Russian roulette
Question: I was recently introduced to a fun mathematical trick called
Russian multiplication.
First take two numbers that you want to multiply and place them at the top of
two adjacent columns. Divide the number in the left-hand column by two. Ignore
any remainder and write the new number beneath the first. Repeat the step until
you are left with 1.
Now double the number in the right-hand column and write the answer below it.
Keep doubling all the way down until there are equal numbers of figures in both
of the columns.
Advertisement
Delete the numbers in the right-hand column that are next to even numbers in
the left-hand column.
Now add up the remaining numbers in the right-hand column and you will find
that the total equals the result of multiplying the two original numbers at the
head of the lists. Why?
For example:
9 脳 8 = 72
4 鈥�16 (delete)
2 鈥� 32 (delete)
1 鈥� 64
Total for right-hand column: 72
Answer: Russian multiplication is a variant on multiplication presented in
the Rhind Papyrus, an ancient Egyptian mathematical treatise dating
from the 18th century BC. A description of the Rhind papyrus, including
multiplication and 鈥淓gyptian fractions鈥� can be found in volume 1 of James
Newman鈥檚 The World of Mathematics.
Both methods essentially involve representing the first number in binary
form, and adding together the corresponding doublings of the second number.
If you represent the first number in binary form, it is easy to see that the
鈥渙dd numbers鈥� in the 鈥渄ivision by two and throw away the remainder鈥� of the first
number correspond to the binary representation of the first number. By using the
鈥渙dd鈥� numbers to pick up the corresponding doublings of the second number, you
produce the correct product.
For instance, in the example given, 9 is 1001 in binary:
(1 脳 8) + (0 脳 4) + (0 脳 2) + (1 脳 1).
By adding together the first and fourth rows in the successive doubling, we
are essentially representing 9 脳 8 as (1 + 8) 脳 8,
or (1 脳 8) + (8 脳 8).
The Egyptian method does not use the successive halving to 鈥減ick out鈥� the
entries that add up to the first number. Instead, you are instructed simply to
pick out (as shown by **) the numbers in column 1 that add up to the first
number, as shown below.
If we want to multiply 9 脳 8, write 1 in the left-hand column and double it
until you reach the highest number that is less than 9. In the right-hand
column, start with 8 and double it to match the number of rows in the left-hand
column.
1 ** 8
2 鈥� 16
4 鈥� 32
8 ** 64
From the left-hand column, select 1 and 8, which add up to 9. The
corresponding numbers in the right-hand column are 8 and 64. Add them together
to get the answer, 72.
Janet Gunn
Nokesville, Virginia
Answer: The (binary) principle underpinning this process was apparently used
for performing long multiplication in number systems without place value, such
as Roman numerals. For example, the sum XVII multiplied by LVII would have been
started by writing each number at the top of a column:
XVII 鈥� LVII
If you expand the left number to make it easier to halve:
VVIIIIIII
And then double the right, and halve the left (ignoring remainders) you
get:
VIII 鈥� LLVVIIII
You can then rationalise the right so that it becomes CXIIII and repeat
successively. Without the intermediate steps, this could have been written:
XVII 鈥� LVII
VIII 鈥� CXIIII
IIII 鈥� CCXXVIII
II 鈥� CCCCLVI
I 鈥� DCCCCXII
Adding all the 鈥渙dd鈥� entries (numbers in the right-hand column that are on
the same row as an odd number on the left-hand column) gives us the answer:
XVI 脳 LVII = LVII + DCCCCXII
= DCCCCLXVIIII or = CMLXIX
Because multiplication is commutative regardless of the technique that is
used to perform it, the skilled Roman numeral operator would learn to spot
numbers which halve easily and put those on the left, whereas we usually put the
smallest number on the bottom of a long multiplication.
If the same number appears in several multiplication sums, a single doubling
table can be constructed for frequent use, making the process even more
efficient. Given a little practice, almost anyone can become proficient at
multiplying Roman numerals.
Barrie Wells
Conwy, Gwynedd
Tight squeeze
Question: When you squeeze the end of a hosepipe to make the opening smaller,
the water comes shooting out much faster. Why doesn鈥檛 this happen with a tap?
Instead, it just slowly reduces to a dribble.
(continued)
Answer: The two explanations that were given previously left out an important
observation. If the tap valve were right at the end of the outlet then you would
get a faster flow as the valve closed.
However, if the valve is back from the exit and the remainder of the pipe
full of water (as it should be) then the emerging water flow will be slower,
although inside, at the valve itself, the speed is increased.
Most firefighters鈥擨 was an industrial fireman for nearly 20
years鈥攌now this and have to act accordingly.
Ed Stefaniuk
Yarm, Cleveland
This week鈥檚 questions
Low gravity lager: Apparently NASA is aiming to brew a beer in space. The
yeast can neither sink nor rise as in traditional beer production and any carbon
dioxide that is produced will not rise to the surface.
So how will the beer ferment and will the final product be anything like
Earth beer?
Roger Cryne
Oldham, Lancashire
Worm count: My 7-year-old recently stumped me with a typical child鈥檚
question: are there more worms in the world than people?
I even struggled to find a way of making any assumptions to base an estimate
on. Can anyone help?
Philip Pluckrose
Lysterfield, Victoria
Stop thief: I have noticed that electrical goods stores use a form of
adhesive metal coil as a security tag. How do these tags set off the door alarms
and how are they deactivated before a legitimate customer leaves the shop?
Neil Yeomans
Solihull, West Midlands