The Simplified Math that Matters - Fractions, Decimals, Percentages, Averages, Ratios and Proportions

Glories to Duranteshwar Mahadev and Aadyanagha Mahadevi 🙏!

Well! Many of you all must have had a phobia for mathematics (or some of you still have it). Right?

This blog aims at removing that phobia for mathematics. We have curated the content to simplify mathematical concepts so that you all do not have to spend extra money in coaching classes! That too free of cost!

This blog aims at focusing on :

1. Fractions

2. Decimals

3. Percentages

4. Averages

5. Ratios and Proportions.

Do refer to our introductory article on mathematical tricks for further assistance.

Mathematical Tricks by The Aadyanagha Foundation

Chapter 1 - Fractions

Think that it is your birthday and you have been gifted a chocolate cake. Of course you will not be able to eat the full cake as you will be sharing the cake among your friends, family and other guests in the birthday party. 

A small slice cut from a whole cake

So the cake is cut into 24 equal parts. Out of that, you eat 5 slices. So you ate only a portion of the cake.

Representation of a Fraction

Fraction = A part of the whole

Numerator = Upper Part of the Fraction

Denominator = Lower Part of the Fraction; It is actually the total number of parts the whole is divided into.

Whole, Equal and Unequal Parts

Now think of a situation. Can you say that the yellow circle is divided into equal parts? Can you say that the lower and upper parts of the circle are 1/2? No.

Hence, to determine a fraction correctly, the given whole must be divided into equal parts.

Proper, Improper and Mixed Fractions

i. Proper Fraction: When numerator is less than the denominator. Example - 2/3, 3/4, 4/5, 1/10, etc.

ii. Improper Fraction: When the numerator is more than the denominator. Example - 5/4, 4/3, 3/2, 23/12, 45/32, etc

iii. Mixed Fraction: Representation of an improper fraction into a whole number and proper fraction.

Examples – 

5/4 is represented as (1 + 1/4 = 4/4 + 1/4).

3/2 is represented as (1 + 1/2 = 2/2 + 1/2).

Like and Unlike Fractions

Let us take a situation. There are three best friends. Mickey, Minnie and Jerry. They are very happy that they passed their examinations. They all go to a pizza shop and each buys an equal sized pizza for himself / herself.

Minnie's pizza is divided into 8 equal parts. Mickey's pizza is divided into 8 equal parts. Jerry's pizza is divided into 4 equal parts.

Minnie ate 3 slices from her pizza while Mickey ate 5 slices from his pizza. Since both the pizzas were cut into an equal number of slices, the slices were of the same size. Hence, it was easy to tell that Mickey is less hungry than Minnie now. The fractions here are like fractions as the denominators are same in both the fractions.

Now. Jerry too ate 3 slices from his pizza but he is less hungry than Minnie. How is that possible? Both ate 3 slices. Since both the pizzas were cut into an unequal number of slices, the slices were of a different size. Since Jerry's pizza was cut into lesser number of slices, his slices were of a larger size. The fractions here are unlike fractions as the denominators are different in both the fractions.

Like Fraction: When denominator of the set of fractions is same. Example - 2/5, 3/5 and 4/5.

When the denominator of given fractions is the same, then the fraction with the higher numerator has the higher value.

For example, 2/5 < 3/5 < 4/5.

Unlike Fraction: When denominator of the set of fractions is different. Example - 2/3 and 3/4. To compare them, we have to create equivalent fractions where the lowest common multiple of the denominators of both the fractions is the final denominator of both the fractions.

Equivalent and Simplified Fractions

Equivalent Fraction: When the numerator and denominator are multiplied by the same number, they cause such fractions. Example - Equivalent Fractions of 2/3 are 4/6, 6/9, 8/12, 10/15 and so on. Equivalent fractions of 3/4 are 6/8, 9/12, 12/16, 15/20 and so on.

Illustrations:

2/3 = (2 × 2)/(3 × 2) = 4/6

2/3 = (2 × 3)/(3 × 3) = 6/9

2/3 = (2 × 4)/(3 × 4) = 8/12

2/3 = (2 × 5)/(3 × 5) = 10/15

3/4 = (3 × 2)/(4 × 2) = 6/8

3/4 = (3 × 3)/(4 × 3) = 9/12

3/4 = (3 × 4)/(4 × 4) = 12/16

3/4 = (3 × 5)/(4 × 5) = 15/20

When the numerator of given fractions is same, then the fraction with the lower denominator has the higher value.

For example, there are two pizzas of equal sizes. One pizza is divided into 2 parts and the other pizza is divided into 4 parts.

The 1 slice of the first pizza would be bigger in size than the 1 slice of the other pizza. Hence, 1/2 is greater than 1/4.

Visual illustration of an equivalent fraction

Simplified Fraction: When the numerator and denominator are divided by their highest common factor (greatest common divisor), they cause such fractions. 

Illustrations :

1. 30/35 = (30 ÷ 5)/(35 ÷ 5) = 6/7

2. 42/48 = (42 ÷ 6)/(48 ÷ 6) = 7/8

3. 56/63 = (56 ÷ 7)/(63 ÷ 7) = 8/9

4. 50/55 = (50 ÷ 5)/(55 ÷ 5) = 10/11

Addition and Subtraction of Like Fractions

1. 4/7 + 2/7 = (4 + 2) / 7 = 6/7 (Denominator is the same)

2. 6/11 – 4/11 = (6 – 4)/11 = 2/11 (Denominator is same)

3. 7/9 + 2/9 = (7 + 2) / 9 = 9 / 9 = 1 (Since numerator is equal to denominator, the fraction becomes a complete whole; hence it becomes 1)

Addition and Subtraction of Unlike Fractions

1. 4/5 + 2/3 

= (4 × 3) / (5 × 3) + (2 × 5) / (3 × 5) 

= 12/15 + 10/15 = 22/15 

= (15 + 7)/15 

= 15/15 + 7/15 

= 1 + 7/15

2. 4/5 - 2/3 

= (4 × 3) / (5 × 3) - (2 × 5) / (3 × 5) 

= 12/15 - 10/15 

= 2/15

In the given numerical, we see that the denominators are different for both 4/5 and 2/3. So, we are using the concept of equivalent fractions to convert both fractions into like fractions by making the denominators same. The lowest common multiple would be the denominator of both fractions. 

Step 1: We multiply 4 and 5 by 3 and get 12/15. Similarly, we multiply 2 and 3 by 5 and get 10/15. 

Step 2:  After adding 12/15 and 10/15, we see an improper fraction is formed (numerator > denominator). We convert it into a mixed fraction (representing an improper fraction as a whole number and proper fraction). Since the denominator is 15, we split 22 as 15 + 7. Then we show 22/15 as 15/15 + 7/15. In 15/15, 15 gets cancelled out and we get the number as 1. Hence 22/15 becomes 1 + 7/15.

Let’s solve 4/15 + 3/20 

We first create an equivalent fraction for both to convert to them like fractions

We know that 60 is the lowest common multiple for both 15 and 20.

