Mental calculation - AbsoluteAstronomy.com

Mental calculation comprises arithmetic

Arithmetic

Arithmetic or arithmetics is the oldest and most elementary branch of mathematics, used by almost everyone, for tasks ranging from simple day-to-day counting to advanced science and business calculations. It involves the study of quantity, especially as the result of combining numbers...

al calculation

Calculation

A calculation is a deliberate process for transforming one or more inputs into one or more results, with variable change.The term is used in a variety of senses, from the very definite arithmetical calculation of using an algorithm to the vague heuristics of calculating a strategy in a competition...

s using only the human brain

Human brain

The human brain has the same general structure as the brains of other mammals, but is over three times larger than the brain of a typical mammal with an equivalent body size. Estimates for the number of neurons in the human brain range from 80 to 120 billion...

, with no help from calculator

Calculator

An electronic calculator is a small, portable, usually inexpensive electronic device used to perform the basic operations of arithmetic. Modern calculators are more portable than most computers, though most PDAs are comparable in size to handheld calculators.The first solid-state electronic...

s, computer

Computer

A computer is a programmable machine designed to sequentially and automatically carry out a sequence of arithmetic or logical operations. The particular sequence of operations can be changed readily, allowing the computer to solve more than one kind of problem...

s, or pen

Pen

A pen is a device used to apply ink to a surface, usually paper, for writing or drawing. Historically, reed pens, quill pens, and dip pens were used, with a nib of some sort to be dipped in the ink. Ruling pens allow precise adjustment of line width, and still find a few specialized uses, but...

and paper

Paper

Paper is a thin material mainly used for writing upon, printing upon, drawing or for packaging. It is produced by pressing together moist fibers, typically cellulose pulp derived from wood, rags or grasses, and drying them into flexible sheets....

. People use mental calculation when computing tools are not available, when it is faster than other means of calculation (for example, conventional methods as taught in educational institutions), or in a competition context. Mental calculation often involves the use of specific techniques devised for specific types of problems.

Many of these techniques take advantage of or rely on the decimal

Decimal

The decimal numeral system has ten as its base. It is the numerical base most widely used by modern civilizations....

numeral system. Usually, the choice of radix

Radix

In mathematical numeral systems, the base or radix for the simplest case is the number of unique digits, including zero, that a positional numeral system uses to represent numbers. For example, for the decimal system the radix is ten, because it uses the ten digits from 0 through 9.In any numeral...

determines what methods to use and also which calculations are easier to perform mentally. For example, multiplying or dividing by ten is an easy task when working in decimal (just move the decimal point), whereas multiplying or dividing by sixteen is not; however, the opposite is true when working in hexadecimal

Hexadecimal

In mathematics and computer science, hexadecimal is a positional numeral system with a radix, or base, of 16. It uses sixteen distinct symbols, most often the symbols 0–9 to represent values zero to nine, and A, B, C, D, E, F to represent values ten to fifteen...

Casting out nines

Main article: Casting out nines
Casting out nines
Casting out nines is a sanity check to ensure that hand computations of sums, differences, products, and quotients of integers are correct. By looking at the digital roots of the inputs and outputs, the casting-out-nines method can help one check arithmetic calculations...

After applying an arithmetic operation to two operands and getting a result, you can use this procedure to improve your confidence that the result is correct.

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. You now have two one-digit numbers, one condensed from the first operand and the other condensed from the second operand. (These one-digit numbers are also the remainders you would end up with if you divided the original operands by 9; mathematically speaking, they're 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 you 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 isn't guaranteed to be.

Example

Say we've calculated 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's a good chance that we were right that 6338 × 79 equals 500702.

You can use the same procedure with multiple operations just repeat steps 1 and 2 for each operation.

Estimation

When checking the mental calculation, it is useful to think of it in terms of scaling. For example, when dealing with large numbers, say 1531 × 19625, estimation instructs you to be aware of the number of digits expected for the final value. A useful way of checking is to estimate. 1531 is around 1500, and 19625 is around 20000, so a result of around 20000 × 1500 (30000000) would be a good estimate for the actual answer (30045875). So if the answer has too many digits, you know you've made a mistake.

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 211 would be unreasonable. Since 15 was a multiple of 5, so should the product. The correct answer is 210.

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, the problem can sometimes be modified:

If only one digit in b is larger than its corresponding digit in a, diminish the offending digit in b until it is equal to its corresponding digit in a. Then subtract further the amount b was diminished by from a. For example, to calculate 872 − 92, turn the problem into 872 − 72 = 800. Then subtract 20 from 800: 780.

If more than one digit in b is larger than its corresponding digit in a, it may be easier to find how much must be added to b to get a. For example, to calculate 8192 − 732, we can add 8 to 732 (resulting in 740), then add 60 (to get 800), then 200 (for 1000). Next, add 192 to arrive at 1192, and, finally, add 7000 to get 8192. Our final answer is 7460.
It might be easier to start from the left (the big numbers) first.

