

Fast Multiplication Algorithms
Name: Aditya P.
Status: student
Age: N/A
Location: N/A
Country: N/A
Date: 8/29/2004
Question:
Earlier on, when I was in the 6th grade, my history teacher was amazed
at how slow everyone in his class was at multiplication. He went up to
the board and showed the class a method to multiply ANY two numbers in a
fairly short time span (up to 6 digits by 6 digits in less than five
seconds). We were all fairly amazed by his exhibition, which employed
some rule not commonly taught in the classroom. None of us, however,
completely got the hang of his trick, and those of us that did soon
forgot it. Though I am a fairly strong math student, I have been unable
to redevelop his trick through the use of any textbook math up to, and
including, calculus. In addition, none of my recent math teachers know
any such method, and searches on the Internet have all been fruitless. Is
there any general algorithm/trick out there through which such rapid
multiplication is possible? It does not have to be mental math: paper and
pencil is fine, and mathematical complexity is fine (though I do not
think the original trick employed much  we were only 6th graders). I am
interested in hearing of any such algorithms, even those that might take
some practice in order to do rapidly.
Replies:
I am not sure about the algorithm you were shown, but I know of one that
"works" that is based on solid mathematical principles. The "method" is best
shown by an example:
Suppose we want to multiply: 768 x 52 = (7x10^2 + 6x10^1 + 8x10^0) * (5x10^1
+ 2x10^0). Look at the exponents of "10". They are (2,1,0) and (1,0)
respectively. So the "answer" is going to have a maximum exponent of
(2+1=3). Now write the multiplication in "vertical" format, lining up the
various exponents of "10" in the same columns. So we have:
Exponent of "10" > 4
3 2 1 0
7 6 8
5 2
______________________
Now we will multiply FROM LEFT TO RIGHT
placing the result in column that is the SUM of 3 5
the COLUMNS of the two digits multiplied. 3
0
For example, the first term is (7x5=35) which
4 0
will appear in column (2 + 1=3)
1 4
1 2
1 6
___________________________
3 9 9 3 6
If the product of the 2digit multiplication is
greater than 10 the lead digit "spills over" into the
next higher column. For example: 7x5=35
appears in the (2+1=3) column so the "3" in the
answer "35" spills over into column 4, i.e. 10^4.
Now why is this "left to right" multiplication easier?
First: All multiplications are single digit multiplications, the highest
possible one is 9x9
Second: There is no "carrying" to keep track of. Each multiplication step
will only "spill over" to the column just to the left. If the addition of
columns exceeds 10, then it is easy to add a "1" to the first column to the
left.
Third: The problem is laid out for easy reality checks. In this case
(~800x50=40,000) so you know what the answer should be.
With a little practice, you can "unlearn" the traditional algorithm of
multiplying from "righttoleft" and you will find hand calculations much
faster, less chance of error, and based upon the meaning of decimal place
holders.
By the way there is a similar algorithm for long division that
eliminates the necessity of finding the greatest multiple in division. This
one is even "cooler" than the multiplication algorithm.
Vince Calder
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Update: June 2012