4/15 = (4 × 4)/(15 × 4) = 16/60

3/20 = (3 × 3)/(20 × 3) = 9/60

16/60 + 9/60 = 25/60 = (5 × 5)/(12 × 5) = 5/12

Multiplication of Fractions

1. 2/3 × 4/5 = (2 × 4) / (3 × 5) = 8/15

2. 2/5 × 7/8 = (2 × 7) / (5 × 8) = 7 / (5 × 4) = 7/20 (since 2 divides 8, so 2 becomes 1 and 8 becomes 4)

3. 4/15 × 9/10 = (4 × 9) / (15 × 10) = (2 × 3) / (5 × 5) = 6/25 (because 2 is a common factor of 4 and 10 ; while 3 is a common factor of 9 and 15)

Division of Fractions

1. 2/3 ÷ 4/5 = 2/3 × 5/4 = (2 × 5) / (3 × 4) = 5 / (3 × 2) = 5/6

2. 6/7 ÷ 8/9 =  6/7 × 9/8 = (6 × 9) / (7 × 8) = (3 × 9) / (7 × 4) = 27/28

Chapter 2 - Decimals

Let’s take a situation. We know 1 over = 6 balls.

In the first over, India scores 12 runs.

In the second over, India scores 4 runs.

In the third over, India scores 14 runs.

In the fourth over, India scores 8 runs.

In 4 overs = (12 + 4 + 14 + 8) = 38 runs

Run Rate = Runs in a given amount of overs / Number of balls = 38/4 = 9.5

We cannot represent the run rate in the form of a fraction on television. So, we convert it to a decimal. In short, Decimal Point is the tiny dot. Everything left to it is a whole number and to its right is a portion.

Some standard Fractions converted to Decimal equivalents:

1/10 = 0.1

1/100 = 0.01

1/1000 = 0.001

1/10000 = 0.0001

Let’s understand decimals with respect to place value.

Let us write this in expanded form.

987.642 = (9 × 100) + (8 × 10) + 7 + (1/10 × 6) + (1/100 × 4) + (1/1000 × 2) = 900 + 80 + 7 + 0.6 + 0.04 + 0.002 = 987 + 0.642

Let’s understand how to derive decimals from division process.

Let us take an example where we divide 29 by 4.

We all know 29 is not a multiple for 4.

i. The multiple of 4 closest to 29 is 28.

ii. So, the part of the whole number left to the decimal point is 7 (a whole number).

iii. Now we get a remainder 1 (29 – 28 = 1). We need to divide it by 4.

iv. We can also show 29 as 29.0 and 29.00.

v. We bring down an invisible 0 to divide the remainder 1. As a result, it becomes 10. 

vi. The multiple of 4 closest to 10 is 8. We put a 2 after the decimal point.

vii. We again get 2 as a remainder (10 – 8 = 2). We again bring down another 0 as a result of which 2 becomes 20.

viii. We know 20 is a multiple of 4. So, we put 5 after 2.

ix. We get the final answer as 7.25 when we divide 29.00 by 4.

When we represent 29 / 4 as a mixed fraction, we get a mixed fraction as 7 + 1/4.

In the same way when we represent it as a decimal, we get it as 7.25.

Types of Decimals

i. Terminating Decimals: These have a clear ending after a point. 

Examples -

1/2 = 0.5; 1/4 = 0.25; 1/8 = 0.125

ii. Recurring Decimals: No matter how many zeroes you bring down, they do not end.

Examples -

1/3 = 0.333333…. ; 1/6 = 0.166666666…. ; 1/9 = 0.111111….

Mathematical Operations on Decimals

1. Let us take a situation where we purchased cashews worth ₹ 24.35, almonds worth ₹ 26.55 and peanuts worth ₹ 31.21. What is the total bill?

Let us use the Left to Right Addition Method.

Cost of cashews = ₹ (20 + 4 + 0.3 + 0.05)

Cost of almonds = ₹ (20 + 6 + 0.5 + 0.05)

Cost of peanuts = ₹ (30 + 1 + 0.2 + 0.01)

Total Bill = ₹ (20 + 4 + 0.3 + 0.05) + ₹ (20 + 6 + 0.5 + 0.05) + ₹ (30 + 1 + 0.2 + 0.01) 

= ₹ [(20 + 20 + 30) + (4 + 6 + 1) + (0.3 + 0.5 + 0.2) + (0.05 + 0.05 + 0.01)] 

= ₹ (70 + 11 + 1.0 + 0.11) = ₹ 82.11

2. The selling price of a perfume is ₹ 123.45, its discount is ₹5.6 and its cost price is ₹ 98.76. Is it a profit or a loss?

Let us use Zigzag Subtraction Method to calculate the final selling price.

Final selling price = ₹ (123.45 - 5.6) = ₹ (123.45 - 5 - 0.6) = ₹ (118.45 - 0.6) = ₹ 117.85

Since the selling price > cost price, it is a profit. 

Let us use Left to Right Subtraction Method.

Amount of profit = ₹ 117.85 - ₹ 98.76 = ₹ (100 + 10 + 7 + 0.8 + 0.05) - ₹ (90 + 8 + 0.7 + 0.06) 

= ₹ [100 + (10 - 90) + (7 - 8) + (0.8 - 0.7) + (0.05 - 0.06)] = ₹ [(100 - 80 - 1) + (0.1 - 0.01)] = ₹ [19 + 0.09] = ₹ 19.09

3. The speed of a car is 15.4 km/hr. If the time taken is 2.5 hours, what is the distance traveled by the car?

Since Distance = Speed × Time ;

Distance traveled by the car = (15.4 × 2.5) km = [(154/10) × (25/10)] km = [3850/100] km = 38.5 km

4. The selling price of each share is ₹ 2.25. The earnings from all shares is ₹ 69.75. How many shares were sold?

Number of shares sold = 69.75 ÷ 2.25 = (69.75 × 100) ÷ (2.25 × 100) = 6975 ÷ 225 = 31

Chapter 3 - Percentages

You all must have heard of statements like :

i. "I am 100% sure this has happened."

ii. "Wow you got 100% marks."

iii. "Cent per cent."

What does the word percent mean? The word 'cent' means 100. So per cent means per 100. Hence, a percentage (symbolized by '%' symbol) is a fraction where the denominator is 100. The use of percentages increased because keeping 100 as a base was easy for calculations. 

Percentages are commonly used in report cards, profit and loss rates, discount rates, interest rates, investment rates, inflation rates, etc.

Illustration for a percentage

In the given picture, we see 20 small squares out of 100 small squares colored in purple in the first large square ; while in the second large square, we see 50 small squares out of 100 small squared colored in purple.

Relationship between Fractions, Decimals and Percentages

Fractions, Decimals and Percentages converting to each other

To convert a fraction or a decimal into percentage, multiply it by 100 and add a % symbol after it.

Why we do that? Because by multiplying by 100% (i.e. 100/100), we are making it an equivalent fraction where the denominator is a 100. 

Illustration :

1. 1/2 = (1 × 50) / (2 × 50) = 50/100 = 50%

2. 1/4 = (1 × 25) / (4 × 25) = 25/100 = 25%

3. 0.54 = 54/100 × 100/100 = 54/100 × 100% = 54%

4. 0.36 = 36/100 × 100/100 = 36/100 × 100% = 36%

Now let us understand the concept of opposites.

Suppose we take a situation where 94% students passed in the mathematics test. How many failed?

We all know 100% means complete!

So the number of students who failed = 100% - 94% = 6%

Let us understand this from an example.

