Mental calculation consists of
arithmetic
Arithmetic () is an elementary part of mathematics that consists of the study of the properties of the traditional operations on numbers—addition, subtraction, multiplication, division, exponentiation, and extraction of roots. In the 19th c ...
al
calculation
A calculation is a deliberate mathematical process that transforms one or more inputs into one or more outputs or ''results''. The term is used in a variety of senses, from the very definite arithmetical calculation of using an algorithm, to t ...
s using only the
human brain
The human brain is the central organ of the human nervous system, and with the spinal cord makes up the central nervous system. The brain consists of the cerebrum, the brainstem and the cerebellum. It controls most of the activities of the ...
, with no help from any supplies (such as pencil and paper) or devices such as a
calculator
An electronic calculator is typically a portable electronic device used to perform calculations, ranging from basic arithmetic to complex mathematics.
The first solid-state electronic calculator was created in the early 1960s. Pocket-sized ...
. People may use mental calculation when computing tools are not available, when it is faster than other means of calculation (such as conventional educational institution methods), or even in
a competitive context. Mental calculation often involves the use of specific techniques devised for specific types of problems. People with unusually high ability to perform mental calculations are called
mental calculator
Human calculator is a term to describe a person with a prodigious ability in some area of mental calculation (such as adding, subtracting, multiplying or dividing large numbers).
The world's best mental calculators are invited every two ye ...
s or ''lightning calculator''s.
Many of these techniques take advantage of or rely on the
decimal
The decimal numeral system (also called the base-ten positional numeral system and denary or decanary) is the standard system for denoting integer and non-integer numbers. It is the extension to non-integer numbers of the Hindu–Arabic numeral ...
numeral system. Usually, the choice of
radix
In a positional numeral system, the radix or base is the number of unique digits, including the digit zero, used to represent numbers. For example, for the decimal/denary system (the most common system in use today) the radix (base number) is t ...
is what determines which method or methods to use.
Methods and techniques
Casting out nines
After applying an arithmetic operation to two
operand
In mathematics, an operand is the object of a mathematical operation, i.e., it is the object or quantity that is operated on.
Example
The following arithmetic expression shows an example of operators and operands:
:3 + 6 = 9
In the above examp ...
s and getting a result, the following procedure can be used to improve confidence in the correctness of result:
# Sum the digits of the first operand; any 9s (or sets of digits that add to 9) can be counted as 0.
# If the resulting sum has two or more digits, sum those digits as in step one; repeat this step until the resulting sum has only one digit.
# Repeat steps one and two with the second operand. There are two single-digit numbers, one condensed from the first operand and the other condensed from the second operand. (These single-digit numbers are also the remainders one would end up with if one divided the original operands by 9; mathematically speaking, they are the original operands
modulo 9.)
# Apply the originally specified operation to the two condensed operands, and then apply the summing-of-digits procedure to the result of the operation.
# Sum the digits of the result that were originally obtained for the original calculation.
# If the result of step 4 does not equal the result of step 5, then the original answer is wrong. If the two results match, then the original answer may be right, though it is not guaranteed to be.
Example
* Say that calculation results that 6338 × 79 equals 500702
# Sum the digits of 6338: (6 + 3 = 9, so count that as 0) + 3 + 8 = 11
# Iterate as needed: 1 + 1 = 2
# Sum the digits of 79: 7 + (9 counted as 0) = 7
# Perform the original operation on the condensed operands, and sum digits: 2 × 7 = 14; 1 + 4 = 5
# Sum the digits of 500702: 5 + 0 + 0 + (7 + 0 + 2 = 9, which counts as 0) = 5
# 5 = 5, so there is a good chance that the prediction that 6338 × 79 equals 500702 is right.
The same procedure can be used with multiple operations, repeating steps 1 and 2 for each operation.
