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CONTENTS

Topics Page No.

n Syllabus Guidelines 04

n Maths Here and There 06

n Beat The Calculator 07

n Peoples in Mathematics 11

n Why Calculate π 13

n Brain Benders 16

n Digit Sums 17

n Useful Shape 19

n Knowledge of Math - A Reward 21

n Genius Test 22

n Curious Mathematical Calculation 23

n Addition Made Easier 24

n Interactive Activities : Maths Puzzles 26

n Brain Twisters 27

n Trick to Amaze Your Friend 29

n Mathematics to Bring Incredible Results 30

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Based on CBSE, ICSE & GCSE Syllabus

UNIT I: NUMBER SYSTEMS REAL NUMBERS

Review of representation of natural numbers, integers, rational numbers on the number line. Representation of terminating/ non-terminating recurring decimals, on the number line through successive magnification. Rational numbers as recurring/terminating decimals. Examples of nonrecurring / non terminating decimals

such as 2, 3, 5 etc.

Existence of non-rational numbers (irrational numbers)

such as 2, 3 , and their representation on the

number line. Explaining that every real number is represented by a unique point on the number line, and conversely, every point on the number line represents a unique real number.

Existence of x for a given positive real number x

(visual proof to be emphasized). Definition of n th root of a real number. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws). Rationalization (with precise meaning) of real numbers

of the type (& their combinations) 1 1

& a + b x x + y

where x and y are natural numbers and a, b are integers.

UNIT II: ALGEBRA 1. POLYNOMIALS

Definition of a polynomial in one var iable, i ts coefficients, with examples and counter examples, its terms, zero polynomial. Degree of a polynomial. Constant, linear, quadrat ic, cubic polynomials; monomials, binomials, t rinomials. Factors and multiples. Zeros/roots of a polynomial/equation.

Remainder Theorem with examples and analogy to integers. Statement and proof of the Factor Theorem.

Factorization of ≠ 2 ax + bx + c,a 0 where a, b, c are

real numbers, and of cubic polynomials using the Factor Theorem. Recall of algebraic expressions and identities. Further identities of the type

+ + = + + + + + ±

= ± ± ± + + −

2 2 2 2 3

3 3 3 3 3

( ) 2 2 2 ,( )

2 ( ), 3

x y z x y z xy yz zx x y

x y xy x y x y z xyz

= + + + + − − − 2 2 2 ( )( ) x y z x y z xy yz zx and their use in factorization of polynomials. Simple expressions reducible to these polynomials.

2. LINEAR EQUATIONS IN TWO VARIABLES Recall of linear equations in one variable. Introduction to the equation in two variables. Prove that a linear equation in two variables has infinitely many solutions, and justify their being written as ordered pairs of real numbers, plotting them and showing that they seem to lie on a line. Examples, problems from real life, including problems on Ratio and Proportion and with algebraic and graphical solutions being done simultaneously.

UNIT III: COORDINATE GEOMETRY 1. COORDINATE GEOMETRY

The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations, plotting points in the plane, graph of linear equations as examples; focus on linear equations of the type ax + by + c = 0 by writing it as y = mx + c and linking with the chapter on linear equations in two variables.

UNIT IV: GEOMETRY 1. INTRODUCTION TO EUCLID S GEOMETRY

History Euclid and geometry in India. Euclid s method of formalizing observed phenomenon into rigorous mathematics with definitions, common/obvious notions, axioms/postulates, and theorems. The five postulates of Euclid. Equivalent versions of the fifth postulate. Showing the relationship between axiom and theorem.

1. Given two distinct points, there exists one and only one line through them.

2. Two distinct lines cannot have more than one point in common.

2. LINES AND ANGLES 1. If a ray stands on a line, then the sum of the two

adjacent angles so formed is 180° and the converse. 2. If two lines intersect, the vertically opposite angles are

equal. 3. Results on corresponding angles, alternate angles,

interior angles when a transversal intersects two parallel lines.

4. Lines, which are parallel to a given line, are parallel. 5. The sum of the angles of a triangle is 180°. 6. If a side of a triangle is produced, the exterior angle so

formed is equal to the sum of the two interiors opposite angles.

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3. TRIANGLES 1. Two triangles are congruent if any two sides and the

included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).

2. Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence).

3. Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).

4. Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle.

5. The angles opposite to equal sides of a triangle are equal.

6. The sides opposite to equal angles of a triangle are equal.

7. Triangle inequalities and relation between angle and facing side inequalities in triangles.

4. QUADRILATERALS 1. The diagonal divides a parallelogram into two congruent

triangles. 2. In a parallelogram opposite sides are equal, and

conversely. 3. In a parallelogram opposite angles are equal and

conversely. 4. A quadrilateral is a parallelogram if a pair of its opposite

sides is parallel and equal. 5. In a parallelogram, the diagonals bisect each other and

conversely. 6. In a triangle, the line segment joining the mid points of

any two sides is parallel to the third side and its converse.

5. AREA Review concept of area, recall area of a rectangle.

1. Parallelograms on the same base and between the same parallels have the same area.

2. Triangles on the same base and between the same parallels are equal in area and its converse.

6. CIRCLES Through examples, arrive at definitions of circle related concepts, radius, circumference, diameter , chord, arc, subtended angle.

1. Equal chords of a circle subtend equal angles at the center and its converse.

2. The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn

through the center of a circle to bisect a chord is perpendicular to the chord.

3. There is one and only one circle passing through three given non-collinear points.

4. Equal chords of a circle (or of congruent circles) are equidistant from the center(s) and conversely.

5. The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

6. Angles in the same segment of a circle are equal. 7. If a line segment joining two points subtends equal angle

at two other points lying on the same side of the line containing the segment, the four points lie on a circle.

8. The sum of the either pair of the opposite angles of a cyclic quadrilateral is 180 0 and its converse.

7. CONSTRUCTIONS 1. Construction of bisectors of line segments & angles,

60°, 90°, 45° angles etc, equilateral triangles. 2. Construction of a triangle given its base, sum/difference

of the other two sides and one base angle. 3. Construction of a triangle of given perimeter and base

angles.

UNIT V: MENSURATION 1. AREAS

Area of a triangle using Hero s formula (without proof) and its application in finding the area of a quadrilateral.

2. SURFACE AREAS AND VOLUMES Surface areas and volumes of cubes, cuboids, spheres (including hemispheres) and right circular cylinders/ cones.

UNIT VI: STATISTICS & PROBABILITY 1. STATISTICS

Introduction to Statistics: Collection of data, presentation of data tabular form, ungrouped/grouped, bar graphs, histograms (with varying base lengths), frequency polygons, qualitative analysis of data to choose the correct form of presentation for the collected data. Mean, median, mode of ungrouped data.

2. PROBABILITY His tory , Repeated exper iments and observed frequency approach to probability. Focus is on empirical probability. (A large amount of time to be devoted to group and to individual activities to motivate the concept; the experiments to be drawn from real - life situations, and from examples used in the chapter on statistics).

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When you buy a car, follow a recipe, or decorate your home, you’re using math principles. People have been using these same principles for thousands even millions of years, across countries and continents. Whether you’re sailing a boat off the coast or building a house, you’re using math to get things done. How can math be so universal? First, human beings didn’t invent math concepts; we discovered them. Also, the language of math is numbers, not English or German or Hindi. If we are well versed in this language of numbers, it can help us make important decisions and perform everyday tasks. Math can help us to shop wisely, buy the right insurance, remodel a home within a budget, understand population growth, or even can understand the mathematical trick behind any game. Mathematics is the only language shared by all human beings regardless of culture, religion, or gender. Pi is still 3.14159 regardless of what country you are in. Adding up the cost of a basket full of groceries involves the same math process regardless of whether the total is expressed in dollars, rubles, or yen. With this universal language, all of us, no matter what our unit of exchange, are likely to arrive at math results the same way. Very few people, if any, are literate in all the world’s tongues—English, Chinese, Arabic, Bengali, and so on. But virtually all of us possess the ability to be “literate” in the shared language of math. This math literacy is called numeracy, and it is this shared language of numbers that connects us with people across continents and through time. It is what links ancient scholars and medieval merchants, astronauts and artists, peasants and presidents. With this language we can explain the mysteries of the universe or the secrets of DNA. We can understand the forces of planetary motion, discover cures for catastrophic diseases, or calculate the distance from Boston to Bangkok. We can make chocolate chip cookies or save money for retirement. We can build computers and transfer information across the globe. Math is not just for calculus majors. It’s for all of us. And it’s not just about pondering imaginary numbers or calculating difficult equations. It’s about making better daily decisions and, hopefully, leading richer, fuller lives.

Maths Here and There

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Squaring numbers in the 600s (upto 609) 1. Choose a number in the 600s (practice with smaller numbers,

then progress to larger ones). 2. The first two digits of the square are 36 : 3 6 _ _ _ _ 3. The next two digits will be 12 times the last 2 digits: _ _ X X _ _ 4. The last two places will be the square of the last two digits: _ _ X X Example: If the number to be squared is 607: 1. The first two digits are 36 : 3 6 _ _ _ 2. The next two digits are 12 times the last 2 digits : 12×07= 84: _ _ 8 4 _ _ 3. Square the last 2 digits: 7×7=49 :_ _ _ _ 4 9 So 607 × 607 = 368,449. For larger numbers reverse the steps: If the number to be squared is 625 : 1. Square the last two digits (keep carry): 25 × 25=625 (keep 6): _ _ _ _ 25 2. 12 times the last 2 digits + carry: 12 × 25 +6 (keep carry)

= 300 + 6 = 306 : _ _ 0 6 _ _ 3. 36 + carry: 36 + 3 = 39 : 39 _ _ _ _000 So 625 × 625 = 390,625.

Squaring numbers in the 700s

1. Choose a number in the 700s (practice with smaller numbers, then progress to larger ones).

2. Square the last two digits (keep the carry) : _ _ _ _ X X 3. Multiply the last two digits by 14 and add the carry : _ _ X X _ _ 4. The first two digits will be 49 plus the carry: X X _ _ _ _ Example: If the number to be squared is 704 : 1. Square the last two digits : 4 × 4 = 16 : _ _ _ _ 1 6 2. Multiply the last two digits by 14 : 14 × 4 = 56: _ _ 5 6 _ _ 3. The first two digits will be 49 : 49 _ _ _ _

Beat The Calculator

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4 . So 704 × 704 = 495,616. See the pattern? If the number to be squared is 725 : 1. Square the last two digits (keep the carry): 25×25= 625: _ _ 2 5 (keep 6) 2. Multiply the last two digits by 14 and add the carry :

14 × 25+6= 350 + 6 = 356 : _ _ 5 6 _ _ (keep 3) 3. The first two digits will be 49 plus the carry: 49 + 3 = 52 : 52 _ _ _ _ 4 . So 725 × 725 = 525,625.

Squaring numbers between 800 and 810

1. Choose a number between 800 and 810. 2. Square the last two digits: _ _ _ _ X X 3. Multiply the last two digits by 16 (keep the carry): _ _ X X _ _ 4. Square 8, add the carry: X X _ _ Example: If the number to be squared is 802 : 1. Square the last two digits : 2 × 2 = 4 : _ _ _ _ 0 4 2. Multiply the last two digits by 16: 16 × 2 = 32 : _ _ 3 2 _ _ 3. Square 8 : 6 4 _ _ _ _

So 802 × 802 = 643,204. See the pattern? If the number to be squared is 807: 1. Square the last two digits : 7 × 7 = 49: _ _ _ _ 4 9 2. Multiply the last two digits by 16 (keep the carry): 16 × 7 =112 : _ _ 1 2 _ _ 3. Square 8, add the carry (1) : 6 5 _ _ _

So 807 × 807 = 651, 249.

Squaring numbers in the 900s

1. Choose a number in the 900s ‐ start out easy with numbers near 1000; then go lower when expert.

2. Subtract the number from 1000 to get the difference. 3. The first three places will be the number minus the difference: X X X _ _ _. 4. The last three places will be the square of the difference: _ _ _ X X X

(if 4 digits, add the first digit as carry).

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Example: If the number to be squared is 985 : 1. Subtract 1000 – 985 = 15 (difference) 2. Number – difference: 985 – 15 = 970 : 9 7 0 _ _ _ 3. Square the difference: 15 × 15 = 225 : _ _ _ 2 2 5

So 985 × 985 = 970225. See the pattern? If the number to be squared is 920: 1. Subtract 1000 – 920 = 80 (difference) 2. Number – difference: 920 – 80 = 840 : 840 _ _ _ 3. Square the difference: 80 × 80 = 6400 : _ _ _ 4 0 0 4. Carry first digit when four digits : 8 4 6 _ _ _

So 920 × 920 = 846400. Squaring numbers between 1000 and 1100

1. Choose a number between 1000 and 1100. 2. The first two digits are : 1,0 _ _, _ _ _ 3. Find the difference between your number and 1000. 4. Multiply the difference by 2 : 1,0 X X, _ _ _ 5. Square the difference : 1,0 _ _, X X X Example: If the number to be squared is 1007 : 1. The first two digits are : 1,0 _ _, _ _ _ 2. Find the difference : 1007 – 1000 = 7 3. Two times the difference: 2 × 7 = 14 : 1,0 1 4, _ _ _ 4. Square the difference: 7 × 7 = 49 : 1,0 1 4, 0 4 9

So 1007 × 1007 = 1,014,049. See the pattern? If the number to be squared is 1012: 1. The first two digits are : 1,0 _ _, _ _ _ 2. Find the difference : 1012 – 1000 = 12 3. Two times the difference : 2 × 12 = 24 : 1,0 2 4, _ _ _ 4. Square the difference: 12 × 12 = 144 : 1,0 2 4, 1 4 4

So 1012 × 1012 = 1,024,144.

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Start with lower numbers and then extend your expertise to all the numbers between 1000 and 1100. Remember to add the first digit as carry when the square of the difference is four digits.

Squaring numbers between 2000 and 2099

1. Choose a number between 2000 and 2099. (Start with numbers below 2025 to begin with, then graduate to larger numbers.)

2. The first two digits are: 4 0 _ _ _ _ _ 3. The next two digits are 4 times the last two digits: 4 0 X X _ _ _ 4. For the last three digits, square the last two digits in the number chosen

(insert zeros when needed) : 4 0 _ _ X X X Example: If the number to be squared is 2003: 1. The first two digits are: 4 0 _ _ _ _ _ 2. The next two digits are 4 times the last two: 4 × 3 =12 : _ _ 1 2 _ _ _ 3. For the last three digits, square the last two : 3 × 3 =9 :_ _ _ _0 0 9

So 2003 × 2003 =4,012,009. See the pattern? For larger numbers, reverse the order: 1. If the number to be squared is 2025: 2. For the last three digits, square the last two:25×25=625: _ _ _ _ 6 2 5 3. The middle two digits are 4 times the last two (keep the carry):

4 × 25 = 100 (keep carry of 1): _ _ 0 0 _ _ _ 4. The first two digits are 40 + the carry: 40 + 1 = 41 : 4 1 _ _ _ _ _

So 2025 × 2025 = 4,100,625.

Now its your chance Square the following numbers. 1. 631  2. 605  3. 650  4. 675  5. 685 Square the following numbers. 1. 710  2. 724  3. 734  4. 770  5. 790 Square the following numbers. 1. 801  2. 805  3. 802  4. 804  5. 808 Square the following numbers. 1. 915  2. 927  3. 1015  4. 2045  5. 2066

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Thales Thales (640 ‐ 546 B.C.) was a philosopher, who studied mathematics, astronomy, physics and other sciences. He was born in Greece but went to Egypt to study. He measured the height of a pyramid using the idea of similarity and predicted the date of an eclipse of the sun. He is sometimes called the father of mathematics and astronomy.

Pythagoras Pythagoras (580‐ 500 B.C.) was a philosopher, who studied mathematics, music and other subjects. He is famous because of the theorem named him.

Will iam Jones Willam jones gave the symbol of  π to denote the ratio of the circumference to the diameter of a circle.

Eucl id Euclid, the Greek mathematician who lived in Alexandria is famous for geometry. He wrote 13 volumes on geometry, which became the most important works in the study of geometry and have been used throughout the world. He was born in 300 BC in Greece.

Arch imedes Archimedes (287‐212 B.C.) was a famous Greek physicist and mathematician. He made important discoveries in geometry, hydrostatics and mechanics. He also invented the principle of lever and gave the concept of density.

Leonardo da Vinci He used perspective to paint solid figures on a plane canvas. Perspective is a way of showing three dimensional objects on paper. His enquiring scientific mind led him to investigate every aspect of natural world from anatomy to aerodynamics.

Nicolaus Copernicus Copernicus (1473‐1543) was a famous astronomer, mathematician and physicist of Poland. His famous idea was that the sun is the centre of the universe and the earth revolves round the sun and

Peoples in Mathematics

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became prime founder of modern astronomy. Gal i leo

Galileo Galilei (1564‐1652) was the famous mathematician, physicist and astronomer of Italy. He revolutionized scientific thought in his day and prepared the ground for Newton. Before Galileo, people believed that the speed of a falling body depends on its weight. Galileo showed for the first time by dropping two metal balls of different weights from the top of the Leaning Tower of Pisa that the speed of falling objects does not depend on their weight. He argued convincingly that free‐falling bodies, heavy or light, had the same constant acceleration and that a body moving on a perfectly smooth horizontal surface would neither speed up nor slow down. Galileo also contributed to the discovery of pendulum clocks. He made a telescope and studied several heavenly bodies. He supported the Copernicus’ ideas that the earth revolves round the sun.

Descartes Rene Descartes was a French mathematician who lived from 1596 to 1650. He is regarded as the discoverer of analytical geometry and founder of the science of optics. He developed the coordinate system for representing points in a plane and space. He was the first person to have used letters of the alphabet to represent numbers.

Pascal At the age of 12, he discovered that the sum of interior angles of a triangle is always 180 0 . He contributed to the development of calculus and probability theory. At the age of 19, Blaise Pascal invented the first calculating machine, which used gear wheels.

Sek i He invented a method of measuring area of figures bounded by curves and the volume of irregular solid figures.

Tadataka He was a Japanese mathematician; who lived from 1745 to 1818.

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His main contribution was to the making of the map of his country. John Dalton

Dalton (1766‐1844) was an English mathematician and chemist. He proposed atomic theory and prepared table of atomic weights.

Johann Gauss Johann Gauss was a German mathematician, who lived from 1777 to 1855. Gauss as one of the world’s greatest mathematician contributed a lot in the fields of astronomy and surveying. He also worked on the theory of numbers, non‐Euclidian geometry and on the mathematical development of electric and magnetic theory.

Simon Steven He advocated the use of decimal numbers in mathematics.

Ptolemy Claudius Ptolemy (100‐178 A.D.) calculated the value of  π for the first time and told that its value was 22/7. His greatest work was known as the Almagest in which he developed the theory of Aristotle.

Why Calculate π is the constant ratio between the diameter of a circle and the circumference of that circle. But why would anyone want to calculate π beyond the first few digits? The ten digits (3.141592654) that are built into most scientific calculators are sufficient for nearly any real‐ world calculation. Mathematician Hermann Schubert offers an example to show the uselessness of hundreds of digits of π .

“Conceive a sphere constructed with the earth at its center, and imagine its surface to pass through Sirius, which is 8.8 light years distant from the earth [that is, light, travelling at velocity of 186,000 miles per second, takes 8.8 years to cover this distance]. Then imagine this enormous sphere to be packed with microbes that in every cubic millimeter millions of million of these diminuitive animacula are present. Now distributed singly along a straight line that every two microbes are as far distant from each other as Sirius from us, 8.8 light years.

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Conceive the long line thus fixed by all the microbes as the diameter of a circle, and imagine this circumference to be calculated by multiplying its diameter by π to 100 decimal places. Then, in the case of a circle of this enormous magnitude even, the circumference so calculated would not vary from the real circumference by a millionth part of millimeter.”

This example will suffice to show that the calculation of π to 100 or 500 decimal places is wholly useless.

Pi is fundamental to the way in which our universe functions. Symbolically, π ’s irrationality denotes an irrationality to the universe that we do not like. Pi represents an omniscience which we will never posses, but that we can nudge closer and closer to as we approach its true value. Calculating π is a quest parallel to trying to fully understand our universe. It is for this reason that we wish to calculate π to millions of places and beyond.

To summarise, according to the

• Egyptians value of ( ) π = = 2 16 3.160 9 • Archimedes < π < 10 10 3 3 71 70

• Ptolemy π = 17 3 120 • Aryabhatta π = 3.1416

• Bhaskara π = 3.1416 • S. Ramanujam π = 355 113

One man’s definition of Pi : 1. The Greek letter π for Pi, corresponding to the roman π . 2. A number, represented by letter Pi, expressing the ratio of the circumference of a perfect circle to its diameter. The value of pi has been calculated to many millions per decimal places, to no readily apparent purpose: no perfect circles or spheres exist in nature, since matter is composed of atoms and therefore lumpy, not smooth. Nature herself sometimes takes to rounding off the more extreme decimals of numbers when they get sufficiently small.

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Do you know the answers of following question

Q 1 . Among the digits of pi currently known, the concentrations of each of the digits 0 ‐ 9 are pretty much equal. However, in the first 30 digits of pi’s decimal expansion, one number is conspicuously missing. Which number is it? A. 7 B . 2 C. 0 D . 8 E. 6

Q2. How does one convert pi in base 10 to base 2?

A . Keep only the ‘0’ and the ‘1’ in the decimal expansion. B . It is impossible because pi > 2. C . Replace each digit of pi in base 10 with a 0 if it is divisible by

2 and with and 1 if it is not. D . Successively multiply pi by 2 and put a ‘1’ when it is greater

than 1 and a ‘0’ when it is smaller than 1. Repeat this step after having kept only the fractional part of the result.

E . Divide pi in base 10 by 5.

Q3 . It has been proven impossible to “square the circle.” What does it mean to square the circle? A . multiply a circle by itself. B . use a straightedge and compass to construct a square equal

in area to a given circle. C . construct a square that perfectly circumscribes a circle. D . determine the value of 2 π . E . draw a circle with area equal to 2 π .

Q4 . What is the formal definition of pi? A . the ratio of a circle’s circumference to its diameter. B . 3.14159 C . the radius of a unit circle D . the surface area of a sphere of diameter 22/7 E . a delicious dessert, especially if it contains cherries

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1 . Use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 using each only once, divide them into two groups of five, and then arrange the digits in each group of five to make a multiplication sum. It could be done like this 0, 1, 5, 6, 9 used to make 106 x 59 (which is 6254) 2, 3, 4, 7, 8 used to make 437 x 82 (which is 35834) and the two multiplication sums have different answers. Find a way that makes the answers to both multiplication sums the same.

2 . A Magic Square is set of numbers arranged in the form of a square so that the total of every row, column and diagonal is the same. On the right is a Magic Square with 3 rows, 3 columns, and 2 diagonals, so that there are 8 totals to be found, each of which adds up to 15.

8 + 3 + 4 = 15 3 + 5 + 7 = 15 6 + 5 + 4 = 15 and so on.

An Anti‐magic Square has the same arrangement but the total of every row, column and diagonal is different. Use the same numbers (1, 2, 3, 4, 5, 6, 7, 8 and 9; once and only once) as in the example above to make an Anti‐magic Square with the added restriction that none of the 8 different totals must be greater than 18.

3 . Mr Khattar was exactly 5 years older than his wife and they had two children, who were twins and went to the local school. One evening, one of the twins (who had just been given a calculator) announced that if you multiplied together the parents’ ages, and then added together the twins’ ages, and divided one result by another, the answer was 57 exactly. How old was Mr Khattar?

Brain Benders

8 3 4 1 5 9 6 7 2

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The digit sum of a number is found by adding the digits in a number and adding again if necessary until a single figure is reached.

So, for example, the digit sum of 23 is 5 as 2 + 3 = 5.

Similarly for 21302 we get 2 + 1 + 3 + 0 + 2 = 8. And for 76 we first get 7 + 6 = 13, but 1 + 3 = 4, so 76 → 4. For 123456 we get 21 which gives 3, so 123456 → 3. So every number, no matter how long, can be reduced to a single figure, its digit sum. Practice Find the digit sum of the following numbers :

a. 42 b. 47 c. 32101 d. 777 e. 2468 f. 991 g. 837364

We can understand how these digit sums work by using the nine points circle in which the number from 1 to 9 are drawn round a circle as shown below. This circle is useful in a number of ways, as we will see. Continuing to number round the circle it seems reasonable to put 10 after the 9 and then 11 at the same place as 2, and so on. So all whole numbers find their place on the circle. Now if we add the digits of 10 we get 1, which is the number beside it. And if we add the digits of 11 we get 2 which is the number beside it, and so on. This means that all numbers with a digit sum of 2 will be on the 2‐branch, and similarly for the other branches. This also shows that adding 9 to a number does not affect its digit sum : whatever branch a number is on, adding 9 (or any number of 9’s, or subtracting 9’s) to it will not change the branch it is on. And this means that if we have to find the digit sum of a number with a nine in it we can delete the 9.

Vedic Maths : Digit Sums ‐ The Basis of Numerology

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The digit sum of 49 for example is just 4 (because 49 and 40

will be on the same branch) : we simply delete the 9. Similarly for the digit sum of 39979 we ignore the 9’s, add the 3 and 7 and give the digit sum as 1 (10 → 1). This can be taken further : any pair or group of digits which add up to 9 can be deleted.

For 3725 we can delete the 72 in the centre because they

add up to 9. We just add 3 and 5 to get the digit sum of 8. For 613954 we see 6 and 3 which make 9, 5 and 4 which make 9 and also a 9. Deleting all these we have only 1 left so 613954 → 1. The number 3251 also contains a 9 because 3 + 5 + 1= 9: deleting these 3251 → 2. Practice Find the digit sum :

a . 392 b . 93949 c . 789789 d . 456 e . 81187 f . 62371 g . 59479 h . 1999 i . 480417 j . 13579 k . 24680 l . 123456789 

For Workshop on Vedic Mathematics  contact National Coordinator, EduHeal Foundation, 103, Ground Floor, Taj Apartment, Near VMMC & Safdarjung Hospital, Ring Road, New Delhi ­ 110029. Ph. 011­26197342. The workshop will help you to acquire skills in quick maths calculations.

Mathematically inspired cats and kitten

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You have 100 meters of fencing with which to enclose a field for some animals to run around in. Because of the way in which the fence pieces are constructed your enclosure must be in the shape of a rectangle. You want to give your pets as much room as possible. What are the dimensions of your enclosure? What is its area? What if you used a different kind of fencing material that could be built in any shape? To get the maximum area for your animals, what would your field look like? What area does the fence enclose? What you can guess is a rectangle that is 24 by 26 meters. The area is 24m × 26m = 624 m 2

We pointed out that if you increased the short side a bit and decreased the long side by the same amount the area would be slightly larger. For example, 24.5m × 25.5m = 624.75 m 2

In fact, you could continue to do this: 24.9m × 25.1m = 624.99 m 2 getting ever closer to the the correct answer, which is 25m × 25m = 625 m 2

“But,” a question arises here that it is not a rectangle, it is a square!!” A square encloses the largest area of all rectangles with a given perimeter but it is a square not a rectangle. The Dictionary defines a rectangle as “a flat shape with four 90 degree angles and four sides, with opposite sides of equal length.” It does not say that the adjacent sides must necessarily be different. So the answer is 25 × 25 cm = 625 cm. If we do follow general definition of a rectangle, and exclude squares, then for any rectangle we can always find another where the short side is slightly longer and the long side is slightly shorter, and the area slightly larger. We approach, as a limit, a square with sides of 25m and an area approaching 625m 2 .

Useful Shape

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For the second part of the problem we asked what would happen if you used a different kind of fencing material that could be built in any shape? To get the maximum area for your animals, what would your field look like? What area does the fence enclose? The shape with a perimeter of 100 meters that has the largest possible area is a circle. The circumference is 100, so π times the diameter equals 100. The diameter is 2 times the radius, so 2 π r = 100m r = 100m / 2 π r = 50m / π The area of a circle is π r 2 , so for our circular fence A = π ( 50m / π ) 2

= π (2500m 2 / π 2 ) = 2500m 2 / π π is approximately equal to 3.14 so 2500m 2 / π = 2500m 2 / 3.14 = 796.18m 2

This is quite a bit larger than the area of a square field enclosed by the same 100 meters.

The Lighter Side of Mathematics Teacher: What is 2k + k ?

Student: 3000! Teacher: “Who can tell me

what 7 times 6 is?” Student: “It’s 42!” Teacher: “Very good! ‐ And who can tell me what 6 times 7 is?” Same student: “It’s 24!”

What does a mathematician present to his fiancée when he wants to propose? A polynomial ring!

The chef instructs his apprentice: “You take two thirds of water, one third of cream, one

third of broth...” The apprentice: “But that makes four thirds already!” “Well ‐ just take a larger pot!”

What do you get if you add two apples and three apples? A high school math problem!

What is the difference between a mathematician and a philosopher? The mathematician only needs paper, pencil, and a trash bin for his work ‐ the philosopher can do without the trash bin...

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A king was driving along a winding mountain road with his son. The foolish prince was not wearing his seat belt, so when the king had to swerve to avoid hitting an elephant that was strolling across the road, the passenger side door opened and the prince was thrown out. He tumbled to the side of the road and was left hanging by his fingertips over the valley several thousand meters below. Fortunately, a young peasant girl happened by just in time and pulled the prince to safety. The grateful king invited her home to the palace for dinner and a game of chess. After the game the king offered the girl a reward of 1,000,000 Rupees for saving the prince. The girl looked at the chess board and suggested an alternative reward: “Put one Rupee on the first square of the chess board. Tomorrow, remove the Rupee from the first square and put two Rupees on the second square. The day after tomorrow, remove the two Rupees and put four Rupees on the third square. The next day, put eight Rupees on the fourth square, and so on each day. On the 64th day, I will leave with whatever amount is on last square of the board.” The king, who was not a very good math student, thought this was a very modest request. He agreed to the girl’s wishes, thinking he would spend quite a bit less than his original offer. Can you guess how much money did the girl end up with? Was it more or less than 1,000,000 Rupees? Make a guess before you try to calculate it.

Knowledge of Math ‐ A Reward

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The king went broke long before the 64 th day. The girl took over the palace and declared herself Queen of his kingdom. She ended up with (18 and a half million trillion rupees) 18,446,7 44,073,70 9,551,616 rupee. If the king had looked at the pattern that was emerging, he would have seen that he was headed for big trouble: 1, 2, 4, 8, 16, 32, … These are the powers of 2. The sequence may be written: 2 0 , 2 1 , 2 2 , 2 3 , 2 4 , 2 5 , … After 64 days, we have 2 64 = 18,446,744,073,709,551,616 rupee Actually, the girl was being modest in her request. In another version of this puzzle, the money is left on each square instead of being re‐ moved. In this case the total would equal: 2 0 + 2 1 + 2 2 + 2 3 + 2 4 + 2 5 + … + 2 62 + 2 63 + 2 64 . Can you calculate how much it would be. The moral of the story is: Beware of exponential deals.

This is a test to gauge your mental flexibility & creativity. Most people got some correct, but no one got them all right. Some get answers long after the test had been set aside, at unexpected moments when their minds were relaxed, & some reported solving it over a period of several days how many can you get right Example : 16 O in a P. Answer : 16 Ounces in a Pound Now the test ..........Good luck ! 1 . 26 L of the A 2 . 7D of the W 3 . 1001 A N 4 . 12 S of the Z 5 . 54 C in a D (with J) 6 . 9 P in the S S 7 . 88 P K 8 . 3 S and IW on the I F 9 . 32 D at which W F 10 . 18 H on a G C 1 1 . 90 D in a R A 12 . 8 S on a S S 13 . 4 Q in a G 14 . 24 H in a D 15 . 1 W on a U 16 . 6 D in a P C 17 . 11 P on a F T 18 . 29 D in F in a L Y 19 . 64 S on a C 20. 21 D on a D 21 . 7 W of the A W

GENIUS TEST

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Interesting products for various multiplication problems. Not very useful, but certainly curious.

3 x 37 = 111 and 1 + 1 + 1 = 3 6 x 37 = 222 and 2 + 2 + 2 = 6 9 x 37 = 333 and 3 + 3 + 3 = 9 12 x 37 = 444 and 4 + 4 + 4 = 12 15 x 37 = 555 and 5 + 5 + 5 = 15 18 x 37 = 666 and 6 + 6 + 6 = 18 21 x 37 = 777 and 7 + 7 + 7 = 21 24 x 37 = 888 and 8 + 8 + 8 = 24 27 x 37 = 999 and 9 + 9 + 9 = 27

1 x 1 = 1 11 x 11 = 121 111 x 111 = 12321 1111 x 1111 = 1234321 11111 x 11111 = 123454321 111111 x 111111 = 12345654321 1111111 x 1111111 = 1234567654321 11111111 x 11111111 = 123456787654321 111111111 x111111111=12345678987654321

1 x 9 + 2 = 11 12 x 9 + 3 = 111 123 x 9 + 4 = 1111 1234 x 9 + 5 = 11111 12345 x 9 + 6 = 111111 123456 x 9 + 7 = 1111111 1234567 x 9 + 8 = 11111111 12345678 x 9 + 9 = 111111111 123456789 x 9 +10 = 1111111111

Curious Mathematical Calculation 9 x 9 + 7 = 88

98 x 9 + 6 = 888

987 x 9 + 5 = 8888

9876 x 9 + 4 = 88888

98765 x 9 + 3 = 888888

987654 x 9 + 2 = 8888888

9876543 x 9 + 1 = 88888888

98765432 x 9 + 0 = 888888888

1 x 8 + 1 = 9

12 x 8 + 2 = 98

123 x 8 + 3 = 987

1234 x 8 + 4 = 9876

12345 x 8 + 5 = 98765

123456 x 8 + 6 = 987654

1234567 x 8 + 7 = 9876543

12345678 x 8 + 8 = 98765432

123456789 x 8 + 9 = 987654321

7 x 7 = 49

67 x 67 = 4489

667 x 667 = 444889

6667 x 6667 = 44448889

66667 x 66667 = 4444488889

666667 x 666667= 444444888889

6666667 x 6666667 = 44444448888889

4 x 4 = 16

34 x 34 = 1156

334 x 334 = 111556

3334 x 3334 = 11115556

33334 x 33334 = 1111155556

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Do you face problem in long column addition. There is a trick by using which you can do long column addition with speed as well as with accuracy. Let do it having an example.

Add 83 + 7631 + 23125 + 67215 + 15276 Step ‐ I : Write the numbers in column.

8395 7631

23215 67215 15276

Step ‐ II : Now start adding column wise. In this method we never count higher than 10. That is when the running total becomes greater than 10, we reduce it by 10 and go ahead with the reduced figure. As we do so, we mark a small check mark besides the number that made our total higher than 10

5 + 1 = 6 6 + 5 = 11, reduce it by 10 and put a mark on 5. 1 + 5 = 6 6 + 6 = 12 reduce it by 10 and make a mark as 6

8 3 9 5 7 6 3 1

2 3 2 1 5’ 6 7 2 1 5 1 5 2 7 6’

2

Addition Made Easier

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Step 3 : Now count the ticks (these are the tens which we dropped out while adding first column.

2 + 9 =11 reduce it by 10 1 + 3 =4 4 + 1 =5 5 + 1 =6 6 + 7 =13

Moving on in the same fashion we will get the answer. 8 3 9′ 5 7′ 6′ 3 1

2 3 2 1 5’ 6′ 7′ 2 1 5

1 5′ 2 7′ 6 1 2 1 7 3 2

PRACTICE TIME Add the following sums in 10 minutes

1 . 73215 + 79315 + 31215 + 11617 + 21312 2 . 17681 + 96192 + 85197 + 159 + 7231

3 . 81257 + 76512 + 66123 + 5121 + 3191

4 . 17218 + 31219 + 82325 + 761 + 321 5 . 22395 + 31679 + 44316 + 53129 + 231

6 . 76721 + 33671 + 71659 + 71111 + 771

7 . 22315 + 47165 + 99936 + 85391 8 . 23176 + 71676 + 616316 + 712371

9 . 11732 + 77321 + 45271 + 58173

10 . 32179 + 72596 + 33172 + 82174

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Solved Puzzle Equation Making : Combine the three numbers in each group to get the same result in each of the three groups. You can use addition, subtraction, multiplication, division, and exponentiation. Here’s an example of a solved puzzle:

