" and "Hey, weren't you really multiplying after all in Steps 2, 4 & 5?"

     To which I say, "Well, yes and no." In Steps 2 & 4, we were perhaps multiplying a little, but we could also say that we were dividing by 2, something many people can do mentally with very little effort. (On another page I show you something called Russian Peasant Multiplication which uses this idea.) So the ease of this operation justifies its usage. In Step 5, all the multiplication to get the squares has been done earlier and by "experts"; all you do is just select the numbers from the table. So you didn't really multiply here either.

     The value of this activity is the same, practically speaking, as that of the 3M Game; so see the reasons given there, if you wish. Plus it is fun sometimes just to do regular things in a different way.

Extension Activity

     For students who are, or have been, in Algebra I, it would be a good challenge to prove why the algorithm works all the time.

     [HINT: Let a = one factor and b = the other factor. Then proceed in a logical, algebraic way.]

Table of Squares
1  126  676512601765776
2  427  729522704775929
3  928  784532809786084
41629  841542916796241
52530  900553025806400
63631  961563136816561

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