There are 200 students in the class. 188 students passed. 

So logically, the number of students who failed = 200 - 188 = 12

% of students who passed = 188/200 × 100% = 94%

% of students who failed = 12/200 × 100% = 6%

% of all students = 94% + 6% = 100%

Let us try some numerical problems.

1. The speed of a bike is 40 km/hr. The total distance is 200 km. The distance traveled by now is 60 km. The speed increases by 25% and the remaining distance is covered. What is the total time taken?

Increase in the speed = (25/100 × 40) km/hr = 10 km/hr

Remaining distance = (200 - 60) km = 140 km

Speed for remaining distance = (40 + 10) km/hr = 50 km/hr

Time taken to cover 60 km = (60 ÷ 40) hours = 1.5 hours

Time taken to cover 140 km = (140 ÷ 50) hours = 2.8 hours

Total time taken to travel 200 km = (1.5 + 2.8) hours = 4.3 hours

2. A customer buys three items. A shampoo worth ₹ 180 ; a body soap worth ₹ 40 ; a hair oil worth ₹ 120 ; a face cream worth ₹ 60. The discount is of 10%. The tax on the bill is 5%. What is the total value of the bill?

Cost of all products = ₹ (180 + 40 + 120 + 60) = ₹ 400

Discount Amount = ₹ (10% × 400) = ₹ 40

Tax Amount = ₹ (5% × (400 + 40)) = ₹ (5% × 440) = ₹ 22

Final Value of the Bill = ₹ (440 + 22) = ₹ 462

Chapter 4 - Averages

There are four friends. Donald, Daisy, Daffy and Melissa. They see some chocolate bars. Donald takes 8 bars, Daisy takes 6 bars, Daffy takes 9 bars and Melissa takes 5 bars. Don't you think there is an inequality? Yes!

The Chocolate Angel steps in and asks them to give her back all the chocolate bars so that all the bars are distributed equally.

Total number of chocolate bars = 8 + 6 + 9 + 5 = 28

How many bars will each friend get? 

Since there are 4 friends, we divide the number of chocolate bars by the number of friends.

Number of chocolate bars each friend will get = 28 ÷ 4 = 7

Now each friend gets a 'fair share'. This 'fair share' is known as an average.

You all must have heard of statements like :

1. "The average Indian spends 10 hours at work."

2. "The average crime rates in India are high and unreported, etc."

Remember the first example we said with respect to the run rate (in the Chapter 2 (Decimals) in this blog)? The run rate is nothing but the average runs scored in an over.

Now lets take a situation.

In the first over, India scores 12 runs.

In the second over, India scores 4 runs.

In the third over, India scores 14 runs.

In the fourth over, India scores 8 runs.

In the fifth over, India scores a certain number of runs because of which the run rate becomes 11. How many runs did India score in the fifth over?

Total runs in 5 overs = Number of overs × run rate = (5 × 11) runs = 55 runs

Runs in the fifth over = 55 - (12 + 4 + 14 + 8) = 55 - 38 = 17 runs

Chapter 5 - Ratios and Proportions

Now take a situation where you make a chocolate - date - walnut cake. It involves the following ingredients :

5 cups of flour ; 4 cups of milk ; 3 cups of chocolate powder ; 2 cups of date paste ; 1 cup of walnuts.

How much will 2 chocolate - date - walnut cakes require?

10 cups of flour ; 8 cups of milk ; 6 cups of chocolate powder ; 4 cups of date paste ; 2 cups of walnuts.

How much will 6 chocolate - date - walnut cakes require?

30 cups of flour ; 24 cups of milk ; 18 cups of chocolate powder ; 12 cups of date paste ; 6 cups of walnuts.

Just imagine if for 2 cakes we only added more flour and kept other quantities intact, would the cake be tasty? No. Without the addition of the milk, the cakes will not be soft. Without the increase in chocolate powder, the cakes would not be chocolaty. Without the addition of the date paste the cakes will not be sufficiently sweet. Without the addition of the walnuts, the crunchiness in the cakes would be less.

Here the relationship between the ingredients is called a ratio (symbolized by ':').

Ratio of flour to milk to chocolate powder to date paste to walnuts = 5 : 4 : 3 : 2 : 1

When you maintain the ratio with the making of every cake, you ensure the taste is good. Keeping that rule the same is called proportion.

Let us take some more examples :

1. There are 6 tigers and 7 lions needed for a circus show.

Ratio of tigers to lions = 6 : 7

Now imagine if we would write the ratio of lions to tigers, would we still write 6 : 7? No. We would write it as 7 : 6. The order of the numbers must match the order of the words!

2. There are 2 puppies, 3 rabbits and 4 hamsters in a family.

Ratio of hamsters to rabbits to puppies = 4 : 3 : 2

Simplified Ratios

Think of a situation where you are making a pudding that requires 24 almonds and 27 cashews, what would be the ratio of almonds to cashews?

Ratio of almonds to cashews = 24 : 27 = (24 ÷ 3) : (27 ÷ 3) = 8 : 9

Why did we do this? In real life situations, such big numbers will only make calculations harder. Hence, we divide the terms of the ratio by their highest common factor to make the ratio look simpler.

Equivalent Ratios

Think of the example of the tigers and lions. Suppose there are three circus shows, each require these number of tigers and lions, what would be the ratio?

Ratio of tigers to lions = (6 × 3) : (7 × 3) = 18 : 21

This kind of a ratio is known as an equivalent ratio where all the terms of the ratio are multiplied by the same number. 

These are the secrets behind proportions.

Relationship between Ratios, Fractions and Percentages

Suppose there are 6 roses and 4 lotuses in a pot.

Total number of flowers in the pot = (6 + 4) flowers = 10 flowers

Portion of Roses in the pot = 6/10 = (3 × 2) / (5 × 2) = 3/5

Portion of Lotuses in the pot = 4/10 = (2 × 2) / (5 × 2) = 2/5

% of Roses in the pot = 3/5 × 100% = 60%

% of Lotuses in the pot = 2/5 × 100% = 40%

Ratio of Roses to Lotuses in the Pot = 6 : 4 = (3 × 2) : (2 × 2) = 3 : 2

Understanding the concept of equivalent ratios, it can also be written as 60 : 40.

Further, it can also be written as 3/5 : 2/5.

If we notice carefully, the numerators in both the fractions are the terms of the ratio. On the other hand, the denominator of the fractions is the sum of the terms of the ratio (3 + 2 = 5).

Product of Means and Extremes in Proportions

As mentioned before, equivalent ratios and simplified ratios are the secret behind proportions.

1. Let us first take the example of almonds and cashews

8 : 9 :: 24 : 27

Here 8 and 27 are the extremes while 9 and 24 are means.

Product of Means = 24 × 9 = 216

Product of Extremes = 27 × 8 = 216

2. Now let us take the example of tigers and lions

6 : 7 :: 18 : 21

Product of Means = 18 × 7 = 126

Product of Extremes = 21 × 6 = 126

From these examples, we understand Product of Means = Product of Extremes.

Let us solve some examples.

1. The ratio of mangoes to papayas is 3 : 4. If we have 9 mangoes, how many papayas will we have?

Let the number of papayas be 'x'.

