Mental calculation

The mental calculation consists of performing mathematical calculations using only the brain, without the help of other instruments such as calculators or even pencil and paper or fingers to count easily. Mental math often involves the use of specific techniques designed for particular types of problems.

Mental calculation. In public school S. A. RachinskiNikolái Bogdánov-Belski's 1895 picture in which children try to mentally calculate the result of the expression on the board, 102+112+122+132+142365{displaystyle {frac {10^{2}+11^{2}+12^{2} +13^{2}+14^{2}}{365}}}}}}.

People with an unusually high ability to perform mental calculations are called mental calculators. Some calculators can perform very complex mathematical operations (such as products of numbers with 5 or more digits) using mental calculation. However, the best mathematicians often do not coincide with the best calculators. Likewise, the best calculators are not the ones with the best memories, since the techniques of mental calculation and memory enhancement techniques are different. The world champions and those who appear in the Guinness book of records for both specialties (calculation and memory) are usually always different.

Practicing mental arithmetic helps the student put various strategies into play. Among its benefits are: development of numerical sense and intellectual abilities such as attention and concentration, as well as a taste for mathematics. For your teaching it is advisable to show the discovery of mnemonic rules.^{[citation required]}

Strategies

Addition and subtraction

If there are no carries, that is, if no partial sum is greater than 9, the sums can be performed directly. The same goes for subtractions.

Otherwise, you have to know how to model the numbers you have, sometimes converting a sum of two numbers into a simpler sum of more addends, and something similar for subtraction. Calculists like Alberto Coto and Jorge Arturo Mendoza Huertas propose to always do the sums from left to right, even if there are carries.

Examples:

• Calculate 456 + 155:

456 + 155 = 461 + 150 = 511 + 100 = 611 (traditional method, adding from right to left)

455 + 5 + 151 = 460 + 40 + 111 = 500 + 111 = 611 (with the first adding to the top ten, to the top hundred... to finish making a simpler sum equivalent to the first)

456 + 155 = 556 + 55 = 606 + 5 = 611 (from left to right)

• Calculate 876 - 98:

876 - 98 = 868 - 90 = 778 (traditional method, from right to left)

876 - 98 = 876 - (100 - 2) = 876 - 100 + 2 = 776 + 2 = 778 (depending on the proximity of the subtraction (98) to one that facilitates the subtraction (100)

876 - 98 = 786 - 8 = 778 (resting from left to right)

• Calculate 634 - 256:

634 - 256 = 434 - 56 = 384 - 6 = 378 (from left to right)

Multiplications and Divisions

Duplication and mediation

Multiplying by 2 is the same as adding the same number to the starting number. Duplication and mediation are a fundamental pillar of Egyptian mathematics.

Example: multiply 173 × 16:

This can be done by successive duplications: 173 × 16 = 346 × 8 = 692 × 4 = 1384 × 2 = 2768.

Multiplication and mediation are used, in general, to calculate the product of any number by the product of powers of 2 and 5. Multiplying by 5 is the same as calculating half the initial number multiplied by 10, which sometimes it's easier to find

Example: multiply 376 × 125

Like 125 = 53 = 103/23, you can find the solution by adding the corresponding three zeros and dividing the result three times by 2.

376 × 125 = 376000/8 = 188000/4 = 94000/2 = 47000.

• 324 x 125 = 324000/8 = 162000/4 = 81000/2 = 40500.

It is useful to know some powers of 2 and 5 to perform these operations fluently.

You can also use this method to multiply by other numbers that are sums of (few) powers of 2 or 5, such as 12 (8 + 4), 130 (125 + 5), 18 (16 + 2), etc

Multiplying by numbers close to powers of 10

Multiplying by 9, 11, 99, 101..., that is, by a power of 10 minus 1 (or plus 1), can be done mentally with a little practice by adding (or subtracting) 10 ⁿ times the initial number plus (or subtracts) the initial number. However, it is easy to make mistakes when adding or subtracting by mixing, for example, ones with tens.

