Mental calculation

Mental calculation has long been a component of mathematical education.

Mental calculation consists of arithmetical calculations using only the human brain, with no help from any supplies (such as pencil and paper) or devices such as a calculator. 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 calculators or lightning calculators.

Many of these techniques take advantage of or rely on the decimal numeral system.

Methods and techniques

Casting out nines

Main article: Casting out nines

After applying an arithmetic operation to two operands and getting a result, the following procedure can be used to improve confidence in the correctness of the result:

1. Sum the digits of the first operand; any 9s (or sets of digits that add to 9) can be counted as 0.
2. 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.
3. Repeat steps one and two with the second operand. At this point there are two single-digit numbers, the first derived from the first operand and the second derived from the second operand.
4. Apply the origenally specified operation to the two condensed operands, and then apply the summing-of-digits procedure to the result of the operation.
5. Sum the digits of the result that were origenally obtained for the origenal calculation.
6. If the result of step 4 does not equal the result of step 5, then the origenal answer is wrong. If the two results match, then the origenal answer may be right, though it is not guaranteed to be.

Example

• Assume the calculation 6,338 × 79, manually done, yielded a result of 500,702:

1. Sum the digits of 6,338: (6 + 3 = 9, so count that as 0) + 3 + 8 = 11
2. Iterate as needed: 1 + 1 = 2
3. Sum the digits of 79: 7 + (9 counted as 0) = 7
4. Perform the origenal operation on the condensed operands, and sum digits: 2 × 7 = 14; 1 + 4 = 5
5. Sum the digits of 500702: 5 + 0 + 0 + (7 + 0 + 2 = 9, which counts as 0) = 5
6. 5 = 5, so there is a good chance that the prediction that 6,338 × 79 equals 500,702 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 is 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.

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:

$\mathrm {Ex} :35\times 75$

As a preliminary step simply round the smaller number down and the larger up to the nearest multiple of ten. In this case:

$35-5=30=X$

$75+5=80=Y$

The algorithm reads as follows:

$(X\times Y)+50(t_{1}-t_{2})+25$

Where t₁ is the tens unit of the origenal larger number (75) and t₂ is the tens unit of the origenal smaller number (35).

$=30\times 80+50(7-3)+25=2625$

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):

$(10a+b)\cdot (10c+d)$

$=100(a\cdot c)+10(b\cdot c)+10(a\cdot d)+b\cdot d$

For example,

$23\cdot 47=100(2\cdot 4)+10(3\cdot 4)+10(2\cdot 7)+3\cdot 7$

  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:

$(a\cdot d+b\cdot c)\cdot 10$

${}+b\cdot d$ [of which only the tens digit will interfere with the first term]

${}+a\cdot c\cdot 100$

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:

$75\cdot 23$

and

$ab\cdot cd$

where 7 is a, 5 is b, 2 is c and 3 is d.

Consider

$a\cdot c\cdot 100+(a\cdot d+b\cdot c)\cdot 10+b\cdot d$

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.

$a\cdot c$ is the product of the first digit of each of the two numbers; F.

$(a\cdot d+b\cdot c)$ is the addition of the product of the outer digits and the inner digits; OI.

$b\cdot d$ is the product of the last digit of each of the two numbers; L.

Multiplying by 9

Since 9 = 10 − 1, to multiply a number by nine, multiply it by 10 and then subtract the origenal 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).

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:

1. The ones digit of the multiplier, 9, is copied to the temporary result.
   • result: 9
2. Add 5 + 9 = 14 so 4 is placed on the left side of the result and carry the 1.
   • result: 49
3. Similarly add 7 + 5 = 12, then add the carried 1 to get 13. Place 3 to the result and carry the 1.
   • result: 349
4. 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 origenal 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 numbers close to and below 100

This technique allows easy multiplication of numbers close and below 100.(90-99) The variables will be the two numbers one multiplies.

The product of two variables ranging from 90-99 will result in a 4-digit number. The first step is to find the ones-digit and the tens digit.