The proportion is = 3 : 4 :: 9 : x

This means 3/4 = 9/x

Product of Extremes = Product of Means

3x = (9 × 4) = 36

x = 36 ÷ 3 = 12

So, we have 12 papayas.

2. The ratio of marks in English to Mathematics is 8 : 7. If the score in Mathematics is 42, what is the score in English? Also give us the percentage in each subject and average marks scored, considering the full marks in each paper is 50.

Let the marks scored in English be 'x'.

The proportion is = 8 : 7 :: x : 42

This means 8/7 = x/42

Product of Means = Product of Extremes

7x = 42 × 8

x = (42 × 8) ÷ 7 = 6 × 8 = 48

Hence, the marks in English is 48.

% marks in English = 48/50 × 100% = 96%

% marks in Mathematics = 42/50 × 100% = 84%

Average marks scored = (48 + 42) ÷ 2 = 90 ÷ 2 = 45

Conclusion 

And that’s it! We have now covered some of the most important everyday mathematical concepts — Fractions, Decimals, Percentages, Averages, and Ratios & Proportions — in a simple and practical way.

Mathematics is not about memorizing complicated rules or spending huge amounts on coaching classes. It is about understanding the logic behind numbers and using simple tricks to make calculations easier. Whether you are sharing a cake, calculating run rates in cricket, preparing a perfect recipe, or checking your exam marks, these concepts are everywhere in our daily lives.

We hope this blog has helped reduce your fear of mathematics and given you useful tools to solve problems confidently. Remember — practice is the key. Try solving a few problems from your textbook or daily life using the methods we discussed. You will surely see the difference!

Thanks and Regards,

The Aadyanagha Foundation.

The Simplified Math that Matters - Factors, Multiples, Exponents and Roots

Glories to Duranteshwar Mahadev and Aadyanagha Mahadevi 🙏!

Well! Many of you all must have had a phobia for mathematics (or some of you still have it). Right?

This blog aims at removing that phobia for mathematics. We have curated the content to simplify mathematical concepts so that you all do not have to spend extra money in coaching classes! That too free of cost!

This blog aims at focusing on the following topics :

1. Introduction to Factors, Multiples and Factorials

2. Prime Numbers and Composite Numbers

3. Highest Common Factor (HCF)

4. Lowest Common Multiple (LCM)

5. Relationship between HCF and LCM

6. Introduction to Exponents

7. Introduction to Roots

Chapter 1 - Introduction to Factors, Multiples and Factorials

Think of a situation where you have 6 balls. How are you going to arrange them in different rows and columns?

Yes!

Combination 1 - 1 row × 6 columns = 6 balls

Combination 2 - 6 rows × 1 column = 6 balls

Combination 3 - 2 rows × 3 columns = 6 balls

Combination 4 - 3 rows × 2 columns = 6 balls

Factors of 6 = {1, 2, 3, 6}

Now imagine had there been 7 balls.

Combination 1 - 1 row × 7 columns = 7 balls

Combination 2 - 7 rows × 1 column = 7 balls

Factors of 7 = {1, 7}

Now imagine had there been 8 balls.

Combination 1 - 1 row × 8 columns = 8 balls

Combination 2 - 8 rows × 1 column = 8 balls

Combination 3 - 2 rows × 4 columns = 8 balls

Combination 4 - 4 rows × 2 columns = 8 balls

Factors of 8 = {1, 2, 4, 8}

Now imagine had there been 9 balls.

Combination 1 - 1 row × 9 columns = 9 balls

Combination 2 - 9 rows × 1 column = 9 balls

Combination 3 - 3 rows × 3 columns = 9 balls

Factors of 9 = {1, 3, 9}

From here we can understand that a factor is a divisor that divides the dividend without leaving a remainder. (Note : The quotient is also the factor when the remainder is 0.)

On the other hand, a multiple is the product we get when one number is multiplied by another number.

(Note : Here we are purely talking about natural numbers. Not fractions, decimals or negative integers.)

Let us check the following examples:

1. Let us consider the number 10.

Factors of 10 = {1, 2, 5, 10}

Multiples of 10 = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100, ................}

Highest Factor = 10

Lowest Multiple = 10

2. Let us consider the number 24.

Factors of 24 = {1, 2, 3, 4, 6, 8, 12, 24}

Multiples of 24 = {24, 48, 72, 96, 120, 144, 168, 192, 216, 240, ................}

Highest Factor = 24

Lowest Multiple = 24

3. Let us consider the number 35.

Factors of 35 = {1, 5, 7, 35}

Multiples of 35 = {35, 70, 105, 140, 175, 210, 245, 280, 315, 350, ................}

Highest Factor = 35

Lowest Multiple = 35

From the above examples, we can understand the following :

i. The factors of a number are finite in nature while there can be infinite multiples of a number. 

ii. The highest factor of a number is the number itself 

iii. The lowest multiple of a number is the number itself. 

iv. 1 is the lowest factor for all numbers.

Now let us understand Factorials.

A Factorial is the product of an integer and all positive integers below it.

0! = 1

1! = 1

2! = 2 × 1 = 2 × 1! = 2

3! = 3 × 2 × 1 = 3 × 2! = 6

4! = 4 × 3 × 2 × 1 = 4 × 3! = 24

5! = 5 × 4 × 3 × 2 × 1 = 5 × 4! = 120

Now why 0! is equal to 1?

Think of a situation of how we can arrange different of crayons in one straight line.

Think that we have three crayons - Pink, Blue and Yellow

There are different ways in which we can put the crayons.

Combination 1 - Pink, Blue, Yellow

Combination 2 - Pink, Yellow, Blue

Combination 3 - Blue, Yellow, Pink

Combination 4 - Blue, Pink, Yellow

Combination 5 - Yellow, Pink, Blue

Combination 6 - Yellow, Blue, Pink

So there are 6 (which means 3!) ways in which we can put three crayons in a straight line.

Now think if we had only the pink and blue crayons.

Combination 1 - Pink, Blue

Combination 2 - Blue, Pink

So there are 2 (which means 2!) ways in which we can put the two crayons in a straight line.

Now if we had only the blue crayon, we could only put the blue crayon in the straight line. So there would be only 1 (which means 1!) way.

Now think if you had 0 crayons left. How many ways would you arrange them?

It would be only 1 way. You just look at the empty floor. An empty floor is exactly one specific look. If there were 0 ways to do it, you wouldn't be allowed to even have an empty floor.

2. Prime Numbers and Composite Numbers

Let us recollect our initial example where we arranged 6 balls, 7 balls, 8 balls and 9 balls in different rows and columns.

Did you notice that for 7 balls, there were only two arrangements? Here all 7 balls would be in either one row or in one column.

What do we understand from this?

Prime Number is a number which has no factors other than 1 and itself.

On the other hand, Composite Numbers are numbers having factors other than 1 and itself.

1 is neither a prime nor a composite number. Because it has no factor other than itself.

Below is a picture of Eratosthenes Sieve to determine prime and composite numbers till 100.

Prime Factorization Method - As per the Fundamental Theorem of Arithmetic, every integer greater than 1 is either a prime number, or itself can be expressed as the product of prime numbers.

Let us see some illustrations.

1. 2 = 2 × 1

2. 4 = 2 × 2

3. 6 = 2 × 3

4. 9 = 3 × 3

5. 12 = 2 × 2 × 3

6. 42 = 2 × 3 × 7

7. 45 = 3 × 3 × 5

Chapter 3 - Highest Common Factor

We have already discussed about factors. Now let us learn something fun.