Factors
When multiplying, a useful thing to remember is that the factors of the operands still remain. For example, to say that 14 × 15 was 201 would be unreasonable. Since 15 is a multiple of 5, the product should be as well. Likewise, 14 is a multiple of 2, so the product should be even. Furthermore, any number which is a multiple of both 5 and 2 is necessarily a multiple of 10, and in the decimal system would end with a 0. The correct answer is 210. It is a multiple of 10, 7 (the other prime factor of 14) and 3 (the other prime factor of 15).
Calculating differences: ''a'' − ''b''
Direct calculation
When the digits of ''b'' are all smaller than the corresponding digits of ''a'', the calculation can be done digit by digit. For example, evaluate 872 − 41 simply by subtracting 1 from 2 in the units place, and 4 from 7 in the tens place: 831.
Indirect calculation
When the above situation does not apply, there is another method known as indirect calculation.
Look-ahead borrow method
This method can be used to subtract numbers left to right, and if all that is required is to read the result aloud, it requires little of the user's memory even to subtract numbers of arbitrary size.
One place at a time is handled, left to right.
Example:
4075
− 1844
------
Thousands: 4 − 1 = 3, look to right, 075 < 844, need to borrow.
3 − 1 = 2, say "Two thousand".
One is performing 3 - 1 rather than 4 - 1 because the column to the right is
going to borrow from the thousands place.
Hundreds: 0 − 8 = negative numbers not allowed here.
One is going to increase this place by using the number one borrowed from the
column to the left. Therefore:
10 − 8 = 2. It's 10 rather than 0, because one borrowed from the Thousands
place. 75 > 44 so no need to borrow,
say "two hundred"
Tens: 7 − 4 = 3, 5 > 4, so 5 - 4 = 1
Hence, the result is 2231.
Calculating products: ''a'' × ''b''
Many of these methods work because of the
distributive property
In mathematics, the distributive property of binary operations generalizes the distributive law, which asserts that the equality
x \cdot (y + z) = x \cdot y + x \cdot z
is always true in elementary algebra.
For example, in elementary arithmeti ...
.
Multiplying any two numbers by attaching, subtracting, and routing
Discovered by Artem Cheprasov, there is a method of multiplication that allows the user to utilize 3 steps to quickly multiply numbers of any size to one another via three unique ways.
First, the method allows the user to attach numbers
to one another, as opposed to adding or subtracting them, during intermediate steps in
order to quicken the rate of multiplication. For instance, instead of adding or subtracting intermediary results such as 357 and 84, the user
could simply attach the numbers together (35784) in order to simplify and
expedite the multiplication problem. Attaching numbers to one another helps to bypass unnecessary steps found in traditional multiplication techniques.
Secondly, this method uses negative numbers as necessary, even
when multiplying two positive integers, in order to quicken the rate of multiplication via subtraction. This means two positive integers can be multiplied together to get negative intermediate steps, yet still the correct positive answer in the end. These negative numbers are actually automatically derived
from the multiplication steps themselves and are thus unique to a particular
problem. Again, such negative intermediate steps are designed to help hasten the mental math.
Finally, another unique aspect of using this method is that the
user is able to choose one of several different “routes of multiplication” to
the specific multiplication problem at hand based on their subjective preferences or strengths and weaknesses with particular integers.
Despite the same starting integers, the different multiplication
routes give off different intermediate numbers that are automatically derived for
the user as they multiply. Some of these
intermediaries may be easier than others (e.g. some users may find a route that uses a negative 7, while another route uses a 5 or a 0, which are typically easier to work with
mentally for most people, but not in all instances).
If one “route” seems to be harder for one student vs. another route and its intermediate numbers, that student can simply choose another simpler route of multiplication for themselves even though it's the
same original problem.
= The "Ends of Five" Formula
=
For any 2-digit by 2-digit multiplication problem, if both numbers end in five, the following algorithm can be used to quickly multiply them together:
:
As a preliminary step simply round the smaller number down and the larger up to the nearest multiple of ten. In this case:
:
:
The algorithm reads as follows:
:
Where t
1 is the tens unit of the original larger number (75) and t
2 is the tens unit of the original smaller number (35).
