Name: Lisa G.
My fourth-grade students are currently learning to
multiply two-digit by two-digit numbers. Is there a non-standard
algorithm that I can teach my accelerated/gifted students? One that does
not involve addition of several columns to find the product?
Yes, there are and it is not only for accelerated/gifted students, i would
not call it different algorithm but it is faster, and it is the one taught
at Brazilian schools.
I am a Brazilian ,but I know quite well the way American schools
teach multiplication at school,
Since my 2 daughters went to elementary school in America.
I can only communicate with you by e-mail, but I am going to look
for a Brazilian site where even you do not understand Portuguese
you will be able to catch it. It is very easy and less troublesome
than the other one, even I would say yours is easier to learn.
I hope to find and to write to you again.
And thanks for asking NEWTON! Tell your friends about us!
(Dr. Mabel Rodrigues)
Hi Lisa! It is me again, from Brazil...
Try looking at
It is a site for questions and answers and somebody is exactly asking
about multiplication algorithm. Even it is written in Portuguese (do not
I think you can easily follow the graphics ( after all, our numbers are the
If a student can learn to multiply two-digit by two-digit, that same student
can learn to multiply billions by billions. Do not look for an algorithm.
Look to provide an understanding of the process. Show how carrying can
extend to 3-digit, 4-digit, and beyond. Let them explore more than 2
digits. See whether they can figure out how to do more than 2 digits, now
that they know how to carry. Perhaps the accelerated/gifted students would
like to explain to other students how these new uses for carrying. Do not
teach the students to look for a quick algorithm for each circumstance.
Teach them to look for significant understanding of the few rules they have,
allowing them to expand these rules to many new situations.
Dr. Ken Mellendorf
Illinois Central College
I am not sure that what follows is "non-standard", but I think fourth graders
who know how to multiply single digit integers (1 through 9) can catch on
quickly to this algorithm that invokes the definition of "base 10" numbers.
Consider the following two numbers: AB and XY. What does this mean? It
means: 10*A +1*B = AB and 10*X + 1*Y respectively. Written as:
(10*A + 1*B)
(10*X + 1*Y)
Verbally that is: 100*(product of the first two vertical digits), 10*(sum of
the products of the "cross" digits (AY+BX) + 1*(product of the last two
vertical digits (BY).
So, an example: 72 = (10*7+1*2)
x47 = (10*4+1*7)
72 * 47 = 100*(7*4) + 10*(7*7+2*4) + (2*7) = 2800 + 570 + 14 = 3384
vertical cross vertical SUM
This algorithm has several teaching advantages:
1. It focuses on what we mean by a number fundamentally: 345 = 100*3 + 10*4
2. It involves only single digit multiplications. In fact, it will only
involve single digit multiplication no matter how big the numbers, it's the
10,100,1000,... that determines the digit placement.
3. It can be extended to numbers in other bases than "base 10".
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Update: June 2012