The "3M" Game

     I well remember my wonder that I experienced when I learned about logarithms in Advanced Algebra. The thing that most caught my attention was that you could "multiply two numbers" through a not-so- simple addition process. To do this, you first find the logarithm of each of the two numbers, add them, then reconvert that back to a number that would be the product required. Sounds complicated, but it works!

(For those readers who have not reached this level of math yet, do not turn away from this page until you've read a little more.)

     In algebra, we write the basic logarithmic property this way:

log (a × b) = log a + log b

     Of course, this is just a "taste" of a much larger and grander topic in mathematics, but it is meant to give a connection to what I want to show you this time, namely how you can "MULTPLY by ADDING", using a much simpler, arithmetic basis.

Triangular Numbers

     Before we can proceed, we must introduce what to some persons may be a new topic, that of Triangular Numbers. Square numbers have been discussed frequently in this website, but now we are going to look at a family of numbers that form triangles. Observe:

10 is a triangular number, because 10 things can be arranged in a triangular array like this:

				* *
				* * *
				* * * *
     From this sort of picture it is easy to form and determine many other triangular numbers.
		*		*		*
				* *		* *
						* * *
     Here we see that 1, 3, and 6 are the first three triangular numbers. And OF COURSE with these few examples we can see a short cut for finding other triangular numbers.
		10 = 1 + 2 + 3 + 4
		 6 = 1 + 2 + 3
		 3 = 1 + 2
		 1 = 1
     So going in the higher direction, we have 15, 21, 28, 36, 45, and 55, thus giving us the first ten triangular numbers.

     To make discussing a particular triangular number a bit more convenient, we will use the following notation [using the 4th one as an example]:

T4 = 10

The First 50 Triangular Numbers
T1 =   1T11 =   66T21 = 231T31 = 496T41 =   861
T2 =   3T12 =   78T22 = 253T32 = 528T42 =   903
T3 =   6T13 =   91T23 = 276T33 = 561T43 =   946
T4 = 10T14 = 105T24 = 300T34 = 595T44 =   990
T5 = 15T15 = 120T25 = 325T35 = 630T45 = 1035
T6 = 21T16 = 136T26 = 351T36 = 666T46 = 1081
T7 = 28T17 = 153T27 = 378T37 = 703T47 = 1128
T8 = 36T18 = 171T28 = 406T38 = 741T48 = 1176
T9 = 45T19 = 190T29 = 435T39 = 780T49 = 1225
T10 = 55T20 = 210T30 = 465T40 = 820T50 = 1275

Magical Multiplying Method

     Finally we're ready to show you the magical way to multiply without multiplying anything. I call it the "Magical Multiplying Method" (or the "3M" way). First, I must confess we will do a little subtracting, too. Second, you will need a copy of the table above. I hope you don't mind too much.

     Let's take as an example 15 × 9.

  1. Take the larger factor 15 and find T15 in the table above. It is 120.

  2. Subtract 1 from 9, the smaller factor, getting 8. Find T8 in the table. It is 36.

  3. Subtract the two factors, 15 - 9; that's 6. Find T6. It is 21.

  4. Add the results of Steps #1 & #2, then subtract the result from Step #3. That's your product!

120 + 36 - 21 = 135

     [Well, I never said it was going to be easier, shorter, or anything like that. It's just an interesting idea by itself, don't you agree? And with practice, it's not too difficult.]

     We can summarize this by a little algebra. This is the formula for the steps given above.

a × b = Ta + Tb-1 - Ta-b

     Doesn't look so bad now, or does it?


     The reasons why I am presenting this activity are various:

  1. It's a different way to do a common thing.

  2. It's a good experience in "following steps" to achieve a certain goal.

  3. It shows a connection between advanced math (logarithms) and simple arithmetic, both of which are being used to multiply two numbers.

  4. It provides an experience in looking up information in tables to use for a certain purpose.

  5. It show how an algebraic formula can show the various steps in compact style.

Extension Activity

     For students who are, or have been, in Algebra I, it would be a good challenge to prove that the formula indeed works for ALL numbers a and b. To do so, however, one needs to know the general formula for the triangular numbers. I will not show how it is derived, rather just state it here.

Tn = ½n(n+1)

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