Think that you are throwing a party. You have 12 Milk Chocolate Bars and 18 Dark Chocolate Bars. You want to make identical goody bags for your friends. Every bag must have the exact same combination of both types of chocolate, so that there are no leftovers. What is the largest number of goody bags you can make?

Lets get factors of both 12 and 18.

Factors of 12 = {1, 2, 3, 4, 6, 12}

Factors of 18 = {1, 2, 3, 6, 9, 18}

We see that 12 and 18 are divisible by 2. We also see that 12 and 18 are divisible by 3. Here 2 and 3 are common factors of 12 and 18. But if we notice carefully, 6 is the number which divides 12 and 18. After 6, 12 and 18 is not divisible by any number. So 6 is the Highest Common Factor (HCF).

Hence, the maximum number of bags that can be made is 6.

Now how many chocolates will each bag have?

Number of Milk Chocolate Bars = 12 ÷ 6 = 2

Number of Dark Chocolate Bars = 18 ÷ 6 = 3

Highest Common Factor is also known as Greatest Common Divisor.

Let us take up some more numericals.

1. There are 24 people who know French, 28 people who know German, 32 people who know Chinese and 36 people who know Japanese? How many identical groups of people can we make so that no person is left out?

First we will find a list of factors of 24, 28, 32 and 36. We will underline the factors that are common for all the given numbers.

Factors of 24 = {1, 2, 3, 4, 6, 8, 12, 24}

Factors of 28 = {1, 2, 4, 7, 14, 28}

Factors of 32 = {1, 2, 4, 8, 16, 32}

Factors of 36 = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Common Factors of 24, 28, 32 and 36 = {1, 2, 4}

Highest Common Factor = 4

We can also use the Prime Factorization method to calculate the HCF.

24 = 2 × 2 × 2 × 3

28 = 2 × 2 × 7

32 = 2 × 2 × 2 × 2 × 2

36 = 2 × 2 × 3 × 3

Highest Common Factor = 2 × 2 = 4

So 4 identical groups can be made.

Number of people in each group :

People knowing French = 24 ÷ 4 = 6

People knowing German = 28 ÷ 4 = 7

People knowing Chinese = 32 ÷ 4 = 8

People knowing Japanese = 36 ÷ 4 = 9

There will be 6 people knowing French in each group, 7 people knowing German in each group, 8 people knowing Chinese and 9 people knowing Japanese.

2. There are 20 Jains, 25 Buddhists, 30 Muslims, 35 Sikhs and 40 devotees of Hindus. A secular devotional committee is to be made for opening centers inculcating spiritual values and national unity. How many identical groups can be made so that all devotees are included in the committee?

Let us use prime factorization method.

20 = 2 × 2 × 5

25 = 5 × 5

30 = 2 × 3 × 5

35 = 5 × 7

40 = 2 × 2 × 2 × 5

Highest Common Factor = 5

There will be 5 identical groups. Each group will have 4 Jains (20 ÷ 5), 5 Buddhists (25 ÷ 5), 6 Muslims (30 ÷ 5), 7 Sikhs (35 ÷ 5) and 8 Hindus (40 ÷ 5).

3. There are 8 singers and 9 dancers. How many identical groups can be made having all artists?

Factors of 8 = {1, 2, 4, 8}

Factors of 9 = {1, 3, 9}

Here we see only 1 as the common factor for 8 and 9. Hence, only 1 group can be made.

8 and 9 have 1 as the highest common factor proving that they are co-prime numbers.

Chapter 4 - Lowest Common Multiple

Think of a scenario.

A red car completes a lap at every 4 minutes. A blue car completes a lap at every 6 minutes. If they start at the same time, after how many minutes will they pass the finishing line together for the first time?

As we know multiplication means repeated addition, we will use the concept of multiples.

Multiples of 4 = {4, 8, 12, 16, 20, 24, 28, 32, 36, ..........}

Multiples of 6 = {6, 12, 18, 24, 30, 36, .............}

We see that both the red and the blue car hit the finishing line together and 12 minutes, 24 minutes, 36 minutes and so on. But for the first time it happens at 12 minutes. Here 12 is the Lowest Common Multiple (LCM). 

Let us take up some numericals.

1. A tiger comes to the forest in every 6 days and a lion comes to the forest in every 8 days. If they start on the same day, in how many days are they going to meet next?

Number of days in which the tiger comes to the forest = Multiples of 6 = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, .........}

Number of days in which the lion comes to the forest = Multiples of 8 = {8, 16, 24, 32, 40, 48, 56, 64, 72, ...........}

Common Multiples of 6 and 8 = {24, 48, 72, .........}

Lowest Common Multiple = 24

The tiger and lion are going to meet in the upcoming 24 days.

2. You are hosting a party. Palak Roti comes in packs of 8 while Dal Makhani comes in packs of 12. You want a perfect 1-to-1 match so that every pack of Palak Roti has a pack of Dal Makhani with 0 leftovers.

Successive Packs of Palak Roti = Multiples of 8 = {8, 16, 24, 32, 40, 48, 56, 64, 72, ..........}

Successive Packs of Dal Makhani = Multiples of 12 = {12, 24, 36, 48, 54, 60, 72, .......}

Lowest Common Multiple = 24

In 24 packs for both Palak Roti and Dal Makhani, they come together because of which we can prepare a 1-to-1 match.

To get a perfect match, how many Palak Roti packs and Dal Makahni packs you need:

Packs of Palak Roti = 24 ÷ 8 = 3

Packs of Dal Makhani = 24 ÷ 12 = 2

3. An arcade lets you trade in two types of tickets for prizes. Small tokens are worth 15 points each, and big medals are worth 20 points each. You want to trade in a bunch of tokens and a bunch of medals so that the total points from your tokens exactly equals the total points from your medals. What is the smallest equal point value you can hit?

Token Points =  Multiples of 15 = {15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, ........}

Medal Points = Multiples 20 = {20, 40, 60, 80, 100, 120, 140, 160, 180, .............}

Lowest Common Multiple = 60 

60 is the smallest point value that can be hit which is possible from 4 small tokens and 3 big medals.

Chapter 5 - Relationship between HCF and LCM

Calculation of HCF and LCM using Division Method

HCF of two numbers × LCM of two numbers = Product of both the numbers

Let us take numbers 24 and 36.

Prime Factorization of :

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

Highest Common Factor = 2 × 2 × 3 = 12

24 = 12 × 2

36 = 12 × 3

Multiples of 24 = {24, 48, 72, 96, 120, 144, 168, 192, 216, ........}

Multiples of 36 = {36, 72, 108, 144, 180, 216, .......}

Lowest Common Multiple = 72 = 12 × 2 × 3 

LCM = HCF × Remaining Uncommon Factors

This demonstrates that when calculating the LCM, we include the HCF only once—multiplied by the remaining unique factors—to avoid duplicating shared values.

Now let us see :

72 × 12 = 24 × 36

(12 × 2 × 3) × 12 = (12 × 2) × (12 × 3)

12² × 2 × 3 = 12² × 2 × 3 (Equal on both sides)

We will understand this further in the next chapter with respect to exponents.

Chapter 6 - Introduction to Exponents

Let us see the example above and try to use prime factorization method.

72 × 12 = 24 × 36

(2 × 2 × 2 × 3 × 3) × (2 × 2 × 3) = (2 × 2 × 2 × 3) × (2 × 2 × 3 × 3)

Don't you think it is going to get complicated?

Just the way multiplication is repeated addition, in the same way we use exponential form when a number is multiplied by itself.

(2 × 2 × 2 × 3 × 3) × (2 × 2 × 3) = (2 × 2 × 2 × 3) × (2 × 2 × 3 × 3)

(2³ × 3²) × (2² × 3) = (2³ × 3) × (2² × 3²)

(2³ × 2²) × (3² × 3) = (2³ × 2²) × (3² × 3)

2⁵ × 3³ = 2⁵ × 3³

Now let us understand further.

2³ × 2² = 2^3 × 2^2 = (2 × 2 × 2) × (2 × 2) = 2^(3 +2) = 2^5 = 2⁵

In 2⁵, the 5 here is known as an exponent or power.

This proves that when aᵐ is multiplied by aⁿ, we get aᵐ⁺ⁿ. This is known as the Product Rule.

1. 7⁵ × 7³ =  (7 × 7 × 7 × 7 × 7) × (7 × 7 × 7) = 7⁵⁺³ = 7⁸

2. 14⁴ × 14³ = (14 × 14 × 14 × 14) × (14 × 14 × 14) = 14⁴⁺³ = 14⁷

3. 21⁷ × 21² = (21 × 21 × 21 × 21 × 21 × 21 × 21) × (21 × 21) = 21⁷⁺² = 21⁹

In the same way, when aᵐ is divided by aⁿ, we get aᵐ⁻ⁿ. This is known as the Quotient Rule.

1. 7⁸ ÷ 7⁵ = (7 × 7 × 7 × 7 × 7 × 7 × 7 × 7) ÷ (7 × 7 × 7 × 7 × 7) = (7 × 7 × 7) = 7⁸⁻⁵ = 7³

2. 14⁷ ÷ 14³ = (14 × 14 × 14 × 14 × 14 × 14 × 14) ÷ (14 × 14 × 14) = (14 × 14 × 14 × 14) = 14⁷⁻³ = 14⁴

3. 21⁹ ÷ 21⁷ = (21 × 21 × 21 × 21 × 21 × 21 × 21 × 21 × 21) ÷ (21 × 21 × 21 × 21 × 21 × 21 × 21) = (21 × 21) = 21⁹⁻⁷ = 21²

The Power of a Power Rule is a quick shortcut used when you have an exponential expression that is being raised to another exponent.

Instead of writing everything out, the rule states that you simply multiply the exponents together while keeping the base the same.

This means (aᵐ)ⁿ = aᵐⁿ

1. 4³ = (2²)³ = 2⁶

2. 8³ = (2³)³ = 2⁹

3. 36² = (6²)² = 6⁴

Always remember : Any number to the power of 0 is 1 and any number to the power of 1 is the number itself.

a⁰ = 1

a¹ = a

Chapter 7 - Introduction to Roots

In mathematics, a root (specifically the nth root) is the inverse operation of raising a number to a power. 

A root asks the question: "What number, multiplied by itself a specific number of times, gives me the starting number?"

It is symbolized by the √ sign.

The most common types of roots are square roots and cube roots.

When a number is multiplied by itself twice, it is called a perfect square. When it is multiplied by itself thrice it is, called a perfect cube.

List of perfect square numbers and cube numbers upto 20

Let us see some sample representations.

1. ²√9 = (9)¹ᐟ² = (3²)¹ᐟ² = 3

2. ³√8 = (8)¹ᐟ³ = (2³)¹ᐟ³ = 2

4. ⁴√16 = (16)¹ᐟ⁴ = (2⁴)¹ᐟ⁴ = 2

Let us try some sample numericals.

1. The area of a square is 25 cm². Find the length of its side.

Area of a square = Side × Side

Side = ²√Area = (²√25) cm = 5 cm

2. The volume of a cube is 343 cm³. Find the length of its side.

Volume of a cube = Side × Side × Side

Side = ³√Volume = (³√343) cm = 7 cm

3. The area of a circle is exactly 154 cm². Find the circumference of the garden.

Area of a circle = π × (Radius)²

π = 22/7

Radius = (²√(154 ÷ 22/7)) cm = (²√49) cm = 7 cm

Circumference = 22/7 × Diameter = [22/7 × (2 × 7)] cm = 44 cm

Conclusion

Hereby we conclude this lecture. We have covered the important aspects of calculations - Factors, Multiples, Exponents and Roots.

Mathematics is not about memorizing complicated rules or spending huge amounts on coaching classes. It is about understanding the logic behind numbers and using simple tricks to make calculations easier.

We hope this blog has helped reduce your fear of mathematics and given you useful tools to solve problems confidently. Remember — practice is the key. Try solving a few problems from your textbook or daily life using the methods we discussed. You will surely see the difference!

Thanks and Regards,

The Aadyanagha Foundation.

Mathematical Tricks by The Aadyanagha Foundation

Glories to Duranteshwar Mahadev and Aadyanagha Mahadevi 🙏!

Well! Many of you all must have had a phobia for mathematics (or some of you still have it). Right?

This blog aims at removing that phobia for mathematics. We have launched some shortcuts to simplify mathematical problems so that you all do not have to spend extra money in coaching classes! That too free of cost!

This post specifically aims to focus on addition, subtraction, multiplication and division.

1. Left to Right Addition and Subtraction 

Let's count the circles below:

Yes. There are 5 circles here.

Now we bring similar number of circles. What are we doing?

It is not possible to count each circle every time. So we add.

Addition is the process of bringing many things together to find a total. It is symbolized by a “+” sign. 

For example, there are 4 cookies in a plate. 2 more cookies are added. Hence, there are 6 cookies now.

If we understand with the context of number line, we realize that addition means moving forward in a number line.

Subtraction is the process of taking away something from a given set of things. It is symbolized by a “-” sign.

In a subtraction problem (Z - A = K), the minuend is the first number (Z), representing the total amount you start with, while the subtrahend is the second number (A), representing the quantity to be taken away. The result is called the difference (K).

For example, Jerry had 10 sweets. He gave 4 sweets to his best friend named Mickey. As a result, Jerry now has 4 less sweets. Hence, Jerry is left with 6 sweets.

If we understand with the context of number line, we realise that subtraction means moving backwards in a number line.

Let's read a sentence. "God is great". Did we read it from left to right or right to left?

Yes! We read the sentence from left to right. Now do you think we can add and subtract in the same way? Why not!

Sample Sum for Left to Right Addition

Sample Sum for Left to Right Subtraction

Let's take a sum!

1. 67 + 89

Since 6 & 9 are in the tens place and 7 & 9 are in the ones place, let's understand the breakdown logically!

67 = 6 Tens + 7 Ones = (6 * 10) + (7 * 1) = 60 + 7

89 = 8 Tens + 9 Ones = (8 * 10) + (9 * 1) = 80 + 9

Since we have got a simple breakdown of the numbers, let's solve the sum!

67 + 89 = (60 + 7) + (80 + 9) = 60 + 7 + 80 + 9 = (60 + 80) + (7 + 9) = 140 + 16 = 156 

Wasn't this easier than the traditional method? The traditional method is prone to errors related to carry forward because we are more used to operating from left to right, but the traditional methods use the right to left approach!

Now let's try another sum.

2. 78 - 42

First we break down the numbers using expanded form.

78 = 7 Tens + 8 Ones = (7 * 10) + (8 * 1) = 70 + 8

42 = 4 Tens + 2 Ones = (4 * 10) + (2 *1) = 40 + 2

Now we solve the sum.

78 - 42 = (70 + 8) - (40 + 2) = 70 + 8 - 40 - 2 = (70 - 40) + (8 - 2) = 30 + 6 = 36

In the same way, this too was simpler than the traditional method!

Now let's do some sums using the combination of both methods.

3. 942 + 786 - 351

First we break down the numbers using expanded form.

942 = 900 + 40 + 2

786 = 700 + 80 + 6

351 = 300 + 50 + 1

Now we solve the sum.

942 + 786 - 351 = (900 + 40 + 2) + (700 + 80 + 6) - (300 + 50 + 1) 

= 900 + 40 + 2 + 700 + 80 + 6 - 300 - 50 - 1 

= (900 + 700 - 300) + (40 + 80 - 50) + (2 + 6 - 1) 

= 1300 + 70 + 7 = 1377

Just imagine! If we use these methods, the risk of human error would be minimized.

2. Zigzag Approach

First we try solving the sum below using Left to Right Approach.

Let us solve the same sum using Zigzag Approach.

Let us solve some problems.

1. 74 + 68 - 92

First we break down the numbers using expanded form

74 = 70 + 4

68 = 60 + 8

92 = 90 + 2

Now we solve the sum.

74 + 68 - 92 = (70 + 4) + (60 + 8) - (90 + 2)

= 70 + 4 + 60 + 8 - 90 - 2

= 74 + 60 + 8 - 90 - 2

= 134 + 8 - 90 - 2

= 142 - 90 - 2

= 52 - 2

= 50

2. 942 + 786 - 351

First we break down the numbers using expanded form.

942 = 900 + 40 + 2

786 = 700 + 80 + 6

351 = 300 + 50 + 1

Now we solve the sum.

942 + 786 - 351 = (900 + 40 + 2) + (700 + 80 + 6) - (300 + 50 + 1) 

= 900 + 40 + 2 + 700 + 80 + 6 - 300 - 50 - 1

= 940 + 2 + 700 + 80 + 6 - 300 - 50 - 1

= 942 + 700 + 80 + 6 - 300 - 50 - 1

= 1642 + 80 + 6 - 300 - 50 - 1 

= 1722 + 6 - 300 - 50 - 1 

= 1728 - 300 - 50 - 1 

= 1428 - 50 - 1 

= 1378 - 1 

= 1377

3. Step Addition

Do you think we will be able to use methods like Left to Right Addition or Zigzag Addition method for larger numbers? No. That will be a time taking process. Hence, we have discovered a new method. But before that, let us understand the logic behind this method. 

Understanding the logic behind step operators

Over here, the first thing that we did was add the hundreds, tens, and ones separately.  Since we will be using the Step Operator, we can see the irrelevancy of zero here. And thus, separated them into 10 hundred, 21 tens, and 16 ones.  Finally, we used the Step Operator to get the final answer. 

The idea here is that we have : 

10 hundreds + 21 tens + 16 ones 

= (10 hundreds) + (2 hundreds + 1 ten) + (1 ten + 6 ones) 

= 12 hundreds + 2 tens + 6 ones 

= 1226 

In this sum, we see too many zeroes over here having minimal impact on the final addition. 

Step operator added to remove zeroes

The Step Operator is just a visual wall used to keep different place-values separated so you don't have to write out endless zeros.

​Think of the | bar as a divider between the Left-Hand Side and the Right-Hand Side.

Now if we see carefully. We see 9 Thousands + 12 Hundreds + 06 Tens + 21 Ones. 

Let us further understand this via a simpler sum.

What is close to the step?

As mentioned before, we add the step operator to remove the extra zeroes. So, from here we can understand that 14 step 28 means 140 + 28. So, the 4 in the 14 is actually in the Tens place and the 2 in the 28 is also in the Tens place. Hence, we add the numbers closest to the step.

When you want to remove the bar and get your final answer, you add the two digits that are touching the bar together.

Let us try an example.

1. 24 + 35

2 + 3 └┐ 4 + 5

= 05 └┐ 09 

= 59

Note : The right side of a step operator should always hold a two-digit number. If a calculation gives you a single digit, you must put a 0 in front of it (e.g., 8 becomes 08). Adding two one-digit numbers or multiplying two one-digit numbers lead to a two-digit number at the maximum. (For example : 9 + 9 = 18 and 9 * 9 = 81 - as 9 is the highest one-digit number)

2. 135 + 420

1 + 4 └┐ 3 + 2 └┐ 5 + 0

= 05 └┐ 05 └┐ 05

= 55 └┐ 05

= 555

3. 7624 + 9835

7 + 9 └┐ 6 + 8 └┐ 2 + 3 └┐ 4 + 5 

= 16  └┐ 14 └┐ 05 └┐ 09

= 174 └┐ 05 └┐ 09

= 1745 └┐ 09

= 17459

Another question arises. Suppose you add many 2 digit numbers and you get something like this while adding the tens place and ones place : 24 └┐ 135 - Then how will you solve?

Let us translate the meaning of it. It will mean 24 Tens + 135 Ones = (2 Hundreds + 4 Tens) + (1 Hundred + 3 Tens + 5 Ones) = 375. This means we will add the 24 and 13, maintaining the 5 intact in the ones place.

When you have 24 └┐ 135, look at the digits that are smashed right up against the step operator. In this case, it is the 4 on the left and the 13 on the right. Treat the numbers directly next to the line as your target. You are adding the 24 and the 13 together. Put your the sum (37) next to the untouched digit (5) giving you a final answer of 375.

4. Multiplication using Expanded Form

First let us understand that multiplication is nothing but repeated addition. For example,

3 × 1 = 3

3 × 2 = 3 + 3 = 6

3 × 3 = 3 + 3 + 3 = 9

3 × 4 = 3 + 3 + 3 + 3 = 12

Multiplication was introduced so that the time spent on repeated addition is minimized.

Let us take some illustrations.

1. 24 × 7 

(20 + 4) × 7 = 140 + 28 = 168

2. 946 × 8

(900 + 40 + 6) × 8 = 7200 + 320 + 48 = 7568

3. 1234 × 5

(1000 + 200 + 30 + 4) × 5 = 5000 + 1000 + 150 + 20 = 6170

5. Step Multiplication

Let us try the same examples using step operator.

1. 24 × 7 

2 × 7 └┐ 4 × 7

= 14 └┐ 28

= 168

2. 946 × 8

9 × 8 └┐ 4 × 8 └┐ 6 × 8

= 72 └┐ 32 └┐ 48

= 752 └┐ 48

= 7568

3. 1234 × 5

1 × 5 └┐ 2 × 5 └┐ 3 × 5 └┐ 4 × 5

= 05 └┐ 10 └┐ 15 └┐ 20

= 60 └┐ 15 └┐ 20

= 615 └┐ 20

= 6170

6. Speed Division

Division is the process of grouping things equally.