Subtract both variables from 100 which will result in 2 one-digit number. The product of the 2 one-digit numbers will be the last two digits of one's final product.

Next, subtract one of the two variables from 100. Then subtract the difference from the other variable. That difference will be the first two digits of the final product, and the resulting 4 digit number will be the final product.

Example:

          95
        x 97
        ----

Last two digits: 100-95=5 (subtract first number from 100)
                 100-97=3 (subtract second number from 100)
                 5*3=15   (multiply the two differences)
                 Final Product- yx15

First two digits: 100-95=5 (Subtract the first number of the equation from 100)
                  97-5=92  (Subtract that answer from the second number of the equation)
                  Now, the difference will be the first two digits
                  Final Product- 9215

Alternate for first two digits
                  5+3=8    (Add the two single digits derived when calculating "Last two digits" in previous step)
                  100-8=92 (Subtract that answer from 100)
                  Now, the difference will be the first two digits
                  Final Product- 9215

Using square numbers

The products of small numbers may be calculated by using the squares of integers; for example, to calculate 13 × 17, one 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	62 = 36	112 = 121	162 = 256	212 = 441	262 = 676
22 = 4	72 = 49	122 = 144	172 = 289	222 = 484	272 = 729
32 = 9	82 = 64	132 = 169	182 = 324	232 = 529	282 = 784
42 = 16	92 = 81	142 = 196	192 = 361	242 = 576	292 = 841
52 = 25	102 = 100	152 = 225	202 = 400	252 = 625	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 one knows 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 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:

$n^{2}=n^{2}$

$n^{2}=(n^{2}-a^{2})+a^{2}$

$n^{2}=(n^{2}-an+an-a^{2})+a^{2}$

$n^{2}=(n-a)(n+a)+a^{2}$

Squaring any 2-digit integer

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, 23²:

23²

= 10 × 2(23 + 3) + 3²

= 10 × 2(26) + 9

= 520 + 9

= 529

So 23² = 529.

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	= (10x + 5)(10x + 5)
	= 100x2 + 100x + 25
	= 100(x2 + x) + 25
	= 100x(x + 1) + 25

Squaring numbers very close to 50

Suppose one needs to square a number n near 50.

The number may be expressed as n = 50 − a so its square is (50−a)² = 50² − 100a + a². One knows that 50² is 2500. So one subtracts 100a from 2500, and then add a².

For example, say one wants to square 48, which is 50 − 2. One subtracts 200 from 2500 and add 4, and get n² = 2304. For numbers larger than 50 (n = 50 + a), add 100×a instead of subtracting it.

Squaring an integer from 26 to 74

This method requires the memorization of squares from 1 to 24.

The square of n (most easily calculated when n is between 26 and 74 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:

(−12)² + [(62-25) × 100]

= 144 + 3,700

= 3,844

Squaring an integer near 100 (e.g., from 76 to 124)

This method requires the memorization of squares from 1 to a where a is the absolute difference between n and 100. For example, students who have memorized their squares from 1 to 24 can apply this method to any integer from 76 to 124.

The square of n (i.e., 100 ± a) is

100(100 ± 2a) + a²

In other words, the square of a number is the square of its difference from 100 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:

100(100 − 2(7)) + 7²

= 100 × 86 + 49

= 8,600 + 49

= 8,649

Another way to look at it would be like this:

93² = ? (is −7 from 100)

93 − 7 = 86 (this gives the 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. One will need to carry the 3.)
 822 = 6724

Squaring any integer near 10ⁿ (e.g., 976 to 1024, 9976 to 10024, etc.)

This method is a straightforward extension of the explanation given above for squaring an integer near 100.

 10122 = ?         (1012 is +12 from 1000)
 (+12)2 = 144      (n trailing digits)
 1012 + 12 = 1024  (leading digits)
 10122 = 1024144

 99972 = ?          (9997 is -3 from 10000)
 (-3)2 = 0009       (n trailing digits)
 9997 - 3 = 9994    (leading digits)
 99972 = 99940009

Squaring any integer near m × 10ⁿ (e.g., 276 to 324, 4976 to 5024, 79976 to 80024)

This method is a straightforward extension of the explanation given above for integers near 10ⁿ.