:
The author also outlines another similar algorithm if one wants to round the original larger number down and the original smaller number up instead.
= The "Borrower's" Formula
=
If two numbers are equidistant from the nearest multiple of 100, then a simple algorithm can be used to find the product.
As a simple example:
:
Both numbers are equidistant (33 away) from their nearest multiple of 100 (0 and 100, respectively).
As a preliminary step simply round the smaller number down and the larger up to the nearest multiple of ten. In this case:
:
:
The algorithm reads as follows:
:
Where u
1 is the original larger number's (67) units digit and u
2 is the original smaller number's (33) units digit. T
1 is the original larger number's tens digit and T
2 is the original larger number's tens digit multiplied by their respective power (in this case by 10, for a tens digit).
And so:
:
Multiplying any 2-digit numbers
To easily multiply any 2-digit numbers together a simple algorithm is as follows (where a is the tens digit of the first number, b is the ones digit of the first number, c is the tens digit of the second number and d is the ones digit of the second number):
:
:
For example,
:
800
+120
+140
+ 21
-----
1081
Note that this is the same thing as the conventional sum of partial products, just restated with brevity. To minimize the number of elements being retained in one's memory, it may be convenient to perform the sum of the "cross" multiplication product first, and then add the other two elements:
:
:
f which only the tens digit will interfere with the first term:
i.e., in this example
:(12 + 14) = 26, 26 × 10 = 260,
to which is it is easy to add 21: 281 and then 800: 1081
An easy mnemonic to remember for this would be
FOIL. F meaning first, O meaning outer, I meaning inner and L meaning last. For example:
:
and
:
where 7 is ''a'', 5 is ''b'', 2 is ''c'' and 3 is ''d''.
Consider
:
this expression is analogous to any number in base 10 with a hundreds, tens and ones place. FOIL can also be looked at as a number with F being the hundreds, OI being the tens and L being the ones.
is the product of the first digit of each of the two numbers; F.
is the addition of the product of the outer digits and the inner digits; OI.
is the product of the last digit of each of the two numbers; L.
Multiplying by 2 or other small numbers
Where one number being multiplied is sufficiently small to be multiplied with ease by any single digit, the product can be calculated easily digit by digit from right to left. This is particularly easy for multiplication by 2 since the carry digit cannot be more than 1.
For example, to calculate 2 × 167:
2×7=14, so the final digit is 4, with a 1 carried and added to the 2×6 = 12 to give 13, so the next digit is 3 with a 1 carried and added to the 2×1=2 to give 3. Thus, the product is 334.
Multiplying by 5
To multiply a number by 5,
1. First multiply that number by 10, then divide it by 2. The two steps are interchangeable i.e. one can halve the number and then multiply it.
The following algorithm is a quick way to produce this result:
2. Add a zero to right side of the desired number. (A.)
3. Next, starting from the leftmost numeral, divide by 2 (B.)
and append each result in the respective order to form a new number;(fraction answers should be rounded down to the nearest whole number).
EXAMPLE: Multiply 176 by 5.
A. Add a zero to 176 to make 1760.
B. Divide by 2 starting at the left.
1. Divide 1 by 2 to get .5, rounded down to zero.
2. Divide 7 by 2 to get 3.5, rounded down to 3.
3. Divide 6 by 2 to get 3. Zero divided by two is simply zero.
The resulting number is 0330. (This is not the final answer, but a first approximation which will be adjusted in the following step:)
C. Add 5 to the number that follows any single numeral
in this new number that was odd before dividing by two;
EXAMPLE: 176 (IN FIRST, SECOND THIRD PLACES):
1.The FIRST place is 1, which is odd. ADD 5 to the numeral after
the first place in the new number (0330) which is 3; 3+5=8.