You may guess what is needed, and accumulate your guesses. Your guess is good as long as you haven't gone beyond the "target" number.
8192 − 732, mentally, you want to add 8000 but that would be too much, so we add 7000, then 700 to 1100, is 400 (so far we have 7400), and 32 to 92 can easily be recognized as 60. The result is 7460.

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".
We 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.
We are going to increase this place by using the number we borrowed from the
column to the left. Therefore:
10 − 8 = 2. It's 10 rather than 0, because we borrowed from the Thousands
place. 75 > 44 so no need to borrow,
say "two hundred"

Tens: 7 − 4 = 3, 5 > 4 so no need to borrow, say "thirty"

Ones: 5 − 4 = 1, say "one"

Calculating products: a × b

Many of these methods work because of the distributive property.

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 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 our 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 our answer is not changed. That
final answer is 0880.
The leftmost zero can be omitted, leaving 880.
So 176 times 5 equals 880.

Multiplying by 9

Since 9 = 10 − 1, to multiply by 9, multiply the number by 10 and then subtract the original number from this result. For example, 9 × 27 = 270 − 27 = 243. You can also use this method for eight but you need to double the number.

Using hands: 1–10 multiplied by 9

Hold hands in front of you, palms facing you. Assign the left thumb to be 1, the left index to be 2, and so on all the way to 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 10ⁿ (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

Integer

The integers are formed by the natural numbers together with the negatives of the non-zero natural numbers .They are known as Positive and Negative Integers respectively...

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 + 17 = 187

17 × 11 = 187

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:



(10+a) x (10+b)



100 + 10*(a + b) + a*b

which can be visualized as:



1

xx

 yy



for example:



17 * 16



1

13    (7+6)

 42   (7*6)



 272  (total)

Multiplying any 2-digit numbers

To easily multiply any 2-digit numbers together a simple algorithm is as follows:

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:

[of which only the tens digit will interfere with the first term]

i.e., in this example = 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.

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)

:right index finger (top)

: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:
=10

=10

8

9

7

--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 you join 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

which is the product we seek.

Using square numbers

The products of small numbers may be calculated by using the squares of integers; for example, to calculate 13 × 17, you can remark 15 is the mean of the two factors, and think of it as (15 − 2) × (15 + 2), i.e. 15² − 2². Knowing that 15² is 225 and 2² is 4, simple subtraction shows that 225 − 4 = 221, which is the desired product.

This method requires knowing by heart a certain number of squares:

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
11² = 121
12² = 144
13² = 169
14² = 196
15² = 225
16² = 256
17² = 289
18² = 324
19² = 361
20² = 400
21² = 441
22² = 484
23² = 529
24² = 576
25² = 625
26² = 676
27² = 729
28² = 784
29² = 841
30² = 900

Squaring numbers

It may be useful to be aware that the difference between two successive square numbers is the sum of their respective square roots. Hence if you know that 12 × 12 = 144 and wish to know 13 × 13, calculate 144 + 12 + 13 = 169.

This is because (x + 1)² − x² = x² + 2x + 1 − x² = x + (x + 1)

x² = (x − 1)² + (2x − 1)

Squaring numbers near 50

Suppose we need to square a number x near 50. This number may be expressed as x = 50 − n, and hence the answer x² is (50−n)², which is 50² − 100n + n². We know that 50² is 2500. So we subtract 100n from 2500, and then add n². Example, say we want to square 48, which is 50 − 2. We subtract 200 from 2500 and add 4, and get x² = 2304. For numbers larger than 50 (x = 50 + n), add n a hundred times instead of subtracting it.

Squaring a number ending in 5

1. Take the digit(s) that precede the five: abc5, where a, b, and c are digits
2. Multiply this number by itself plus one: abc(abc + 1)
3. Take above result and attach 25 to the end
- Example: 85 × 85
  1. 8
  2. 8 × 9 = 72
  3. So, 85² = 7,225
- Example: 125²
  1. 12
  2. 12 × 13 = 156
  3. So, 125² = 15,625
- Mathematical explanation

|(10x + 5)²> x + 5)(10x + 5)
>-
| x² + 100x + 25
>-
| x² + x) + 25
>-
|

Squaring an integer from 26 to 75

This method requires the memorization of squares from 1 to 25.