Example: multiply 28 × 99

28 × 99 = 28 × (100 - 1) = 2800 - 28 = 2772

Another example: multiply 37 × 121

121 is the square of 11, so what is requested is the same as multiplying 37 by 11 and the result again by 11: 37 × 121 = 37 ×(10 + 1) × 11 = (370 + 37) × 11 = 407 × 11 = 4477

In addition, multiplying by 11 is easy: the numbers are separated and then the units number is always written and then groups of two consecutive numbers are added, putting the result or the last number of the sum carrying a carry of 1 if the sum is greater than 10, and finally the most significant figure is placed, like this:

Multiply:

12345 × 11: 1 unid 5, 5+4=9, 4+3=7, 3+2=5,2+1=3, and finally 1; now place in reverse order: 135795

8946 × 11: 1 unid 6, 6+4=10 (0 and carries 1), 4+9+1(carree)=14 (4 and carries 1), 9+8+1(carree)=18 (8 and carries 1), and finally 8+1(carree)=9; now place in reverse order: 98406

Similarly, this can be applied to multiplications by powers of 2, or 5, plus 1. For example, 26, 17, 124, and 63.

Multiply by 37

First, just remember the following:

• 37 × 3 = 111
• 37 × 27 = (37 × 3) × 9 = 999 = 1000 - 1

The procedure is this:

1. The other factor is divided between 3. Remember the quotient and the rest. If the rest is 1, the final result will have to be added 37; if it is 2, 74 will have to be added.

   Example: in 37 × 94, it is taken 94: 3 = 31, rest 1. Now the product is 111 × 31.

2. The quotient of the previous step is divided between 9. The quotient is multiplied by 999 (= 1000 - 1) and the rest by 111.

   In the previous example, 31: 9 = 3, rest 4. We now have the sum of two products: 999 × 3 (= 2997, or, if preferred, 3000 - 3) and 111 × 4 = 444. As the rest of the first quotient we did was 1, the result will have to be added 37.

3. It adds everything.

   3000 - 3 + 444 + 37 = 3000 + 444 + 37 - 3 (it is often easier to organize the terms in this way, leaving the number remaining at the end) = 3444 + 3478.

A variant is to take by excess and not by default the quotient of the division of the first step. This means that one is added to the quotient and 3 is subtracted from the remainder. Thus, instead of a number of the form 3 × Q + R (where R = 1 or 2) we have one of the form 3 × (Q + 1) + R' (where R' = -2 or -1, respectively), and 74 or 37 will be subtracted from the final result (because the new "remainder" of the division is negative).

More examples:

37 × 54 = 111 × 18 = 999 × 2 = 2000 - 2 = 1998

37 × 79 (usual method) = 111 × 26 + 37 = 999 × 2 + 111 × 8 + 37 = 2000 - 2 + 888 + 37 = 2925 - 2 = 2923

37 × 79 (variant) = 111 × 27 - 74 = 999 × 3 - 74 = 3000 - 3 - 74 = 3000 - 77 = 2923

As you can see, in this case the variant is easier, although it does not have to be always like this. In general, if the factor is one or two less than a multiple of 27 (remember that 37 × 27Q = 999Q), it's easier to go for that 27th multiple.

If one of the factors of the product is not 37 but is a multiple, the multiplication can be reformulated making one of the factors 37. Let's try for example with the following squares:

74 × 74 = 37 × 2 × 74 = 37 × 148 = 111 × 49 + 37 = 999 × 5 + 111 × 4 + 37 = 5000 - 5 + 444 + 37 = 5444 + 32 = 5476

111 × 111 = 37 × 3 × 111 = 37 × 333 = 999 × 12 + 333 = 12000 - 12 + 333 = 12321 (in this case, as we already had 333, the procedure was easier)

148 × 148 = 37 × 4 × 148 = 37 × 592 = 111 × 198 - 74 (in this case the variant is reused because 594 is multiple of 27) = 999 × 22 - 74 = 22000 - 22 - 74 = 21904

Methods like this work when one of the multiplication factors has a multiple that is a concatenation of nines. It is therefore a matter of finding that multiple. Another notable example is the number 142857. Not only is the product of this number times 7 equal to 999999, but its multiplication table is very simple, since in the string 142857142857... it is enough to take six consecutive digits from a given position:

142857 × 1 = 142857

142857 × 2 = 285714

142857 × 3 = 428571

142857 × 4 = 571428

142857 × 5 = 714285

142857 × 6 = 857142

Let's try calculating the square of this six-digit (!) number:

142857 × 142857 = (142857 × 7) × (142857: 7) = 9999 × 20408 + 142857 (Like the rest of 142857: 7 da 1, the result of multiplication is 142857. It is the same as in multiplication by 37) = (1,000.000 - 1) × 20,408 + 128,857 = 20.408,000.000 - 20,408 + 142857 = 20.408,000.000 + 122,449 = 20,408,122.449

Notable equalities and calculus of squares

The so-called notable equalities can be applied to mental calculation:

1. (a + b2 = a2 + 2ab + b2
2. (a - b2 = a2 - 2ab + b2
3. (a + b) (a - b) = a2 - b2

Calculation of the square of any two-figure number

The first two identities can be applied to calculating perfect squares. Suppose we want to calculate 52². 52 = 50 + 2, so we apply the corresponding identity to the square of the sum, where a = 50 and b = 2.

(50 + 2)2 = 502 + 2 × 50 + 22 = 2500 + 200 + 4 = 2704

More examples:

172 = (10 + 7)2 = 102 + 2 × 10 + 72 = 100 + 140 + 49 = 289

762 = (70 + 6)2 = 702 + 2 × 6 × 70 + 62 = 4900 + 840 + 36 = 5776

952 = (90 + 5)2 = 902 + 2 × 5 × 90 + 52 = 8100 + 900 + 25 = 9025

With this method it is also easy to calculate the square of a number with a whole number and a decimal, you just have to remember the place that each number occupies:

2.42 = (2 + 0.4)2 = 0.12 × 142 = 0.01 × (202 + 2 × 4 × 20 + 42) = 0.01 × (400 + 160 + 16) = 0.01 × 576 = 5,76

Algorithm to square a two-digit number starting with 4: (4*10+u)^2 = (15+u) and (10-u)^2 Example: 47^2= (15+7) and (10-7)^2 = 22 and 09 =2209, since 47^2= 40x40 + 40x7x2 + 7x7 = 1600 + 560 + 49 = 2209.

Idem algorithm, for those that start with 5.- (5*10+u)^2 =(25+u) and u^2; example: 53^2= (25+3) and 3^2 = 2809

Idem algorithm, for those that start with 9.- (9*10+u)^2= (80+2u)y(10-u)^2; example: 96^2=(80+2*6)y(10-6)^2= 92y16= 9216

Idem algorithm, for those with three digits that start with 10.- (10*10+u)^2= (100+2u)y u^2; example 108^2= (100+2*8)y8^2 = 116y64= 11664

Some calculators know the multiplication tables from 1 to 100 by heart, so they can easily use this method to find the square of a number of four digits or more. This is only achieved after a lot of training, but it greatly simplifies the calculation as can be seen:

57822 = (5700 + 82)2 = 57002 + 2 × 82 × 5700 + 822 = 32.490.000 + 934.800 + 6.724 = 33.431.524

Product of two numbers equidistant from a number whose square is known

The number whose square is known will generally be one ending in 0. For example, when calculating 58 × 62 we will rely on 60, since both are at the same distance (2 units) from 60. Here we You can use the third identity, that of the product of sum by difference, where a = 60 and b = 2.

(60 + 2) (60 - 2) = 602 - 22 = 3600 - 4 = 3596

More examples:

77 × 83 = (80 - 3) (80 + 3) = 6400-9= 6391

95 × 105 = (100 - 5) (100 + 5) = 10000-25= 9975

128 × 152 = (140 - 12) (140 + 12) = 19600-144= 19456

Square of a number ending in 5

Calculating the square of a number ending in 5 can be simplified by using the third identity. Here a will be the starting number (for example, 65), and b = 5:

(a + 5) (a - (5) = a2 - 25

Therefore, you have to:

(a + 5) (a - 5) + 25 = a2

If a = 65, the result is as follows:

652 = 70 × 60 + 25 = 4200 + 25 = 4225.

More examples:

35 × 35 = 40 × 30 + 25 = 1225

105 × 105 = 110 × 100 + 25 = 11025

255 × 255 = 260 × 250 + 25 = 65025

In the latter case, to calculate 260 × 250 you can choose to formulate it this way: 260 × 250 = (250 + 10) × 250 = 2502 + 2500, and we already know how to easily calculate 2502, so it would remain 62500 + 2500 + 25 = 65025.

Cubes and higher powers

Calculating cubes and higher powers using notable equalities is progressively more difficult, and it is often easier to find the fourth power of a number as the square of its square:

95⁴ = (952)2 = (+34) 90252 = (9000 + 25)2 = 90002 + 2 × 25 × 9000 + 252 = 81,000.000 + 450.000 + 625 = 81.450.625 (It makes it very easy to calculate the fact that the second figure of 9025 is zero)

Roots

Square Root Approximation

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

root known square root− − square known− − Square unknown2× × known square root{displaystyle {text{raíz }}simeq {text{ square root}-{frac {{text}{text}}-{text {unknown}}}}{text}{2times {text}{known square phrase}}}}}}}}{,}

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

root 4− − 16− − 152× × 4 4− − 0.125 3.875{displaystyle {begin{aligned}{text{raíz}}}{simeq 4-{frac {16-15}{2times 4}}}{ exposesimeq 4-0.125 fakesimeq 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 keep in mind is that no matter what the original assumption was, the estimated answer will always be greater than the actual answer due to the inequality of the arithmetic and geometric means. Therefore, one should try to round the estimated answer down.

Please note that if n² is the perfect square closest to the starting number x and d = x - n² is its difference, it is more convenient to express this approximation in the form of mixed fraction as nd2n{displaystyle n{tfrac {d}{2n}}}. Thus, in the previous example, the square root of 15 is 4− − 18.{displaystyle 4{tfrac {-1}{8}}}}. !. As another example, the square root of 41 is 6512=6.416{displaystyle 6{tfrac {5}{12}}}=6.416} while the real value is 6.4031...

Demo

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

r2=x{displaystyle mathrm {r} ^{2}=x,!}

Then the root is redefined

r=a− − b{displaystyle mathrm {r} =a-b,!}

where a is a known root (4 from the previous example) and b is the difference between the known root and the answer being searched for.

(a− − b)2=x{displaystyle (a-b)^{2}=x,!}

Expanding the above formula

a2− − 2ab+b2=x{displaystyle a^{2}-2ab+b^{2}=x,!}

If 'a' is near the wanted number, 'b' will be a small enough number for the element +b2{displaystyle} {+b^{2},} of the equation is insignificant. Therefore, it can be deleted +b2{displaystyle} {+b^{2},} and reorganize the equation so that

b a2− − x2a{displaystyle bsimeq {frac {a^{2}-x}{2a},!}

and therefore

root a− − a2− − x2a{displaystyle {text{raíz}}simeq a-{frac {a^{2-}x}{2a}}{,!}

which can be reduced to

root a2+x2a{displaystyle {text{raíz}}simeq {frac {a^{2}+x}{2a}{2a},!}

Extraction of roots of integer powers

Extraction of roots of integer powers is often practiced. The difficulty of the task does not depend so much on the number of digits of the integer power, as on the precision, that is, on the number of digits of the root. Furthermore, it also depends on the order of the root; Finding exact roots, where the order of the root is coprime with respect to 10, is somewhat easier, since the digits are consistently distributed, as shown in the next section.

Cube Root Extraction

An easy task for the beginner is to extract cube roots of cubes of 2-digit numbers. For example, given 74088, determine what two-digit number, when multiplied by itself once and then multiplied again, gives 74088. Anyone familiar with the method will quickly know that the answer is 42, since 42³ = 74088.

Before learning the procedure, the performer is required to memorize the number cubes 1 through 10:

Notice that there is a pattern in the rightmost digit: add and subtract with 1 or 3. Starting at zero:

• 0³ = 0
• 1³ = 1 (has uploaded 1 since 0)
• 2³ = 8 (has dropped 3 since 11)
• 3³ = 27 (has dropped 1 since 8)
• 4³ = 64 (has dropped 3 since 7)
• 5³ = 125 (has climbed 1 since 4)
• 6³ = 216 (has climbed 1 since 5)
• 7³ = 343 (has dropped 3 since 6)
• 8³ = 512 (down from 3)
• 9³ = 729 (has dropped 3 since 12)
• 10³ = 1000 (has climbed 1 since 9)

There are two steps to extracting the cube root of the cube of a two-digit number. For example, to extract the cube root of 29791, first determine the ones place 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, your cubic root should end in 0.
• If the perfect cube ends in 1, your cubic root must end in 1.
• If the perfect cube ends in 2, your cubic root must end in 8.
• If the perfect cube ends in 3, your cubic root should end in 7.
• If the perfect cube ends in 4, your cubic root must end in 4.
• If the perfect cube ends in 5, your cubic root should end in 5.
• If the perfect cube ends in 6, its cubic root must end in 6.
• If the perfect cube ends in 7, its cubic root must end in 3.
• If the perfect cube ends in 8, your cubic root should end in 2.
• If the perfect cube ends in 9, your cubic root must end in 9.

Note that each digit corresponds to itself except 2, 3, 7, and 8, which are simply subtracted from ten to get 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, the last three digits of the given cube (29791 → 29) are removed and the largest cube that is greater is found (this is where you need to know the cubes of numbers 1 through 10). Here, 29 is greater than 1 cubed, greater than 2 cubed, greater than 3 cubed, but not greater than 4 cubed. The first largest cube is greater than 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 cubic root of 456533.
• The cubic root ends in 7.
• After removing the last three digits, there's 456 left.
• 456 is greater than all cubes up to 7 to the cube.
• The first digit of the cubic root is 7.
• The cubic root of 456533 is 77.

This process can be extended to find cube roots of 3-digit length using arithmetic module 11.

This type of trick can be used on any root where the order of the root is coprime with respect to 10; therefore, it does not work with the square root, since the index of the power, 2, is a divisor of 10. In contrast, 3 is not a divisor of 10, so the method can be applied to cube roots.

Calculation of logarithms (base 10)

To approximate the common or base 10 logarithm to one or two significant figures requires knowing some properties of logarithms and memorizing some logarithms. In particular, you need to know the following:

• log(ab) = log(a) + log(b)
• log(a: b) = log(a) - log(b)
• log(0) if
• log(1) = 0
• log(2) ~ 0.33

• log(3) ~ 0,48
• log(7) ~ 0.85
• log(10) = 1
• Yeah. a ▪ bforcibly log(a) /2005 log (b). In mathematical language, it is said that the logarithm function is growing.

From this information, you can calculate the logarithm of any number from 1 to 9:

• log(1) = 0
• log(2) ~ 0.30
• log(3) ~ 0,48
• log(4) = log(2 × 2) = log(2) + log(2) ~ 0.60
• log(5) = log(10: 2) = log(10) - log(2) ~ 0.70
• log(6) = log(2 × 3) = log(2) + log(3) ~ 0,78
• log(7) ~ 0.85
• log(8) = log(2 × 2) = log(2) + log(2) + log(2) ~ 0.90
• log(9) = log(3 × 3) = log(3) + log(3) ~ 0,96 (actually closer to 0,95)
• log(10) = 1

The first step in approximating the common logarithm of a number is to express that number in scientific notation. For example, the number 45 in scientific notation is 4.5 × 10¹. In general, we will have a number of the form a × 10^b, where a is a number between 1 and 10. The second step is to use what is called linear interpolation to estimate the logarithm that we want to calculate from two already known logarithms. In the example of 45 (= 4.5 × 10), we assume that log(4) ~ 0.60 and log(5) ~ 0.70, and since 4.5 is halfway between 4 and 5, log(4.5) will be approximately halfway between log(4) and log(5), so it will be approximately 0.65. In reality, the correct result is always slightly larger than expected, in fact, log(4,5) = 0.6532125... The third and final step, once log(a) is obtained, is to add b to it to obtain the desired logarithm. In this case, since log(4.5) ~ 0.65, just add 1 to get log(45) ~ 1.65. The actual value is log(45) ~ 1.6532125...

The same process can be used to calculate the logarithm of a number between 0 and 1. For example, 0.045 in scientific notation is expressed as 4.5 × 10^-2. You have to be careful with this exponent, which is negative. This will give the result log(0.045) ~ 0.65 - 2 = -1.35.

Another method is to calculate the logarithm of the number from a factorization of numbers whose logarithms are known. In the example above, 45 = 9 × 5, so log(45) = log(9) + log(5) ~ 0.96 + 0.70 = 1.66.

Check the result

There are several ways to check if the result is correct:

• Order of magnitude: If, after multiplying two numbers less than 100, the result is greater than 10,000, surely there is some problem. In a multiplication of two factors, it must be verified that the result has an equal number of figures, or a larger unit (according to the case) than the sum of the factors figures. Often errors in the order of magnitude are due to a bad position of one of the numbers when adding partial products. For example, multiply 65 × 205 instead of 65 × 25, or vice versa.
• Number of units: It consists in checking that the last number of the result is correct view the last number of each of the numbers with which it is split. For example, 73 × 64 should end in 2, as 3 × 4 = 12. This verification allows us to know a certain number.
• Proof of the nine: This verification is based on the sum of the numbers of each factor and the result until there are only numbers of one figure. For example, if we have 73 × 64 = 4662, we can check if it is true by adding the figures for each of the numbers:

7 + 3 = 10, 1 + 0 = 1

6 + 4 = 10, 1 + 0 = 1

4 + 6 + 6 + 2 = 18, 1 + 8 = 9

However, 1 × 1 is not equal to 9, so the result is not correct. The multiplication should be revised again or performed again. (The correct result is 4672) This method is good for detecting stroke errors.

Mental arithmetic as a psychological skill

Physical activity at an adequate level, carried out in advance, can lead to an increase in the performance of an intellectual task, such as doing mental calculations. However, it has been shown that during high levels of physical activity there is an effect negative on the performance of mental tasks. This means that too much physical work can decrease the accuracy and performance of mental mathematical calculations. Physiological measurements, specifically electroencephalography (EEG), have also been shown to be useful in assessing mental workload. Using an EEG as a measure of mental workload after different levels of physical activity can help to determine the level of physical exertion that will be most beneficial for mental performance. Previous work done at Michigan Technological University by Ranjana Mehta includes a recent study that involved subjects engaged in performing mental and physical tasks. The study investigated the effects of mental activity on physical performance at different levels of physical exertion, and finally found a decrease in physical performance when mental tasks were completed at the same time, with more statistical significance at the highest level of physical workload. The Brown–Peterson task, a widely known procedure employing mental arithmetic used primarily in cognitive experiments, suggests that mental subtraction is useful for testing the effects that mental exercise may have on the persistence of short-term memory.

Mental arithmetic competitions

World Mental Calculation Championships

The first World Championship of Mental Calculations took place in 1997. This event is repeated every year, and consists of a series of different tasks, such as the addition of ten ten-digit numbers, the multiplication of two eight-digit numbers, the calculation of square roots, the calculation of the days of the week for given dates, the calculation of cube roots and some surprise miscellaneous tasks.

Mental Calculation World Cup

The first World Mental Calculation Championships (World Cup of Mental Calculation) took place in 2004, and is held every two years. It consists of six different tasks: addition of ten ten-digit numbers, multiplication of two eight-digit numbers, calculating square roots and calculating days of the week for given dates, calculating cube roots, and some surprise miscellaneous tasks.

Memoriad - World Memory, Mental Calculation and Speed Reading Olympics

Memoriad is the first test that combines contests of "mental calculation", "memory" and "photographic reading". Games and competitions are held in the year of the Olympic Games, every four years. The first Memoriad was held in Istanbul, Turkey, in 2008. The second edition took place in Antalya, Turkey, from November 24-25, 2012. 89 competitors from 20 countries participated. Trophies and cash prizes were awarded for 10 categories in total; of which 5 categories had to do with mental calculation (mental addition, mental multiplication, mental square roots (not integers), mental calendar date calculation and Flash Anzan).