2.The number in the second place of 176, 7, is also odd. The
corresponding number (0 8 3 0) is increased by 5 as well;
3+5=8.
3.The numeral in the third place of 176, 6, is even, therefore
the final number, zero, in the answer is not changed. That
final answer is 0880.
The leftmost zero can be omitted, leaving 880.
So 176 times 5 equals 880.
EXAMPLE: Multiply 288 by 5.
A. Divide 288 by 2. One can divide each digit individually to get 144. (Dividing smaller number is easier.)
B. Multiply by 10. Add a zero to yield the result 1440.
Multiplying by 9
Since 9 = 10 − 1, to multiply a number by nine, multiply it by 10 and then subtract the original number from the result. For example, 9 × 27 = 270 − 27 = 243.
This method can be adjusted to multiply by eight instead of nine, by doubling the number being subtracted; 8 × 27 = 270 − (2×27) = 270 − 54 = 216.
Similarly, by adding instead of subtracting, the same methods can be used to multiply by 11 and 12, respectively (although simpler methods to multiply by 11 exist).
= Using hands: 1–10 multiplied by 9
=
To use this method, one must place their hands in front of them, palms facing towards them. Assign the left thumb to be 1, the left index to be 2, and so on all the way to the right thumb is ten. Each ", " symbolizes a raised finger and a "−" represents a bent finger.
1 2 3 4 5 6 7 8 9 10
, , , , , , , , , ,
left hand right hand
Bend the finger which represents the number to be multiplied by nine down.
Ex: 6 × 9 would be
, , , , , − , , , ,
The right little finger is down. Take the number of fingers still raised to the left of the bent finger and prepend it to the number of fingers to the right.
Ex: There are five fingers left of the right little finger and four to the right of the right little finger. So 6 × 9 = 54.
5 4
, , , , , − , , , ,
Multiplying by 10 (and powers of ten)
To multiply an integer by 10, simply add an extra 0 to the end of the number. To multiply a non-integer by 10, move the decimal point to the right one digit.
In general for base ten, to multiply by 10
''n'' (where ''n'' is an integer), move the decimal point ''n'' digits to the right. If ''n'' is negative, move the decimal , ''n'', digits to the left.
Multiplying by 11
For single digit numbers simply duplicate the number into the tens digit, for example: 1 × 11 = 11, 2 × 11 = 22, up to 9 × 11 = 99.
The product for any larger non-zero
integer
An integer is the number zero (), a positive natural number (, , , etc.) or a negative integer with a minus sign (−1, −2, −3, etc.). The negative numbers are the additive inverses of the corresponding positive numbers. In the language o ...
can be found by a series of additions to each of its digits from right to left, two at a time.
First take the ones digit and copy that to the temporary result. Next, starting with the ones digit of the multiplier, add each digit to the digit to its left. Each sum is then added to the left of the result, in front of all others. If a number sums to 10 or higher take the tens digit, which will always be 1, and carry it over to the next addition. Finally copy the multipliers left-most (highest valued) digit to the front of the result, adding in the carried 1 if necessary, to get the final product.
In the case of a negative 11, multiplier, or both apply the sign to the final product as per normal multiplication of the two numbers.
A step-by-step example of 759 × 11:
# The ones digit of the multiplier, 9, is copied to the temporary result.
#* result: 9
# Add 5 + 9 = 14 so 4 is placed on the left side of the result and carry the 1.
#* result: 49
# Similarly add 7 + 5 = 12, then add the carried 1 to get 13. Place 3 to the result and carry the 1.
#* result: 349
# Add the carried 1 to the highest valued digit in the multiplier, 7 + 1 = 8, and copy to the result to finish.
#* Final product of 759 × 11: 8349
Further examples:
* −54 × −11 = 5 5+4(9) 4 = 594
* 999 × 11 = 9+1(10) 9+9+1(9) 9+9(8) 9 = 10989
** Note the handling of 9+1 as the highest valued digit.
* −3478 × 11 = 3 3+4+1(8) 4+7+1(2) 7+8(5) 8 = −38258
* 62473 × 11 = 6 6+2(8) 2+4+1(7) 4+7+1(2) 7+3(0) 3 = 687203
Another method is to simply multiply the number by 10, and add the original number to the result.
