DEFINING NORMALCY
Defining Normalcy may look normal,till one attempts to define,What is normal?,Who is normal?, Are there Limitations to what normality means?


In the early twentieth century, President Warren Harding popularized the word normalcy with his slogan: “return to normalcy," about getting back to normal life after the war. Since then, the word has been used often instead of "normality."

Normalcy, pronounced "NOR-mal-see," is a state of being normal, usual, or expected. When something returns to a state of normalcy, it is returning to the way that it ought to be. It is another word for normality. Some people believe that normalcy should not be used because the word normality is more formally correct, but in the U.S., you'll often see normalcy used. When it comes to choosing one, take your pick.

Even in the process of trying to arrange a page discussing normalcy, There are various options but only one Font says "Normal", Funny?

Every person is distinctive, a particular individual with his own ideas and his own ways of doing things. The mentally ill seem special only in that they are more distinctive. They are idiosyncratic or eccentric, even peculiar; yet in their strangeness there is nothing unrecognizable. They experience no impulse nor longing that is foreign to a normal person, and they suffer no illusion that a normal person has not known. The symptoms of mental illness are embedded in, and grow out of, the normal personality. Since life is varied and complex anyway, it is hard to determine where normal behavior leaves off and abnormal behavior begins. In retreat from this tantalizing ambiguity, some psychiatrists have chosen to take the position that there is no such thing as mental illness. In similar argument, one might contend that since orange blends closely into red, there is no such thing as orange.
A medical student assigned to a psychiatric ward evaluated his first patient, an eighteen-year-old girl who had been admitted to the hospital because her mother had complained that there was something the matter with her. Following an argument with a teacher, the girl had become withdrawn and preoccupied with religion. She began eating poorly and leaving her room in a mess, which was out of character, for she was usually very neat. Then on the day before her admission she was found sitting on a subway platform, dangling her feet over the edge.
The medical student, after speaking to her on a few occasions, told the director of the ward that in his opinion she was not really ill. All he noticed about her was a kind of flightiness of thought and a somewhat depressed mood, which he thought was not out of keeping with the unpleasant circumstances of being on a psychiatric ward, especially for a sensitive young girl. Although his patient had little to say about the episode in the subway station, he explained it as an adolescent prank. Adolescence is a time when one is given to silly and impulsive behavior, he said, and to preoccupations with religion, for that matter. He went on to say that there was a time during his own adolescence when he himself was concerned with religious questions. And as for messiness, his room was even now, a mess. He said in conclusion that if she was emotionally sick, he thought that perhaps he himself, and certainly some of his classmates, were sick also.
This medical student felt a rapport with his patient which allowed him to see the world through her eyes. Nevertheless, he was wrong about her. She was psychotic. He could not yet tell where normal behavior came to an end and something else began. He had not seen orange and red set up against each other often enough to tell one from the other. As for his classmates—it is likely that he was right and that among them there were a few who were indeed mentally ill, for emotional disturbances are common, and no one is immune.
If the manifestations of mental illness can only be seen in relief against normal behavior, what then, after all, is normal behavior? What sort of person is a normal person? He is someone, first of all, who feels happy a considerable part of the time. Not all the time, of course. He is angry when he is frustrated, disappointed when he fails. He grieves when he has lost someone. Sometimes he is frightened. But he is not characteristically in any single mood, for there is an aptness to his feelings, a fit between them and the circumstances of his life. Since the circumstances of every person’s life are varied, so are his feelings. Still, in general he thinks of himself as being happy. He can relate to other people, being assertive when appropriate and conciliatory at other times. He may not feel comfortable with everyone, but he does feel secure within his family and among his friends.
And there are groups --religious, perhaps, or social-with whose members he feels an identity And he can love, which means he can be trusting and affectionate. And he can make love without feeling oppressed or uncomfortable. Such a person often finds himself committed to other people, or even to an idea, to a degree where he loses himself in a larger purpose. Consequently when he has children, he becomes part of them and continues on in their lives, so that personal death becomes less real and less frightening.
 A normal person can work with satisfaction, if it is decent work; and he can relax when he is not working. He enjoys play. In fact he enjoys himself much of the time, even if he is only talking to a friend or watching television. But probably the defining quality of a normal person is a kind of flexibility that allows him to adapt to the different demands and stresses that life imposes upon him. He can tolerate conflict and frustration and loneliness-to some extent. And he is able to seize upon life, too. He is successful at work and with people and within his family-to some extent. Probably it is possible to go on endlessly sketching the form of an ideal human being, but as he is drawn more and more precisely, he less and less resembles anyone in the real world. And yet there are normal people.
 Perhaps it is easier to say what is not meant by normal than what is. Normal is not average. If someone could be average in every conceivable way, he could still be disturbed, although a perfectly average man is no more real than someone who is perfectly normal. Conversely, being atypical does not mean someone is sick. A man may still be normal who every day sleeps only five hours, works an extra full-time job, and takes three showers. But knowing just how someone deviates from the average is important in understanding him, for three reasons:
1. Although a particular behavior may not be in itself abnormal, it may be part of a pattern that reflects an abnormal process. sleeping less than average, for instance, is sometimes associated with severe depressions and other psychoses. Also, if someone is extremely far from average in some respect of behavior or attitude, it is likely he will turn out to be emotionally ill by some other criteria.
2. Someone who is significantly different from other people may be under special strain as a result, for in order to be with people, it is necessary to do pretty much the things other people do. A man who works two full-time jobs, for instance, is not likely to be at home when his family is. If he is away from the important people in his life most of the time, to some extent, inevitably, he will be emotionally deprived.
3. But most important, an individual is most himself at just those points where he is different from others. Consequently these are points of departure for a search into an understanding of him. Human behavior is always hard to understand, for it is an expression for the most part of unconscious wishes and conflicts; nevertheless, personality becomes tangible in these out-of-the-ordinary habits and attitudes. One may ask why, for instance, a man would take three showers every day. It may be because he thinks of himself or his body as being unclean; or he might find a sensual pleasure in rubbing himself; or there might be countless other reasons, any of which may illuminate a facet of his personality. Knowing what is special about someone is knowing, at least, what is worth paying attention to for a therapist and what to ask about.
Normal also does not mean adjusted, if that word is used to refer to someone who accepts the dominant ethic of a particular society. For one thing, that ethic is frequently not what it pretends to be. The unwritten rules that govern sexual behavior in this country, for example, are scarcely those that issue from the pulpit. Even such a fundament of civilization as the law, which is spoken of invariably with esteem, is treated actually with bland disregard. People are not really expected to strictly obey the law, any more than they are expected to work a full day or to keep every promise they make.
But even were society more consistent, there would still be room for an unconventional person. At every extreme of social custom, there are normal people, and creative valuable people. Normal does not refer to a state of being at all, but rather a process, the process of growing up and into life. Adolescence has been described as a continual adjustment to the business of growing up, but that adjustment goes on at every age. Somewhere along the way, if a person is healthy, he learns to be himself and fulfill himself, but the process is never complete. He may know himself, but never altogether, for he is always changing. If he is unable to change enough to compensate for new circumstances in this life, then he will develop an emotional disorder. For that reason emotional illness may always be regarded as a failure of adaptation.

So if mental health is a process of growth, mental illness is an inhibition of that process. If being normal is to work and love and feel happy—relatively—being disturbed emotionally is to suffer relative impairment in these capacities. There is not yet a better distinction between health and illness, or between an ordinary person and someone who has become ill. (This passage is excerpted from “Caring: Home Treatment for the Emotionally Disturbed.”) © Fredric Neuman.

Calculators or Brains?

Mental calculation comprises of Arithmetical calculations using only the Human brain with no help from calculators, computers, or pen and paper. 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 competitive 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 numeral system. Usually, the choice of radix 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
 mentawellbeing.blogspot.com
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

While 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 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, 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 are 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

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 a number by nine, multiply it by 10 and then subtract the original number from the result. For example, 9 × 27 = 270 − 27 = 243.

This method can be adjusted to multiply by eight instead of nine, by doubling the number being subtracted; 8 × 27 = 270 − (2×27) = 270 − 54 = 216.

Similarly, by adding instead of subtracting, the same methods can be used to multiply by 11 and 12, respectively (although simpler methods to multiply by 11 exist).

Using hands: 1–10 multiplied by 9

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 can be found by a series of additions to each of its digits from right to left, two at a time.

First take the ones digit and copy that to the temporary result. Next, starting with the ones digit of the multiplier, add each digit to the digit to its left. Each sum is then added to the left of the result, in front of all others. If a number sums to 10 or higher take the tens digit, which will always be 1, and carry it over to the next addition. Finally copy the multipliers left-most (highest valued) digit to the front of the result, adding in the carried 1 if necessary, to get the final product.

In the case of a negative 11, multiplier, or both apply the sign to the final product as per normal multiplication of the two numbers.

A step-by-step example of 759 × 11:

The ones digit of the multiplier, 9, is copied to the temporary result.
result: 9

Add 5 + 9 = 14 so 4 is placed on the left side of the result and carry the 1.
result: 49

Similarly add 7 + 5 = 12, then add the carried 1 to get 13. Place 3 to the result and carry the 1.
result: 349

Add the carried 1 to the highest valued digit in the multiplier, 7 + 1 = 8, and copy to the result to finish.
Final product of 759 × 11: 8349


Further examples:

−54 × −11 = 5 5+4(9) 4 = 594
999 × 11 = 9+1(10) 9+9+1(9) 9+9(8) 9 = 10989
Note the handling of 9+1 as the highest valued digit.

−3478 × 11 = 3 3+4+1(8) 4+7+1(2) 7+8(5) 8 = −38258
62473 × 11 = 6 6+2(8) 2+4+1(7) 4+7+1(2) 7+3(0) 3 = 687203

Another method is to simply multiply the number by 10, and add the original number to the result.

For example:

17 × 11

     17 × 10 = 170

170 + 17 = 187

17 × 11 = 187

One last easy way:

If you have a two-digit number, take it and add the two numbers together and put that sum in the middle, and you can get the answer.

For example: 24 x 11 = 264 because 2 + 4 = 6 and the 6 is placed in between the 2 and the 4.

Second example: 87 x 11 = 957 because 8 + 7 = 15 so the 5 goes in between the 8 and the 7 and the 1 is carried to the 8. So it is basically 857 + 100 = 957.

Or if 43 x 11 is equal to first 4+3=7 (For the tens digit) Then 4 is for the hundreds and 3 is for the tens. And the answer is 473

Multiplying two 2 digit numbers between 11 and 19

To easily multiply 2 digit numbers together between 11 and 19 a simple algorithm is as follows (where a is the ones digit of the first number and b is the ones digit of the second number):

(10+a)×(10+b)
100 + 10×(a+b) + a×b

which can be visualized as three parts to be added:

1
xx
 yy

for example:

17×16

1               = 100
13    (7+6)     = 10×(a+b)
 42   (7×6)     = a×b
272   (total)

Multiplying any 2-digit numbers

To easily multiply any 2-digit numbers together a simple algorithm is as follows (where a is the tens digit of the first number, b is the ones digit of the first number, c is the tens digit of the second number and d is the ones digit of the second number):




For example

  800
 +120
 +140
 + 21
-----
 1081

Note that this is the same thing as the conventional sum of partial products, just restated with brevity. To minimize the number of elements being retained in one's memory, it may be convenient to perform the sum of the "cross" multiplication product first, and then add the other two elements:


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


i.e., in this example

(12 + 14) = 26, 26 × 10 = 260,

to which is it is easy to add 21: 281 and then 800: 1081

An easy mnemonic to remember for this would be FOIL. F meaning first, O meaning outer, I meaning inner and L meaning last. For example:



and



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

Consider



this expression is analogous to any number in base 10 with a hundreds, tens and ones place. FOIL can also be looked at as a number with F being the hundreds, OI being the tens and L being the ones.

is the product of the first digit of each of the two numbers; F.

is the addition of the product of the outer digits and the inner digits; OI.

is the product of the last digit of each of the two numbers; L.

Multiplying any two numbers by attaching, subtracting, and routing

Discovered by Artem Cheprasov, there is a method of multiplication that allows the user to utilize 3 steps to quickly multiply numbers of any size to one another via three unique ways.

First, the method allows the user to attach numbers to one another, as opposed to adding or subtracting them, during intermediate steps in order to quicken the rate of multiplication. For instance, instead of adding or subtracting intermediary results such as 357 and 84, the user could simply attach the numbers together (35784) in order to simplify and expedite the multiplication problem. Attaching numbers to one another helps to bypass unnecessary steps found in traditional multiplication techniques.

Secondly, this method uses negative numbers as necessary, even when multiplying two positive integers, in order to quicken the rate of multiplication via subtraction. This means two positive integers can be multiplied together to get negative intermediate steps, yet still the correct positive answer in the end. These negative numbers are actually automatically derived from the multiplication steps themselves and are thus unique to a particular problem. Again, such negative intermediate steps are designed to help hasten the mental math.

Finally, another unique aspect of using this method is that the user is able to choose one of several different “routes of multiplication” to the specific multiplication problem at hand based on their subjective preferences or strengths and weaknesses with particular integers.

Despite the same starting integers, the different multiplication routes give off different intermediate numbers that are automatically derived for the user as they multiply. Some of these intermediaries may be easier than others (e.g. some users may find a route that uses a negative 7, while another route uses a 5 or a 0, which are typically easier to work with mentally for most people, but not in all instances).

If one “route” seems to be harder for one student vs. another route and its intermediate numbers, that student can simply choose another simpler route of multiplication for themselves even though it’s the same original problem.

The "Ends of Five" Formula

For any 2 digit by 2 digit multiplication problem, if both numbers end in five, the following algorithm can be used to quickly multiply them together:



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





The algorithm reads as follows:



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



The author also outlines another similar algorithm if you want to round the original larger number down and the original smaller number up instead.

The "Borrower's" Formula

If two numbers are equidistant from the nearest multiple of 100, then a simple algorithm can be used to find the product.

As a simple example:



Both numbers are equidistant (33 away) from their nearest multiple of 100 (0 and 100, respectively).

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





The algorithm reads as follows:



Where u₁ is the original larger number's (67) units digit and u₂ is the original smaller number's (33) units digit. T₁ is the original larger number's tens digit and T₂ is the original larger number's tens digit multiplied by their respective power (in this case by 10, for a tens digit).

And so:

Using hands: 6–10 multiplied by another number 6–10

This technique allows a number from 6 to 10 to be multiplied by another number from 6 to 10.

Assign 6 to the little finger, 7 to the ring finger, 8 to the middle finger, 9 to the index finger, and 10 to the thumb. Touch the two desired numbers together. The point of contact and below is considered the "bottom" section and everything above the two fingers that are touching are part of the "top" section. The answer is formed by adding ten times the total number of "bottom" fingers to the product of the number of left- and right-hand "top" fingers.

For example, 9 × 6 would look like this, with the left index finger touching the right little finger:

                               =10==   :right thumb              (top)
                               ==9==   :right index finger       (top)
                               ==8==   :right middle finger      (top)
        left thumb:   =10==    ==7==   :right ring finger        (top)                    
 left index finger:   --9---><---6--   :right little finger      (BOTTOM)  
left middle finger:   --8--                                      (BOTTOM)
  left ring finger:   --7--                                      (BOTTOM)
left little finger:   --6--                                      (BOTTOM)

In this example, there are 5 "bottom" fingers (the left index, middle, ring, and little fingers, plus the right little finger), 1 left "top" finger (the left thumb), and 4 right "top" fingers (the right thumb, index finger, middle finger, and ring finger). So the computation goes as follows: 9 × 6 = (10 × 5) + (1 × 4) = 54.

Consider another example, 8 × 7:

                               =10==   :right thumb              (top)
        left thumb:   =10==    ==9==   :right index finger       (top)
 left index finger:   ==9==    ==8==   :right middle finger      (top)
left middle finger:   --8---><---7--   :right ring finger        (BOTTOM)
  left ring finger:   --7--    --6--   :right little finger      (BOTTOM)
left little finger:   --6--                                      (BOTTOM)

Five bottom fingers make 5 tens, or 50. Two top left fingers and three top right fingers make the product 6. Summing these produces the answer, 56.

Another example, this time using 6 × 8:

 --8---><---6--
 --7--
 --6--

Four tens (bottom), plus two times four (top) gives 40 + 2 × 4 = 48.

Here's how it works: each finger represents a number between 6 and 10. When 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.

Multiplying two numbers close and below 100

This technique allows easy multiplication of numbers close and below 100.(90-99)^[3] The variables will be the two numbers you multiply.

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 your 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 your 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 your equation from 100)
                  97-5=92  (Subtract that answer from the second number of your 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, 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	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 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

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

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 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

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.

Finding roots

Approximating square roots

An easy way to approximate the square root 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

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.

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. 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.

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

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 × 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 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, 0.045 would be written as 4.5 × 10⁻². The only difference is that b is now negative, so when adding you are really subtracting. This would yield the result 0.653 − 2, or −1.347.

Approximating natural exponents (log base e)

Natural exponents are used in many important expressions in modern science and engineering, with applications not limited to quantum mechanics, thermodynamics and signal communications. Using the laws of Natural exponents, memorization of the approximations below, and combination with other mental calculation methods, create a powerful and elegant means for changing complicated problems in the physical sciences into simple sums and products. The laws of Natural exponents (Exponentiation) are:

e^a x e^b = e ^(a+b) and e ^−a = 1/ e^a and also

e^a x e^−b = e ^(a-b) = e^a / e^b

Table of approximations

Where possible, single digit numbers, followed by zeros, are used for ease of memorization, accuracy and to eliminate redundancy: (e^4.5)² = e⁹, e³ x e⁴ = e⁷, and e⁸ are used instead of (e⁴)².

Exponent	Value	Error		Exponent	Value	Error
e3	≈ 20	0.43%+		e23	≈ 1010	0.11%+
e4	≈ 55	0.74%−		e25	≈ 7.2x1010	0.0003%+
e4.25	≈ 70	0.15%−		e50.6	≈ 1022	0.02%+
e4.5	≈ 90	0.02%−		e61	≈ π x 1026	0.02%−
e5.75	≈ 100π	0.01%−		e99	≈ 1043	0.01%−
e8	≈ 3,000	0.64%−		e108	≈ 8x1046	0.0015%−

Optimized Relations

This table presents optimized suggestions derived from the above table.

e10	= e3x e7	≈ 22,000	0.01%+		e0.25	= e4.25 /e4	≈ 70/55	3.5%+
e1	= e10 /e9	≈ 220/81	0.08%+		e0.5	= e4.5 /e4	≈ 90/55	1.5%+
e2	= e9 / e7	≈ 81/11	0.4%−		e0.3	= √(e50.6) /e25 = e25.3 /e25	≈ 100/72	0.5%+
e5	= e9 /e4	≈ 8100/55	0.15%−			better approximations:	e4 ≈ 54.6	e6 ≈ 403 or 404

Physical Science and Communication Approximations

Exponent	Value	Error	Application
eπ	≈ 20+π	0.004%−	Spherical wave attenuation
e274	≈ 10119	0.0003%+	1 °C, 274 Kelvin
e280	≈ 4 x 10121	0.00003%+	e280/e7 = e273 near ambient, Kelvin
e452	≈ 2 x 10196	0.000004%+	Accurate thermal temperature
kbe2	≈ 10−22	1.4%+	kb = Boltzmann's constant

The +/- superscript after the error represents if the actual number value is higher or lower than the approximation; for example the approximation of ln(400) is less than 6. The +/- symbols after errors may also be used to make the approximations more accurate by compensation, for example;

e¹⁶ = e⁸ x e⁸, accurate to 12 parts in 1000, may be made more accurate by using

e⁷ x e⁹ ≈ 1,100 x 8,100 or 8,910,000, accuracy 2 parts in 1000.

Other examples

e^{(9 + π)} = 8,100 x (20 + π) = 162,000 + ~25,500 = 187,500, (accurate to 1 part in 5000)

Mental arithmetic as a psychological skill

Physical assertion 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 1997 at the Mind Sports Olympiad. This event repeats every year. It consists of a range of different tasks such as: addition of ten ten-digit numbers, multiplication of two eight-digit numbers, calculation of square roots, and calculation of weekdays for given dates, calculation of cube roots plus some surprise miscellaneous tasks.

Mental Calculation World Cup

Main article: 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, calculation of cube roots plus some surprise miscellaneous tasks.

Memoriad - World Memory, Mental Calculation & Speed Reading Olympics

Memoriad is the first platform combining "mental calculation", "memory" and "photographic reading" competitions. 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 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).     

                                                  Sourced: Wikipedia(2016).