Group 1: 4, 15, 17 Group 2: 2, 26, 25 Group 3: 42, 44, 24

Group 1: 4 × 15 + 17 = 77 Group 2: 2 × 26 + 25 = 77 Group 3: 42 × 44/24 = 77

solution

1 Group 1: 1, 18, 26 Group 2: 19, 22, 23 Group 3: 20, 45, 50

2 Group 1: 2, 17, 23 Group 2: 13, 18, 26 Group 3: 23, 24, 36

3 Group 1: 4, 5, 9 Group 2: 2, 23, 24 Group 3: 27, 29, 50

4 Group 1: 5, 14, 15 Group 2: 17, 19, 34 Group 3: 20, 34, 37

Interactive Activities : Maths Puzzles Rectangle Puzzle : Draw rectangles along the grid lines so that no two rectangles share a corner or part of an edge, and each number indicates the total area of rectangles it is contained in. A solved example is shown below:

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1 . Jyoti was decorating her store window for a business sale. She wanted to make a figure that looks like the following. The shaded piece is made of a different material. How much material does she need in metres squared?

2 . Ganesh and Prince were playing with their fish tank. They had a difficult time keeping their fish alive. The fish tank is 100cm long, 60 cm wide and 40 cm high. They tilted the tank, as shown, resting on a 60 cm edge, with the water level reaching the midpoint of the base. When they rest the tank down to a horizontal position, what is the depth of the water in cm?

3 . If ABCD is a square and ABE is an equilateral triangle, then what is the measure of ∠ BFC?

4 . A classroom contained an equal number of boys and girls. Eight girls left to play hockey, leaving twice as many boys as girls in the classsroom. What was the original number of students present?

5 . Preeti had an average of 56% on her first 7 exams. What would she have to make on her eighth exam to obtain an average of 60% on 8 exams?

6 . Three grade nine Math students were given the following problem: A three digit number 2A4 is added to 329 and gives 5B3. If 5B3 is divisible by 3, then what is the largest possible value of A? One

Brain Twisters

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student thought A could be 1. Another student thought A was 5. The last student thought A was 4. Who was correct?

7 . Sunita wasn’t very keen on Algebra. Her teacher gave her an Algebra problem and told Sunita to solve it. She was having problems, can you help her?

3x + 7 = x 2 + k = 7x + 15 What is the value of k?

8 . A cubic metre of water weighs 1000 kg. What is the weight of a waterbed mattress that is 2 metres by 3 metres by 20cm if the casing of the mattress weighs 1 kg?

9 . If the figure shown is folded to make a cube, then what is the letter opposite the S?

C

U B

I C

S

10 . If the five expressions 2x + 1, 2x ‐ 3, x + 2, x + 5, and x ‐ 3 can be arranged in a different order so that the first three have a sum 4x + 3, and the last three have a sum 4x + 4, what would the middle expression be?

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This is another good party trick. Take a thick book with plenty of pages. Ask someone to open it at any page and choose a line in the first nine lines and a word in the first nine words of that line. Make sure you are out of the room or have turned your back as they are picking a word and that they let the other people know what it is. Now tell them to double the number of the page, multiply by 5 and 20. They ask them to add in the number of the line, add 5, multiply by 10 and then add the number of the word in the line. Finally ask them to add 1, 111 to the total and let you know the result. You think for a while, take the book, thumb through the pages and then point to a word, asking dramatically. “ Is this the word?” There is a gasp from the audience. You are right. Let’s go through an example : The chosen word is the 9th word on the 8th line of page 251 Double the number of the page 502 Multiply by 5 2510 Add 20 2530 Add the number of the line (8) 2538 Add 5 2543 Multiply by 10 25430 Add number of word in line (9) 25439 Add 111 26550

This is the result you are given. All you do is subtract 1361 from the result, if you can do that in your head ‐ otherwise, write the number down and do your subtraction while you are “studying” the result. You will finish up with the number 25189, which you do not show to anyone. This gives you the page number (251), the line (8) and the word (9).

Trick to Amaze Your Friend

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Mathematics to Bring Incredible Results What Day of the Week Were You Born On? Even though you were there at the moment of your birth, you may not remember exactly what day of the week it was. In fact, not only will this method help you find that out, you can find out the day of the week for any date you want in ANY A.D. year, not just 20th century. Step 1) Write down the last two digits of the year. Call it A. Step 2) Divide that number by four dropping the remainder call this

answer B. Step 3) Find the month number corresponding to the month from

the table below. Call it C. Step 4) The date call it D. Step 5) Take the century number of the year and divide it by four. If the remainder is 0 then E = 6, If the remainder is 1 then E = 4 If the remainder is 2 then E = 2, If the remainder is 3 then E = 0 Step 6) Now add A+B+C+D+E. Divide this sum by 7. The remainder

you get is the key to the day of the week. Step 7) In the table of days below, match the remainder with the

day of week. TABLE OF TABLE OF LEAP YEARS MONTHS DAYS

Sunday = 1 If year divides 400, yes Jan = 1 July = 0 Monday = 2 If year divides 100, no (0 in leap year) Feb = 4 Aug = 3 Tuesday = 3 If year divides 4, yes (3 in leap year) Mar = 4 Sept = 6 Wednesday = 4 If year does not divide 4, no

April = 0 Oct = 1 Thursday = 5 May = 2 Nov = 4 Friday = 6

June = 5 Dec = 6 Saturday = 0