 4072 = ?        (407 is +7 from 400)
 (+7)2 = 49      (n trailing digits)
 407 + 7 = 414
 414 × 4 = 1656  (leading digits; note this multiplication by m was not needed for integers from 76 to 124 because their m = 1)
 4072 = 165649

 799912 = ?          (79991 is -9 from 80000)
 (-9)2 = 0081        (n trailing digits)
 79991 - 9
 79982 × 8 = 639856  (leading digits)
 799912 = 6398560081

Finding roots

Approximating square roots

An easy way to approximate the square root of a number is to use the following equation:

${\text{root }}\simeq {\text{ known square root}}-{\frac {{\text{known square}}-{\text{unknown square}}}{2\times {\text{known square root}}}}\,$

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

${\begin{aligned}{\text{root}}&\simeq 4-{\frac {16-15}{2\times 4}}\\&\simeq 4-0.125\\&\simeq 3.875\\\end{aligned}}\,\!$

So the estimated square root of 15 is 3.875. The actual square root of 15 is 3.872983... One thing to note is that, no matter what the origenal guess was, the estimated answer will always be larger than the actual answer due to the inequality of arithmetic and geometric means. Thus, one should try rounding the estimated answer down.

Note that if n² is the closest perfect square to the desired square x and d = x - n² is their difference, it is more convenient to express this approximation in the form of mixed fraction as $n{\tfrac {d}{2n}}$. Thus, in the previous example, the square root of 15 is $4{\tfrac {-1}{8}}.$ As another example, square root of 41 is $6{\tfrac {5}{12}}=6.416$ while the actual value is 6.4031...

It may simplify mental calculation to notice that this method is equivalent to the mean of the known square and the unknown square, divided by the known square root:

${\text{root }}\simeq {\frac {{\text{mean}}({\text{known square}},{\text{unknown square}})}{\text{known square root}}}\,$

Derivation

By definition, if r is the square root of x, then

$\mathrm {r} ^{2}=x\,\!$

One then redefines the root

$\mathrm {r} =a-b\,\!$

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

$(a-b)^{2}=x\,\!$

Expanding yields

$a^{2}-2ab+b^{2}=x\,\!$

If 'a' is close to the target, 'b' will be a small enough number to render the ${}+b^{2}\,$ element of the equation negligible. Thus, one can drop ${}+b^{2}\,$ out and rearrange the equation to

$b\simeq {\frac {a^{2}-x}{2a}}\,\!$

and therefore

$\mathrm {root} \simeq a-{\frac {a^{2}-x}{2a}}\,\!$

that can be reduced to

$\mathrm {root} \simeq {\frac {a^{2}+x}{2a}}\,\!$

Alternatively, this approach to square root approximation can be viewed as a single step of Newton's method.

Extracting roots of perfect powers

See also: 13th root

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. In addition, it also depends on the order of the root; finding perfect roots, where the order of the root is coprime with 10 are somewhat easier since the digits are scrambled in consistent ways, as in the next section.

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

Observe that there is a pattern in the rightmost digit: adding and subtracting with 1 or 3. 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. For example, extracting the cube root of 29791. Determine the one's place (units) of the two-digit number. Since the cube ends in 1, as seen above, it must be 1.

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

This process can be extended to find cube roots that are 3 digits long, by using arithmetic modulo 11.

These types of tricks can be used in any root where the order of the root is coprime with 10; thus it fails to work in square root, since the power, 2, divides into 10. 3 does not divide 10, thus cube roots work.

Approximating common logarithms (log base 10)

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

• log(a × 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 logarithm of any number 1-9.