The square of n (most easily calculated when n is between 26 and 75 inclusive) is

(50 − n)² + 100(n − 25)

In other words, the square of a number is the square of its difference from fifty added to one hundred times the difference of the number and twenty five. For example, to square 62, we have:

(−12)² + [(62-25) × 100]

= 144 + 3,700

= 3,844

if want to square a two digit number that ends with 5 then it is easy.
Example 25*25=625
the last two digits are _ 25
for the first number multiply the first digit with next number
That is,
2*(2+1)=2*3=6
Hence the answer is 25*25=625
Same way 35*35=1225
45*45=2025
55*55=3025
65*65=4225
75*75=5625
85*85=7225
95*95=9025

Squaring an integer from 76 to 125

This method requires the memorization of squares from 1 to 25.

The square of n (most easily calculated when n is between 76 and 125 inclusive) is

(100 − n)² + 100(100 − 2(100 − n))

In other words, the square of a number is the square of its difference from one hundred added to the product of one hundred and the difference of one hundred and the product of two and the difference of one hundred and the number. For example, to square 93, we have:

7² + 100(100 − 2(7))

= 49 + 100 × 86

= 49 + 8,600

= 8,649

Another way to look at it would be like this:

93² = ? (is −7 from 100)

93 − 7 = 86 (this gives us our first two digits)

(−7)² = 49 (these are the second two digits)

93² = 8649

Another example:
82² = ? (is -18 from 100)
82 - 18 = 64 (subtract. First digits.)
(-18)² = 324 (second pair of digits. We'll need to carry the 3.)
82² = 6724

Squaring any number

Take a given number, and add and subtract a certain value to it that will make it easier to multiply. For example:

492²

492 is close to 500, which is easy to multiply by. Add and subtract 8 (the difference between 500 and 492) to get

492 -> 484, 500

Multiply these numbers together to get 242,000 (This can be done efficiently by dividing 484 by 2 = 242 and multiplying by 1000). Finally, add the difference (8) squared (8² = 64) to the result:

492² = 242,064

The proof follows:

Squaring any 2-digit integers

This method requires memorization of the squares of the one-digit numbers 1 to 9.

The square of mn, mn being a two-digit integer, can be calculated as

10 × m(mn + n) + n²

Meaning the square of mn can be found by adding n to mn, multiplied by m, adding 0 to the end and finally adding the square of n.

For example, we have 23²:
23²
= 10 × 2(23 + 3) + 3²
= 10 × 2(26) + 9
= 520 + 9
= 529

So 23² = 529.

Approximating square roots

An easy way to approximate the square root

Square root

In mathematics, a square root of a number x is a number r such that r2 = x, or, in other words, a number r whose square is x...

of a number is to use the following equation:

The closer the known square is to the unknown, the more accurate the approximation. For instance, to estimate the square root of 15, we could start with the knowledge that the nearest perfect square is 16 (4²).

So we've estimated the square root of 15 to be 3.875. The actual square root of 15 is 3.872983...

Derivation

Say we want to find the square root of a number we'll call x. By definition

We then redefine the root

where a is a known root (4 from the above example) and b is the difference between the known root and the answer we seek.

Expanding yields

And here's the trick. If 'a' is close to your target, 'b' will be a small enough number to render the

element of the equation negligible. So we drop

out and rearrange the equation to

and therefore

that can be reduced to

Extracting roots of perfect powers

Extracting roots of perfect powers is often practiced. The difficulty of the task does not depend on the number of digits of the perfect power but on the precision, i.e the number of digits of the root.

Extracting cube roots

An easy task for the beginner is extracting cube roots from the cubes of 2 digit numbers. For example, given 74088, determine what two digit number, when multiplied by itself once and then multiplied by the number again, yields 74088. One who knows the method will quickly know the answer is 42, as 42³ = 74088.

Before learning the procedure, it is required that the performer memorize the cubes of the numbers 1-10:

1³ = 1
2³ = 8
3³ = 27
4³ = 64
5³ = 125
6³ = 216
7³ = 343
8³ = 512
9³ = 729
10³ = 1000

A neat trick here is that there is a pattern. Remember that the pattern is adding and subtracting. Starting from zero:

0³ = 0
1³ = 1 up 1
2³ = 8 down 3
3³ = 27 down 1
4³ = 64 down 3
5³ = 125 up 1
6³ = 216 up 1
7³ = 343 down 3
8³ = 512 down 1
9³ = 729 down 3
10³ = 1000 up 1

There are two steps to extracting the cube root from the cube of a two digit number. Say you are asked to extract the cube root of 29791. Begin by determining the one's place (units) of the two digit number. You know it must be one, since the cube ends in 1, as seen above.

If perfect cube ends in 0, the cube root of it must end in 0.
If perfect cube ends in 1, the cube root of it must end in 1.
If perfect cube ends in 2, the cube root of it must end in 8.
If perfect cube ends in 3, the cube root of it must end in 7.
If perfect cube ends in 4, the cube root of it must end in 4.
If perfect cube ends in 5, the cube root of it must end in 5.
If perfect cube ends in 6, the cube root of it must end in 6.
If perfect cube ends in 7, the cube root of it must end in 3.
If perfect cube ends in 8, the cube root of it must end in 2.
If perfect cube ends in 9, the cube root of it must end in 9.

