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Can You Multiply Three Numbers Simultaneously?

Date: 03/15/2004 at 13:30:25
From: Michele
Subject: Can you multiply three numbers simultaneously? If not, why?

If I have the problem 24 + 342 + 506 I can line the numbers up and add 
them at the same time.  Can I do this with the problem 24 x 342 x 506? 
If I can, how do you do it?  If I can't, why not?  If multiplication
is fast adding, then multiplying three numbers should work if you can
add three numbers simultaneously.



Date: 03/15/2004 at 14:43:00
From: Doctor Ian
Subject: Re: Can you multiply three numbers simultaneously? If not, why?

Hi Michele,

Good question!  I've never really thought about this before. 

I guess the first thing to note is that you can't actually add more 
than two numbers at a time.  What you can do is do a series of 
additions without writing down any intermediate results:

     [1]
      2 4       Add 4 to 2 to get 6
    3 4 2       Add 6 to 6 to get 12, carry the 1
  + 5 0 6       
  -------       
        2       


     [1]
      2 4       
    3 4 2       
  + 5 0 6       Add 1 to 2 to get 3
  -------       Add 3 to 4 to get 7
      7 2       Add 7 to 0 to get 7, with no carry

and so on.  But that's not the same thing as adding more than two 
things at once.  

Addition and multiplication are both binary operations, which means
you can apply them to only two numbers at a time. 

Now, can you cleverly set up a series of multiplications so that you
appear to be doing them in one shot?  Sure, by using the distributive
property:

  24 * 342 * 506

  = (20 + 4)(300 + 40 + 2)(500 + 6)

  = 20(300 + 40 + 2)(500 + 6) + 4(300 + 40 + 2)(500 + 6)

  = (20*300 + 20*40 + 20*2)(500 + 6) + (4*300 + 4*40 + 4*2)(500 + 6)

  = 500(20*300 + 20*40 + 20*2) + 6(20*300 + 20*40 + 20*2) 
    + 500(4*300 + 4*40 + 4*2) + 6(4*300 + 4*40 + 4*2)

  = 500*20*300 + 500*20*40 + 500*20*2 + 6*20*300 + 6*20*40 + 6*20*2
    + 500*4*300 + 500*4*40 + 500*4*2 + 6*4*300 + 6*4*40 + 6*4*2

  =    5*2*3*10^5
     + 5*2*4*10^4
     + 5*2*2*10^3
     + ...

All you're doing here is finding every combination of digits from
different numbers, multiplying them, and applying the appropriate
power of 10 to the result, which you could do just by counting up
decimal places: 

  24 * 342 * 506

  Combination     Product        Power of 10
  
  2, 3, 5           30           1 + 2 + 2 = 5  [*]
  2, 3, 0            0           1 + 2 + 1 = 4
  2, 3, 6           36           1 + 2 + 0 = 3
  4, 3, 5           60           0 + 2 + 2 = 4
  4, 3, 0            0            etc.
  4, 3, 6           72
  2, 4, 5           60
  2, 4, 0            0
  2, 4, 6           48
   etc.

                                 [*] Here, the 2 comes from 24, so
                                     it represents 20, which is 
                                     2*10^1; the 3 comes from 342,
                                     so it represents 300, which
                                     is 3*10^2; and the 5 comes
                                     from 506, so it represents
                                     500, which is 5*10^2.  To get
                                     the final exponent, we just 
                                     add the individual exponents.

Now we're doing a bunch of one-digit multiplications and adding up the
intermediate products... which isn't really any different from doing a
bunch of one-digit additions and adding up the intermediate sums.  

If we really wanted to pursue this, we might do something like this: 
Looking at our three numbers, 

  24, 342, 506

we could realize that the largest possible exponent would be 

  1 + 2 + 2 = 5

and so we could set up columns for each of the possible exponents;
then we could write our intermediate products in the appropriate columns:

    Exponent:   5       4       3        2        1      0

              2,3,5   2,3,0   2,3,6
                      4,3,5   4,3,0    4,3,6
                      2,4,5   2,4,0    2,4,6
                              4,4,5    4,4,0     4,4,6

and so on.  Then we could find the intermediate products; add the
products in each column, and multiply by 10 raised to the appropriate
exponent; and then add the totals of the columns.  

The main difference is that when we include a new addend in a series
sum, the new number of additions is the old number PLUS the number of
digits in the new addend.  When we include a new multiplicand in a
series product, the new number of multiplications is the old number
TIMES the number of digits in the new multiplicand.  This very quickly
increases to the point where it's easier to just do two full
multiplications at a time, rather than try to keep track of all the
combinations of digits. 

Does that make sense? 

But suppose we want to compute something like 3 * 4 * 5 in one shot. 
We could convert it to the equivalent addition, 

   5  \             \
   5   |             |
   5   |  4 * 5      |
   5   |             |
   5  /              |
   5  \              |
   5   |             |
   5   |  4 * 5      |  3 * (4 * 5)
   5   |             |
   5  /              |
   5  \              |
   5   |             |
   5   |  4 * 5      |
   5   |             |
 + 5  /             /
 ---

and now, since you can add as many numbers as you want "at the same
time", this should be much faster than doing the multiplications,
right?  But of course, it's not, which suggests that perhaps you
_can't_ add more than two numbers at a time.  

The most you can do, either with addition or multiplication, is
arrange things so that you're never adding or multiplying by anything
larger than a single digit.  It just turns out that when you try to do
this for multiplication, the cost exceeds the benefit in most cases.  

Thanks for asking such an interesting question.  Write back if you'd
like to talk more about this, or anything else. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
Elementary Addition
Elementary Multiplication

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