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**Mental calculation** is the practice of doing mathematical calculations using only the human brain, with no help from any computing devices. It is practiced as a sport in the Mind Sports Olympiad. Mental calculation is said to improve mental capability, speed of response, memory power and concentration.^{[How to reference and link to summary or text]}

Practically, mental calculations are not only helpful when computing tools are not available, but they also can be helpful in situations where it is beneficial to calculate with speed. When a method is much faster than the conventional methods (as taught in school), it may be called a shortcut. Although used to aid or quicken tedious computation, many also practice or create such tricks to impress their peers with their quick calculating skills. Almost all such methods make use of the fact that we use the base 10 system.

There are many different techniques for performing mental calculations, many of which are specific to a type of problem.

## Cognitive psychology and mental calculation[]

## Sanity tests[]

*Main article: Sanity test*

A quick test to further increase confidence that the correct answer to a calculation has been found.

### Casting out nines[]

*Main aricle: Casting out nines*- Sum the digits of the two operands separately, any 9s can be counted as 0
- Repeat step one until both operands degenerate to one digit
- Sum the digits of the supposed answer as in step one
- Apply the same operation to the two degenerated operands and then apply the same summing procedure
- If the result of step 4 does not equal the result of step 3, the answer is wrong

- Example
- 632 times 7 equals 4424

- 6 + 3 + 2 = 9 + 2 --> 0 + 2 =
**2**, 7 =**7** - One digit already
- 4 + 4 + 2 + 4 = 14, 1 + 4 = 5
- 2 × 7 = 14, 1 + 4 = 5
- 5 = 5, so now we can say with confidence that 632 × 7 = 4424

### 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, it, 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 therefore a result of around 20000X1500 (30000000) would be a good estimate. 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 of 5, so should the product. The correct answer is 210.

## Calculating differences: *a* − *b*[]

### Direct calculation[]

When the digits of *b* are all smaller than the 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.

### 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" Hundreds: 0-8=negative numbers not allowed here, 10-8=2, 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[]

In this case, the product can be essentially calculated digit by digit. This is not exactly the case because it is possible to have a remainder, but if there is a remainder, it is always 1, which simplifies things greatly. Still, the product must be calculated from right to left: 2 × 167 is by 4 with a remainder, then a 2 (so 3) with another remainder, then a 2 (so 3). Thus, we get 334.

### Multiplying by 5[]

To multiply a number by 5, first multiply that number by 10, then divide it by 2. The following algorithm is a quick way to produce this result: First, append a zero to right side of the desired number. Next, starting from the leftmost numeral, divide by 2 and append each result in the respective order to form a new number; fraction answers should be rounded down to the nearest whole number. For example, if you intended to multiply 176 by 5, you would first append a zero to 176 to make 1760. Next, divide 1 by 2 to get .5, rounded down to zero. Divide 7 by 2 to get 3.5, rounded down to 3. Divide 6 by 2 to get 3. Zero divided by two is simply zero. The resulting number is 0330. The final step involves adding 5 to the number that follows any single numeral in this new number that was odd before dividing by two; this is better understood through the example. In the original number, 176, the first place is 1, which is odd. Therefore, we add 5 to the numeral after the first place in our newly constructed number (0330), which is 3; 3+5=8. The numeral in the second place of 176, 7, is also odd. Therefore the number-place after the corresponding numeral in the constructed number (0830) is increased by 5 as well; 3+5=8. 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.

#### 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 append 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 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:

1a x 1b 100 + 10 * (a+b) + a*b which can be visualized as: 1 xx yy for example: 17 * 16 1 13 42 272

### Multiplying Any 2 digit Numbers Together[]

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

ab * cd 100*(a*c) + 10*(b*c) + 10*(a*d)+ b*d for example 23 47 800 120 140 21 1081

### 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 to desired numbers together. The point of contact and below is considered the "below" section and everything above the two fingers that are touching are part of the "above" section. For example, 6 × 9 would look like this:

-10-- --9-- --8-- (above) -10-- --7-- ==================== --9-- --6-- left index and right little finger are touching --8-- (below) --7-- --6-- (9 × 6)

-10-- -10-- --9-- --9-- --8-- --8-- --7-- --7-- --6-- --6--

Here are two examples:

- 9 × 6

above:

-10-- --9-- --8-- -10-- --7--

below:

--9-- --6-- --8-- --7-- --6--

- 5 fingers below make 5 tens - 4 fingers above to the right - 1 finger above to the left

the result: 9 × 6 = 50 + 4 × 1 = 54

- 6 × 8

above:

-10-- --9-- --8-- -10-- --7-- --9--

below:

--6-- --8-- --7-- --6--

- 4 fingers below make 4 tens - 2 fingers above to the right - 4 fingers above to the left

result: 6 × 8 = 40 + 2 × 4 = 48

How it works: each finger represents a number (between 6 and 10). Join the fingers representing the
numbers you wish to multiply (*x* and *y*). The fingers below give the number of tens, that is (*x* − 5) + (*y* − 5). The digits to the upper left give (10 − *x*) and those to the upper right give (10 − *y*), leading to [(*x* − 5) + (*y* − 5)] × 10 + (10 − *x*) × (10 − *y*) = *x* × *y*.