For example:
17 × 11
17 × 10 = 170
170 + 17 = 187
17 × 11 = 187
One last easy way:
If one has a two-digit number, take it and add the two numbers together and put that sum in the middle, and one can get the answer.
For example: 24 x 11 = 264 because 2 + 4 = 6 and the 6 is placed in between the 2 and the 4.
Second example: 87 x 11 = 957 because 8 + 7 = 15 so the 5 goes in between the 8 and the 7 and the 1 is carried to the 8. So it is basically 857 + 100 = 957.
Or if 43 x 11 is equal to first 4+3=7 (For the tens digit) Then 4 is for the hundreds and 3 is for the tens. And the answer is 473
Multiplying two 2 digit numbers between 11 and 19
To easily multiply 2 digit numbers together between 11 and 19 a simple algorithm is as follows (where a is the ones digit of the first number and b is the ones digit of the second number):
(10+a)×(10+b)
100 + 10×(a+b) + a×b
which can be visualized as three parts to be added:
1
xx
yy
for example:
17×16
1 = 100
13 (7+6) = 10×(a+b)
42 (7×6) = a×b
272 (total)
Using hands: 6–10 multiplied by another number 6–10
This technique allows a number from 6 to 10 to be multiplied by another number from 6 to 10.
Assign 6 to the little finger, 7 to the ring finger, 8 to the middle finger, 9 to the index finger, and 10 to the thumb. Touch the two desired numbers together. The point of contact and below is considered the "bottom" section and everything above the two fingers that are touching are part of the "top" section. The answer is formed by adding ten times the total number of "bottom" fingers to the product of the number of left- and right-hand "top" fingers.
For example, 9 × 6 would look like this, with the left index finger touching the right little finger:
=10 :right thumb (top)
9 :right index finger (top)
8 :right middle finger (top)
left thumb: =10 7 :right ring finger (top)
left index finger: --9---><---6-- :right little finger (BOTTOM)
left middle finger: --8-- (BOTTOM)
left ring finger: --7-- (BOTTOM)
left little finger: --6-- (BOTTOM)
In this example, there are 5 "bottom" fingers (the left index, middle, ring, and little fingers, plus the right little finger), 1 left "top" finger (the left thumb), and 4 right "top" fingers (the right thumb, index finger, middle finger, and ring finger). So the computation goes as follows: 9 × 6 = (10 × 5) + (1 × 4) = 54.
Consider another example, 8 × 7:
=10 :right thumb (top)
left thumb: =10 9 :right index finger (top)
left index finger: 9 8 :right middle finger (top)
left middle finger: --8---><---7-- :right ring finger (BOTTOM)
left ring finger: --7-- --6-- :right little finger (BOTTOM)
left little finger: --6-- (BOTTOM)
Five bottom fingers make 5 tens, or 50. Two top left fingers and three top right fingers make the product 6. Summing these produces the answer, 56.
Another example, this time using 6 × 8:
--8---><---6--
--7--
--6--
Four tens (bottom), plus two times four (top) gives 40 + 2 × 4 = 48.
Here's how it works: each finger represents a number between 6 and 10. When one joins fingers representing ''x'' and ''y'', there will be 10 - ''x'' "top" fingers and ''x'' - 5 "bottom" fingers on the left hand; the right hand will have 10 - ''y'' "top" fingers and ''y'' - 5 "bottom" fingers.
Let
:
(the number of "top" fingers on the left hand)
:
(the number of "top" fingers on the right hand)
:
(the number of "bottom" fingers on the left hand)
:
(the number of "bottom" fingers on the right hand)
Then following the above instructions produces
:
:
:
: