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Mental calculation

Mental calculation comprises 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 competition context. Mental calculation often involves the use of specific techniques devised for specific types of problems.

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

Methods and Techniques

Casting out nines

Main article:Casting out nines

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.
* 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 operands; just repeat steps 1 and 2 for each operand.


When checking the mental calculation, it is useful to think of it in terms of scaling. For example, when dealing with large numbers, say 1531 × 19625, estimation instructs you to be aware of the number of digits expected for the final value. A useful way of checking is to estimate. 1531 is around 1500, and 19625 is around 20000, so therefore 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.


When multiplying, a useful thing to remember is that the factors of the operands still remain. For example, to say that 14 × 15 was 211 would be unreasonable. Since 15 was a multiple of 5, so should the product. The correct answer is 210.

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.


4075 − 1844 ------

Thousands: 4 − 1 = 3, look to right, 075 < 844, need to borrow. 3 − 1 = 2, say "Two thousand"

Hundreds: 0 − 8 = negative numbers not allowed here, 10 − 8 = 2, 75 > 44 so no need to borrow, say "two hundred"

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

Ones: 5 − 4 = 1, say "one"

Calculating products: ''a'' × ''b''

Many of these methods work because of the distributive property.

Multiplying by 2 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 =&nbsp;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

Production system

A production system (or production rule system) is a computer program typically used to provide some form of artificial intelligence, which consists primarily of a set of rules about behavior. These rules, termed productions, are a basic representation found useful in automated planning, expert systems and action selection. A production system provides the mechanism necessary to execute productions in order to achieve some goal for the system.

Productions consist of two parts: a sensory precondition (or "IF" statement) and an action (or "THEN"). If a production's precondition matches the current state of the world, then the production is said to be triggered. If a production's action is executed, it is said to have fired. A production system also contains a database, sometimes called working memory, which maintains data about current state or knowledge, and a rule interpreter. The rule interpreter must provide a mechanism for prioritizing productions when more than one is triggered.

Basic operation

Rule interpreters generally execute a forward chaining algorithm for selecting productions to execute to meet current goals, which can include updating the system's data or beliefs. The condition portion of each rule (left-hand side or LHS) is tested against the current state of the working memory.

In idealized or data-oriented production systems, there is an assumption that any triggered conditions should be executed: the consequent actions (right-hand side or RHS) will update the agent's knowledge, removing or adding data to the working memory. The system stops processing either when the user interrupts the forward chaining loop; when a given number of cycles has been performed; when a "halt" RHS is executed, or when no rules have true LHSs.

Real-time and expert systems, in contrast, often have to choose between mutually exclusive productions --- since actions take time, only one action can be taken, or (in the case of an expert system) recommended. In such systems, the rule interpreter, or inference engine, cycles through two steps: matching production rules against the database, followed by selecting which of the matched rules to apply and executing the selected actions.

Matching production rules against working memory

Production systems may vary on the expressive power of conditions in production rules. Accordingly, the pattern matching algorithm which collects production rules with matched conditions may range from the naive -- trying all rules in sequence, stopping at the first match -- to the optimized, in which rules are "compiled" into a network of inter-related conditions.

The latter is illustrated by the RETE algorithm, designed by Charles L. Forgy in 1983, which is used in a series of production systems, called OPS and originally developed at Carnegie Mellon University culminating in OPS5 in the early eighties. OPS5 may be viewed as a full-fledged programming language for production system programming.

Choosing which rules to evaluate

Production systems may also differ in the final selection of production rules to execute, or fire . The collection of rules resulting from the previous matching algorithm is called the conflict set , and the selection process is also called a conflict resolution strategy.

Here again, such strategies may vary from the simple -- use the order in which production rules were written; assign weights or priorities to production rules and sort the conflict set accordingly -- to the complex -- sort the conflict set according to the times at which production rules were previously fired; or according to the extent of the modifications induced by their RHSs. Whichever conflict resolution strategy is implemented, the method is indeed crucial to the efficiency and correctness of the production system.

Using production systems

The use of production systems varies from simple string rewriting rules to the modeling of human cognitive processes, from term rewriting and reduction systems to expert systems.

A simple string rewriting production system example

This example shows a set of production rules for reversing a string from an alphabet that does not contain the symbols "$" and "*" (which are used as marker symbols).

P1: $$ -> * P2: *$ -> * P3: *x -> x* P4: * -> null & halt P5: $xy -> y$x P6: null -> $

In this example, production rules are chosen for testing according to their order in this production list. For each rule, the input string is examined from left to right with a moving window to find a match with the LHS of the production rule. When a match is found, the matched substring in the input string is replaced with the RHS of the production rule. In this production system, x and y are variables matching any character of the input string alphabet. Matching resumes with P1 once the replacement has been made.

The string "ABC", for instance, undergoes the following sequence of transformations under these production rules:

$ABC (P6) B$AC (P5) BC$A (P5) $BC$A (P6) C$B$A (P5) $C$B$A (P6) $$C$B$A (P6) *C$B$A (P1) C*$B$A (P3) C*B$A (P2) CB*$A (P3) CB*A (P2) CBA* (P3) CBA (P4)

In such a simple system, the ordering of the production rules is crucial. Often, the lack of control structure makes production systems difficult to design. It is, of course, possible to add control structure to the production systems model, namely in the inference engine, or in the working memory.

An OPS5 production rule example

In a toy simulation world where a monkey in a room can grab different objects and climb on others, an example production rule to grab an object suspended from the ceiling would look like:

(p Holds::Object-Ceiling {(goal ^status active ^type holds ^objid &lt;O1&gt;) &lt;goal&gt;} {(physical-object ^id &lt;O1&gt; ^weight light ^at &lt;p&gt; ^on ceiling) &lt;object-1&gt;} {(physical-object ^id ladder ^at &lt;p&gt; ^on floor) &lt;object-2&gt;} {(monkey ^on ladder ^holds NIL) &lt;monkey&gt;} -(physical-object ^on &lt;O1&gt;) --> (write (crlf) Grab &lt;O1&gt; (crlf)) (modify &lt;object1&gt; ^on NIL) (modify &lt;monkey&gt; ^holds &lt;O1&gt;) (modify &lt;goal&gt; ^status satisfied) )

In this example, data in working memory is structured and variables

Agricultural productivity

Agricultural productivity is measured as the ratio of agricultural outputs to agricultural inputs. While individual products are usually measured by weight, their varying densities make measuring overall agricultural output difficult. Therefore, output is usually measured as the market value of final output, which excludes intermediate products such as corn feed used in the meat industry. This output value may be compared to many different types of inputs such as labour and land (yield). These are called partial measures of productivity. Agricultural productivity may also be measured by what is termed total factor productivity (TFP). This method of calculating agricultural productivity compares an index of agricultural inputs to an index of outputs. This measure of agricultural productivity was established to remedy the shortcomings of the partial measures of productivity; notably that it is often hard to identify the factors cause them to change. Changes in TFP are usually attributed to technological improvements.

Importance of agricultural prosuctivity

The productivity of a region's farms is important for many reasons. Aside from providing more food, increasing the productivity of farms affects the region's prospects for growth and competitiveness on the agricultural market, income distribution and savings, and labour migration. An increase in a region's agricultural productivity implies a more efficient distribution of scarce resources. As farmers adopt new techniques and differences in productivity arise, the more productive farmers benefit from an increase in their welfare while farmers who are not productive enough will exit the market to seek success elsewhere.

As a region's farms become more productive, its comparative advantage in agricultural products increases, which means that it can produce these products at a lower opportunity cost than can other regions. Therefore, the region becomes more competitive on the world market, which means that it can attract more consumers since they are able to buy more of the products offered for the same amount of money.

Increases in agricultural productivity lead also to agricultural growth and can help to alleviate poverty in poor and developing countries, where agriculture often employs the greatest portion of the population. As farms become more productive, the wages earned by those who work in agriculture increase. At the same time, food prices decrease and food supplies become more stable. Labourers therefore have more money to spend on food as well as other products. This also leads to agricultural growth. People see that there is a greater opportunity earn their living by farming and are attracted to agriculture either as owners of farms themselves or as labourers.

However, it is not only the people employed in agriculture who benefit from increases in agricultural productivity. Those employed in other sectors also enjoy lower food prices and a more stable food sup. At the same time, they may see their wages rise as well.

Agricultural productivity is becoming increasingly important as the world population continues to grow. India, one of the world's most populous countries, has taken steps in the past decades to increase its land productivity. Forty years ago, North India produced only wheat, but with the advent of the earlier maturing high-yielding wheats and rices, the wheat could be harvested in time to plant rice. This wheat/rice combination is now widely used throughout the Punjab, Haryana, and parts of Uttar Pradesh. The wheat yield of three tons and rice yield of two tons combine for five tons of grain per hectare, helping to feed India's 1.1 billion people.

Agricultural productivity and sustainable development

Increase in agricultural productivity are often linked with questions about sustainability and sustainable development. Changes in agricultural practices necessarily bring changes in demands on resources. This means that as regions implement measures to increase the productivity of their farm land, they must also find ways to ensure that future generations will also have the resources they will need to live and thrive.

Productive farms

Nevertheless, for many farmers (especially in non-industrial countries) agricultural productivity may mean much more. A productive farm is one that provides most of the resources necessary for the farmer's family to live, such as food, fuel, fiber, healing plants, etc. It is a farm which ensures food security as well as a way to sustain the well-being of a community. This implies that a productive farm is also one which is able to ensure proper management of natural resources, such as biodiversity, soil, water, etc. For most farmers, a productive farm would also produce more goods than required for the community in order to allow trade.

Diversity in agricultural production is one key to productivity, as it enables risk management and preserves potentials for adaptation and change. Monoculture is an example of such a nondiverse production system. In a monocultural system a farmer may produce only crops, but no livestock, or only livestock and no crop.

The benefits of raising livestock, among others, are that it provides multiple goods, such as food, wool, hides, and transportation. It also has an important value in term of social relationships (such as gifts in

Product rule

In calculus, the product rule is a formula used to find the derivatives of products of two or more functions. It may be stated thus:

(f\cdot g)'=f'\cdot g+f\cdot g' \,\!

or in the Leibniz notation thus:

\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}.

The derivative of the product of three functions is:

\dfrac{d}{dx}(u\cdot v \cdot w)=\dfrac{du}{dx} \cdot v \cdot w + u \cdot \dfrac{dv}{dx} \cdot w + u\cdot v\cdot \dfrac{dw}{dx}.

Discovery by Leibniz

Discovery of this rule is credited to Gottfried Leibniz (however, Child (2008) argues that it is due to Isaac Barrow), who demonstrated it using differentials. Here is Leibniz's argument: Let u(x) and v(x) be two differentiable functions of x. Then the differential of uv is

\begin{align} d(u\cdot v) & {} = (u + du)\cdot (v + dv) - u\cdot v \\ & {} = u\cdot dv + v\cdot du + du\cdot dv. \end{align}

Since the term du·dv is "negligible" (compared to du and dv), Leibniz concluded that

d(u\cdot v) = v\cdot du + u\cdot dv \,\!

and this is indeed the differential form of the product rule. If we divide through by the differential dx, we obtain

\frac{d}{dx} (u\cdot v) = v \cdot \frac{du}{dx} + u \cdot \frac{dv}{dx} \,\!

which can also be written in "prime notation" as

(u\cdot v)' = v\cdot u' + u\cdot v'. \,\!


  • Suppose one wants to differentiate &fnof;(x) = x2sin(x). By using the product rule, one gets the derivative &fnof;&nbsp;'(x) = 2x sin(x) + x2cos(x) (since the derivative of x2 is 2x and the derivative of sin(x) is cos(x)).
  • One special case of the product rule is the constant multiple rule which states: if c is a real number and &fnof;(x) is a differentiable function, then c&fnof;(x) is also differentiable, and its derivative is (c&nbsp;&times;&nbsp;&fnof;)'(x) = c&times; &fnof;&nbsp;'(x). This follows from the product rule since the derivative of any constant is zero. This, combined with the sum rule for derivatives, shows that differentiation is linear.
  • The rule for integration by parts is derived from the product rule, as is (a weak version of) the quotient rule. (It is a "weak" version in that it does not prove that the quotient is differentiable, but only says what its derivative is if it is differentiable.)

Physics example: rocket acceleration

Consider the vertical acceleration of a model rocket relative to its initial position at a fixed point on the ground. Newton's second law says that the force is equal to the time rate change of momentum. If F is the net force (sum of forces), p is the momentum, and t is the time,

F = \frac{dp}{dt}.

Since the momentum is equal to the product of mass and velocity, this yields

F = \frac{d}{dt}\left( mv \right), where m is the mass and v is the velocity. Application of the product rule gives

F = v\frac{dm}{dt} + m\frac{dv}{dt}.

Since the acceleration a, is defined as the time rate change of velocity, a = dv/dt,

F = v\frac{dm}{dt} + ma.

Solving for the acceleration,

a= \frac{F - v\frac{dm}{dt}}{m}.

Since the rocket is losing mass, dm/dt is negative, and the changing mass term results in increased acceleration.

A common error

It is a common error, when studying calculus, to suppose that the derivative of (uv) equals (u&nbsp;&prime;)(v&nbsp;&prime;) (there is an exaggerated story that Leibniz himself made this error initially); however, there are clear counterexamples to this. For a &fnof;(x) whose derivative is &fnof; '(x), the function can also be written as &fnof;(x)&nbsp;·&nbsp;1, since 1 is the identity element for multiplication. If the above-mentioned misconception were true, (u&prime;)(v&prime;) would equal zero. This is true because the derivative of a constant (such as&nbsp;1) is zero and the product of &fnof;&nbsp;'(x)&nbsp;·&nbsp;0 is also zero.

Proof of the product rule

A rigorous proof of the product rule can be given using the properties of limits and the definition of the derivative as a limit of Newton's difference quotient.


h(x) = f(x)g(x),\,

and &fnof; and g are each differentiable at the fixed number x, then

h'(x) = \lim_{w\to x}{ h(w) - h(x) \over w - x} = \lim_{w\to x}{f(w)g(w) - f(x)g(x) \over w - x}. \qquad\qquad(1)

Now the difference

f(w)g(w) - f(x)g(x)\qquad\qquad(2)

is the area of the big rectangle minus the area of the small rectangle in the illustration.

The region between the smaller and larger rectangle can be split into two rectangles, the sum of whose areas is

f(x) \Bigg( g(w) - g(x) \Bigg) + g(w)\Bigg( f(w) - f(x) \Bigg).\qquad\qquad(3)

Therefore the expression in (1) is equal to

\lim_{w\to x}\left( f(x) \left( {g(w) - g(x) \over w - x} \right) + g(w)\left( {f(w) - f(x) \over w - x} \right) \right).\qquad\qquad(4)

Assuming that all limits used exist, (4) is equal to

\left(\lim_{w\to x}f(x)\right) \left(\lim_{w\to x} {g(w) - g(x) \over w - x}\right)

+ \left(\lim_{w\to x} g(w)\right) \left(\lim_{w\to x} {f(w) - f(x) \over w - x} \right). \qquad\qquad(5)


\lim_{w\to x}f(x) = f(x)\, &nbsp;&nbsp; and &nbsp;&nbsp; \lim_{w\to x} g(w) = g(x)\,

because f(x) and g(x) remain constant as w→ x;


\lim_{w\to x} {f(w) - f(x) \over w - x} = f'(x) &nbsp;&nbsp; and &nbsp;&nbsp; \lim_{w\to x} {g(w) - g(x) \over w - x} = g'(x)

because f and g are differentiable at x;

We conclude that the expression in (5) is equal to

From Yahoo Answers

Question:I need help with algebra 1, how do you find each product or quotient? Ex. 6 1.12

Answers:product means multiply: 6 1.12 = 6.72 quotient means divide

Question:I have a project due monday and a can't find the states amin productions. If any one knows them or knows a website i would be very grateful!

Answers:You are probably going to search each individual state.

Question:Hi I really need help with this geometry 1)the figure is a big rectangle length 17 the width The second figure small rectangle with a width of 3? 2)A big right triangle length 15 the second figure small righ triangle width 3 length 4

Answers:"Similar" means that all corresponding LENGTHS between the two figures are in the same ratio. If the width of the big rectangle corresponds to the width of the small rectangle, then the common ratio is 17/3 (=5.666...). The height of the big rectangle will be 17/3 times the height of the small rectangle, the diagonal of the big rectangle will be 17/3 times the diagonal of the small rectangle, the perimeter of the big rectangle will be 17/3 times the perimeter of the small rectangle, ... Area is calculated from the product of two lengths. Rectangle area is width times height. Triangle area is half width times height. Circle area is pi times radius times radius. For a similar area, each length is multiplied by the common ratio, so the area is multiplied by the square of the common ratio. Your large rectangle area is height times width. That is 17/3 times the small rectangle height times 17/3 times the small rectangle width, or (17/3)^2 times the small rectangle width times the small rectangle height, or (17/3)^2 = 289/9 (=32.111...) times the small rectangle area. For your triangles, decide which side of the small right triangle CORRESPONDS to the length of the big right triangle. Then you can find the common ratio. The area ratio is just the square of that common ratio.

Question:A 1.00-L solution saturated at 25 C with calcium oxalate CaC2O4 contains 0.0061 g of CaC2O4. Calculate the solubility-product constant for this salt at 25 C.

Answers:CaC2O4 (s) Ca++ (aq) + C2O4-- (aq) Ksp = [Ca++] [C2O4--] The molecular mass of CaC2O4 is 128.1 g/mole so the concentration of the saturated solution is: (0.0061 g) / (1.00 L) (128.1 g/mol) = 4.76x10^-5 M. Since one Ca++ ion and one C2O4-- ion dissolve for each CaC2O4 that dissolves, Ksp = [4.76x10^-5] [4.76x10^-5] = 2.27x10^-9.

From Youtube

Finding Products of Polynomials :www.gdawgenterprises.com In this video, a conceptual approach to understanding and performing finding the products of polynomials is shown. Algebra tiles are used and a progression made to using the box method to find products of polynomials. Different examples demonstrate the versatility of the box method. Also, a graphing calculator is shown briefly at the end for the purpose of checking work for accuracy.

Calculs: Find the Root, Then Find the Formula :www.mindbites.com Professor Burger explains how to find the powers and roots of complex numbers. The equation of a complex number is z= r(cosx + isinx). To raise the complex number to a power, n, the equation is z^n = r^n[cos(nx) + isin(nx)]. In general, if you are raising a complex number to the power of n or 1/n (taking the nth root), you will come up with n solutions, as you will always have one solution for each of the degrees of power. When taking the root of a complex number, you will find one solution for each degree of power. To find the nth root of a complex number the equation is n root of z = (n root r) *[cos ((x + 2 Pi K)/n) + 1 sin ((x + 2 Pi K)/n)] where k = 0, 1, 2,...n-1.Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Precalculus. This course and others are available from Thinkwell, Inc. The full course can be found at www.thinkwell.com The full course covers angles in degrees and radians, trigonometric functions, trigonometric expressions, trigonometric equations, vectors, complex numbers, and more. Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College. He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won ...