### 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. 15^{2} − 2^{2}. Knowing that 15^{2} is 225 and 2^{2} 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:

- 1
^{2}= 1 - 2
^{2}= 4 - 3
^{2}= 9 - 4
^{2}= 16 - 5
^{2}= 25 - 6
^{2}= 36 - 7
^{2}= 49 - 8
^{2}= 64 - 9
^{2}= 81 - 10
^{2}= 100 - 11
^{2}= 121 - 12
^{2}= 144 - 13
^{2}= 169 - 14
^{2}= 196 - 15
^{2}= 225 - 16
^{2}= 256 - 17
^{2}= 289 - 18
^{2}= 324 - 19
^{2}= 361

### Squaring numbers[]

Any square number may be easily calculated by adding the previous square number, its positive square root, and the number whose square you wish to know. For example, the square of 13 is 144 + 12 + 13 = 169.

#### 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 *x*^{2} is (50−*n*)^{2}, which is *50 ^{2} − 100n + n^{2}*. We know that 50

^{2}is 2500. So we subtract 100

*n*from 2500, and then add

*n*. Example, say we want to square 48, which is 50 − 2. We subtract 200 from 2500 and add 4, and get

^{2}*x*= 2304. For numbers larger than 50 (

^{2}*x*=50+

*n*), add

*n*a hundred times instead of subtracting it.

#### Squaring a number ending in 5[]

- Take the digit(s) that precede the five -
*abc5*, where*a, b,*and*c*are digits - Multiply this number by itself plus one -
*abc × (abc + 1)* - Take above result and attach
*25*to the end

- Example: 85 × 85
- 8
- 8 × 9 = 72
- So, 85 squared = 7225

- Example: 125^2
- 12
- 12 × 13 = 156
- So, 125 squared = 15625

- Mathematical explanation
*(10x + 5)^2 = 100(x(x + 1)) + 25**(10x + 5)(10x + 5) = 100(x^2 + x)) + 25**100x^2 +100x + 25 = 100x^2 + 100x + 25*

- Take the digit(s) that precede the five -

## Finding roots[]

### Approximating square roots[]

Say we want to find out the square root of a non-square number. Using the formula (*a* − *b*)^{2} = *a*^{2} − 2*ab* + *b*^{2}. If you choose a 'b' value small enough you can get an accurate estimate. For example, if we are asked to find the square root of 15, we could start with the knowledge that the root of 16 is 4. Now we need a 'b' to put into the equation (4 − *b*)^{2} = 15, or thereabouts. Since (4 − *b*)^{2} = 16 − 2 × 4 × *b* roughly, we get *b* = (16 − 15) ÷ (2 × 4), or roughly 0.125. So then an estimation for the square root is 3.875. If you're after a more accurate value, then restart with an estimate of around 3.9. 3.9)^{2} we can work out as 15.21, so we do the same working as before; but end up with (3.9 − *b*)^{2} = 15, getting *b* = (15 − 3.9^{2}) ÷ (2 × 3.9) = (15 − 15.21) ÷ (7.8) = roughly -0.027. The square root of 15 is now estimated as 3.9 − 0.027 or 3.873. (the real square root of 15 is 3.8729833...)

### Extracting roots of perfect powers[]

This is a surprisingly easy task for many higher powers, but not very useful except for impressing friends (practical uses of finding roots rarely use perfect powers). The task not as hard as it sounds mainly because the basic method is to find the last digit using the last digit of the given power and then finding the other digits by using the magnitude of the given power. Such feats may seem obscure but are nevertheless recorded and practiced. See 13th 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 42^{3} = 74088.

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

- 1
^{3}= 1 - 2
^{3}= 8 - 3
^{3}= 27 - 4
^{3}= 64 - 5
^{3}= 125 - 6
^{3}= 216 - 7
^{3}= 343 - 8
^{3}= 512 - 9
^{3}= 729 - 10
^{3}= 1000

There are two steps to extracting the cube root from the cube of a two digit number. Say you 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.

## 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 mathematics
- Trachtenberg system
- Abacus system
- Chisenbop

## Mental Calculation World Cup[]

The first World Mental Calculation Championships (Mental Calculation World Cup) took place 2004. They are repeated every second year. The event of 2006 took place on November 4, 2006 in Giessen, Germany. It consists of four 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 two surprise tasks. It was won by Robert Fountain from England.

The next Championship is scheduled for 2008.

## See also[]

- Mental calculator
- 13th root
- Doomsday rule for calculating the day of the week

## External links[]

- official 13th root site
- Mental Calculation World Cup
- Fast Arithmetic Tips
- Divisibility Criteria
- Let's abolish Paper and Pencil Arithmetic
- MathAbacus.com Learning Mathematics with the Abacus for Kids
- Free Speed Math Program A 26k speed math program made in flash with a google video manual.
- Mental calculation is said to improve mental capability, speed of response, memory power and concentration.
- Mental processes and the creation of a secondary memory to facilitate calculation
- Evidence for Increased Functional Specialization in the Left Inferior Parietal Cortex
- Large EEG waves ellicited by Mental Calculation PDF
- http://mikesmath.com/
- "Dead Reckoning: Calculating Without Instruments" website

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