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Topic: Russian Peasant Multiplication: how does it work?
Replies: 4   Last Post: Mar 25, 2002 4:41 PM

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Kevin Fortin

Posts: 2
Registered: 12/12/04
Re: Russian Peasant Multiplication: Explained!
Posted: Mar 3, 1998 8:45 PM
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Thanks to the sci.math readers whose responses helped me to understand
"Russian Peasant Multiplication"!

Here's my own attempt to explain it:

[Please use a non-proportional font so the columns line up. Also, I
will use the caret (^) to indicate an exponent, e.g. 2^3 = "two cubed".
The procedure is described in Jan Gullberg's "Mathematics" (1997) and
other sources.]

Example: 19 x 54 = X

(H = "halving column", D = "doubling column"; in column H, any remainder
is discarded after each halving.)

_H_ _D_
54 19 (ignored)
27 38
13 76
6 152 (ignored)
3 304
1 608

The even numbers in column H are crossed out, along with the
corresponding entries in column D across the way. The remaining numbers
in column D are added up to give the product:
19 x 54 = (38 + 76 + 304 + 608) = 1026

After the halving is finished (i.e., you have reached a result of 1),
what matters in column H is whether the numbers are even or odd. The
pattern of even and odd numbers in column H corresponds to a binary (or
base 2) representation of the original number:

Place in
binary
notation
54 even 0 (2^0)
27 odd 1 (2^1)
13 odd 1 (2^2)
6 even 0 (2^3)
3 odd 1 (2^4)
1 odd 1 (2^5)

54 base ten, converted to base 2, equals 110110, or (back in base 10):
(2^5 + 2^4 + 2^2 + 2^1) = (32 + 16 + 4 + 2) = 54. [If the first entry
in column H is even, there will always be a zero in the unit (2^0) place
of the binary representation; if the first entry is odd, there will be a
one in the unit place.]

Therefore, 19 x 54 = 19(2^5 + 2^4 + 2^2 + 2^1) = (608 + 304 + 76 + 38) =
1026.

As 2^0 and 2^3 are not part of the sum shown above that produces 54, the
corresponding products in column D (19 x 2^0 and 19 x 2^3) are also
ignored. The surviving elements of column D are added up to give the
product of 54 x 19.

Hoping this helps others puzzled (as I was) by this method of
multiplying,

Kevin








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