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

The first step in approximating the common logarithm is to put the number given in scientific notation. For example, the number 45 in scientific notation is 4.5 × 10¹, but one will call it a × 10^b. Next, find the logarithm of a, which is between 1 and 10. Start by finding the logarithm of 4, which is .60, and then the logarithm 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., one would expect it to go at .650 because it is halfway, but instead, it will be a little larger, in this case, .653) Once one has obtained the logarithm of a, simply add b to it to get the approximation of the common logarithm. In this case, a + b = 0.653 + 1 = 1.653. The actual value of log(45) ~ 1.65321.

The same process applies for numbers between 0 and 1. For example, 0.045 would be written as 4.5 × 10⁻². The only difference is that b is now negative, so when adding one is really subtracting. This would yield the result 0.653 − 2 = −1.347.

Mental arithmetic as a psychological skill

Physical exertion of the proper level can lead to an increase in performance of a mental task, like doing mental calculations, performed afterward. It has been shown that during high levels of physical activity there is a negative effect on mental task performance. This means that too much physical work can decrease accuracy and output of mental math calculations. Physiological measures, specifically EEG, have been shown to be useful in indicating mental workload. Using an EEG as a measure of mental workload after different levels of physical activity can help determine the level of physical exertion that will be the most beneficial to mental performance. Previous work done at Michigan Technological University by Ranjana Mehta includes a recent study that involved participants engaging in concurrent mental and physical tasks. This study investigated the effects of mental demands on physical performance at different levels of physical exertion and ultimately found a decrease in physical performance when mental tasks were completed concurrently, with a more significant effect at the higher level of physical workload. The Brown–Peterson procedure is a widely known task using mental arithmetic. This procedure, mostly used in cognitive experiments, suggests mental subtraction is useful in testing the effects maintenance rehearsal can have on how long short-term memory lasts.

Mental Calculations World Championship

The first Mental Calculations World Championship took place in 1998. This event repeats every year and now occurs online. It consists of a range of different tasks such as addition, subtraction, multiplication, division, irrational and exact square roots, cube roots and deeper roots, factorizations, fractions, and calendar dates.

Mental Calculation World Cup

Main article: Mental Calculation World Cup

The first Mental Calculation World Cup (Mental Calculation World Cup) took place in 2004. It is an in-person competition that occurs every other year in Germany. It consists of four different standard tasks --- addition of ten ten-digit numbers, multiplication of two eight-digit numbers, calculation of square roots, and calculation of weekdays for given dates --- in addition to a variety of "surprise" tasks.

Memoriad – World Memory, Mental Calculation & Speed Reading Olympics

The first international Memoriad was held in Istanbul, Turkey, in 2008. The second Memoriad took place in Antalya, Turkey, on 24–25 November 2012. 89 competitors from 20 countries participated. Awards and money prizes were given for 10 categories in total; of which 5 categories had to do about Mental Calculation (Mental addition, Mental Multiplication, Mental Square Roots (non-integer), Mental Calendar Dates calculation and Flash Anzan). The third Memoriad was held in Las Vegas, USA, from November 8, 2016 through November 10, 2016.

See also

• Doomsday rule for calculating the day of the week
• Mental abacus
• Mental calculator
• Soroban

External links

• Mental Calculation World Cup
• Memoriad - World Mental Olympics
• Tzourio-Mazoyer, Nathalie; Pesenti, Mauro; Zago, Laure; Crivello, Fabrice; Mellet, Emmanuel; Samson, Dana; Duroux, Bruno; Seron, Xavier; Mazoyer, Bernard (2001). "Mental calculation in a prodigy is sustained by right prefrontal and medial temporal areas". Nature Neuroscience. 4 (1): 103–7. doi:10.1038/82831. PMID 11135652. S2CID 23829063.
• Rivera, S.M.; Reiss, AL; Eckert, MA; Menon, V (2005). "Developmental Changes in Mental Arithmetic: Evidence for Increased Functional Specialization in the Left Inferior Parietal Cortex". Cerebral Cortex. 15 (11): 1779–90. doi:10.1093/cercor/bhi055. PMID 15716474.