Note that every digit corresponds to itself except for 2, 3, 7 and 8, which are just subtracted from ten to obtain the corresponding digit.

The second step is to determine the first digit of the two digit cube root by looking at the magnitude of the given cube. To do this, remove the last three digits of the given cube (29791 -> 29) and find the greatest cube it is greater than (this is where knowing the cubes of numbers 1-10 is needed). Here, 29 is greater than 1 cubed, greater than 2 cubed, greater than 3 cubed, but not greater than 4 cubed. The greatest cube it is greater than is 3, so the first digit of the two digit cube must be 3.

Therefore, the cube root of 29791 is 31.

Another example:

Find the cube root of 456533.
The cube root ends in 7.
After the last three digits are taken away, 456 remains.
456 is greater than all the cubes up to 7 cubed.
The first digit of the cube root is 7.
The cube root of 456533 is 77.

Approximating common logs (log base 10)

To approximate a common log (to at least one decimal point accuracy), a few log rules, and the memorization of a few logs is required. One must know:

log(a x b) = log(a) + log(b)
log(a / b) = log(a) - log(b)
log(0) does not exist
log(1) = 0
log(2) ~ .30
log(3) ~ .48
log(7) ~ .85

From this information, one can find the log of any number 1-9.

log(1) = 0
log(2) ~ .30
log(3) ~ .48
log(4) = log(2 x 2) = log(2) + log(2) ~ .60
log(5) = log(10 / 2) = log(10) - log(2) ~ .70
log(6) = log(2 x 3) = log(2) + log(3) ~ .78
log(7) ~ .85
log(8) = log(2 x 2 x 2) = log(2) + log(2) + log(2) ~ .90
log(9) = log(3 x 3) = log(3) + log(3) ~ .96
log(10) = 1 + log(1) = 1

The first step in approximating the common log is to put the number given in scientific notation. For example, the number 45 in scientific notation is 4.5 x 10^1, but we will call it a x 10^b. Next, find the log of a, which is between 1 and 10. Start by finding the log of 4, which is .60, and then the log of 5, which is .70 because 4.5 is between these two. Next, and skill at this comes with practice, place a 5 on a logarithmic scale between .6 and .7, somewhere around .653 (NOTE: the actual value of the extra places will always be greater than if it were placed on a regular scale. i.e., you would expect it to go at .650 because it is halfway, but instead it will be a little larger, in this case .653) Once you have obtained the log of a, simply add b to it to get the approximation of the common log. In this case, a + b = .653 + 1 = 1.653. The actual value of log(45) = 1.65321.

The same process applies for numbers between 0 and 1. For example, .045 would be written as 4.5 x 10^-2. The only difference is that b is now negative, so when adding you are really subtracting. This would yield the result .653-2, or -1.347.

Other systems

There are many other methods of calculation in mental mathematics. The list below shows a few other methods of calculating, though they may not be entirely mental.

Vedic mental mathematics - (Vedic Mathematics page has been co-opted by history of indian mathematics page)
Trachtenberg system
Trachtenberg system
The Trachtenberg System is a system of rapid mental calculation. The system consists of a number of readily memorized operations that allow one to perform arithmetic computations very quickly. It was developed by the Jewish engineer Jakow Trachtenberg in order to keep his mind occupied while being...
Abacus system
Abacus system
The abacus system of mental calculation is a system where users mentally visualize an abacus to do calculations. No physical abacus is used; only the answers are written down....

- As students become used to manipulating the abacus
Abacus
The abacus, also called a counting frame, is a calculating tool used primarily in parts of Asia for performing arithmetic processes. Today, abaci are often constructed as a bamboo frame with beads sliding on wires, but originally they were beans or stones moved in grooves in sand or on tablets of...

with their fingers, they are typically asked to do calculation by visualizing abacus in their head. Almost all proficient abacus users are adept at doing arithmetic mentally.
Chisanbop

Mental Calculation World Cup

The first World Mental Calculation Championships (Mental Calculation World Cup) took place in 2004. They are repeated every second year. It consists of six different tasks: addition of ten ten-digit numbers, multiplication of two eight-digit numbers, calculation of square roots, and calculation of weekdays for given dates, plus some surprise tasks.

MEMORIAD - World Memory, Mental Calculation & Speed Reading Olympics

MEMORIAD is the first and only platform combining "MENTAL CALCULATION", "MEMORY" and "PHOTOGRAPHIC READING" competitions. Games and competitions are held in the olympic games order once every four year.

External links

The source of this article is wikipedia, the free encyclopedia. The text of this article is licensed under the GFDL.