Just the way multiplication is repeated addition, in the same way division is repeated subtraction.

Let us take a situation. There are six friends - Gauri, Shriya, Vani, Siya, Radhika and Meenakshi. They see a box having 42 dolls. They decide to distribute the dolls among each other equally.

42 - 6 = 36

36 - 6 = 30

30 - 6 = 24

24 - 6 = 18

18 - 6 = 12

12 - 6 = 6

6 - 6 = 0

Since one friend picks one doll each in the each attempt, 6 dolls are reduced. So the number of times each friend picks a doll is 7 times. So each friend has 7 dolls. Here 42 is the dividend, 6 is the divisor and 7 is the quotient.

One day the friends see 15 shells. They want to distribute the shells equally.

15 - 6 = 9

9 - 6 = 3

Each friend could collect only 2 shells. The remaining 3 were left on the beach. Here 3 is the remainder.

I. Tabular Division

Here we solve the following sums.

1. 13240 / 5

13240 is the dividend ; 5 is the divisor ; 2648 is the quotient ; 0 is the remainder

2. 968 / 4

968 is the dividend ; 4 is the divisor ; 242 is the quotient ; 0 is the remainder

3. 143622936 / 6

143622936 is the dividend ; 6 is the divisor ; 23937156 is the quotient ; 0 is the remainder

II. Divisibility Test for numbers

1. All even numbers are divisible by 2. Numbers ending with 0, 2, 4, 6 and 8.

2. Any number whose digits sum up to a number that is a multiple of 3, is divisible by 3. For example : the sum of the digits of numbers 12, 24 and 63 is 3, 6 and 9 respectively.

Any even number divisible by 3 is a number divisible by 6.

Any number whose digits sum up to 9 is divisible by 9.

Any even number divisible by 9 is divisible by 18.

3. Any number whose last two digits is divisible by 4 is a number divisible by 4. For example, 124 is 100 + 24. We already know 100 is a multiple of 4. We also know that multiplication is repeated addition. So if you add two multiples of 4, it is automatically divisible by 4. 

In the same way, any number whose last three digits are divisible by 8 is a number divisible by 8. For example, 1024 is 1000 + 24. We already know 1000 is a multiple of 8. So if you add two multiples of 8, it is automatically divisible by 8. 

Any number divisible by 3 and 4 is divisible by 12 ; Simultaneously any number divisible by 3 and 8 is divisible by 24.

4. Any number which ends with 0 or 5 is divisible by 5. Any number ending with 0 is a multiple of 10. In the same way any number which ends with 00, 25, 50 or 75 is divisible by 25.

5. To check whether a number is divisible by 7, we double the last digit, subtract it from the rest of the number. If the result is divisible by 7, the original number is too.

For example, in 343 : 34 - (3 × 2) = 28 which is divisible by 7. 

Let's also see for 14 : 1 - (4 × 2) = -7 which again is divisible by 7.

6. To check whether a number is divisible by 11, we alternatively add and subtract the digits of the number from left to right. If the end result is 0 or a multiple of 11, it is a multiple of 11.

For example, in 1364 = 1 - 3 + 6 - 4 = 0, which proves it is divisible by 11.

III. Division Shortcuts

I. Divisibility by 5 - Dividing by 5 can also be written as multiplying by 1/5. 2/10 is an equivalent fraction of 1/5. So while dividing a number by 5, we can multiply the number by 2 and then divide it by 10.

For example, 165 ÷ 5 = (165 × 2) ÷ 10 = 330 ÷ 10 = 33

II. Divisibility by 4 and 8 - Dividing by 4 is considered as multiplying by a quarter (1/4). A quarter is half of half (1/2).

For example, 28 ÷ 4 = (28 ÷ 2) ÷ 2 = 14 ÷ 2 = 7

In the same way, dividing by 8 is considered as dividing by half of a quarter.

For example, 72 ÷ 8 = (72 ÷ 4) ÷ 2 = ((72 ÷ 2) ÷ 2) ÷ 2 = (36 ÷ 2) ÷ 2 = 18 ÷ 2 = 9

III. Divisibility by 25 - 4/100 is an equivalent fraction of 1/25. So while dividing a number by 25, we can multiply the number by 4 and then divide it by 100.

For example, 625 ÷ 25 = (625 × 4) ÷ 100 = 2500 ÷ 100 = 25

7. Double Step Multiplication

Do you think we can use a step operator while multiplying 2 two-digit numbers? Let us try.

1. 42 × 35

42 × 3 └┐└┐ 42 × 5

= 4 × 3 └┐ 2 × 3 └┐└┐ 4 × 5 └┐ 2 × 5

= 12 └┐ 06 └┐└┐ 20 └┐ 10

= 126 └┐└┐ 210 (we will add the 26 with the 21)

= 1470

2. 98 × 67

98 × 6 └┐└┐ 98 × 7

= 9 × 6 └┐ 8 × 6 └┐└┐ 9 × 7 └┐ 8 × 7

= 54 └┐ 48 └┐└┐ 63 └┐ 56

= 588 └┐└┐ 686

= 6566

8. Optimized Double Step Multiplication

Let us try an alternative method.

1. 42 × 35

42 × 3 └┐└┐ 42 × 5

= 4 × 3 └┐ 2 × 3 └┐└┐ 4 × 5 └┐ 2 × 5

= 12 └┐ 06 └┐└┐ 20 └┐ 10

= 126 └┐└┐ 20 └┐ 10 (Adding 26 and 20) (double step)

= 146 └┐ 10 (Adding 6 and 1) (single step)

= 1470

2. 98 × 67

98 × 6 └┐└┐ 98 × 7

= 9 × 6 └┐ 8 × 6 └┐└┐ 9 × 7 └┐ 8 × 7

= 54 └┐ 48 └┐└┐ 63 └┐ 56

= 588 └┐└┐ 63 └┐ 56 (Adding 88 and 63) (double step)

= 651 └┐ 56 (adding 1 and 5) (single step)

= 6566

9. Cross Multiplication

I. 2 Digit × 2 Digit

Steps :

i. First we calculate the value in the hundreds place by multiplying the digits in the tens place. Tens × Tens = Hundreds.

Targeting the Hundreds Place

ii. Then we put a step operator and calculate the cross products of the digits in tens place with the digits in ones place and sum them up to find the value in the tens place. Tens × Ones = Tens.

Targeting the Tens place

iii. Then we again put a step operator and calculate the value in the ones place by multiplying the digits in the ones place. Ones × Ones = Ones.

Targeting the Ones Place

iv. Conducting a step operation and solving the sum

Performing Step Operation

1. 42 × 35

(4 × 3) └┐ (2 × 3) + (4 × 5) └┐ (2 × 5)

= 12 └┐ 06 + 20 └┐ 10

= 12 └┐ 26 └┐ 10

= 146 └┐ 10

= 1470

2. 98 × 67

(9 × 6) └┐ (8 × 6) + (9 × 7) └┐ (8 × 7)

= 54 └┐ 48 + 63 └┐ 56

= 54 └┐ 111 └┐ 56

= 651 └┐ 56

= 6566 

II. 3 Digit × 3 Digit

Steps :

i. First we calculate the value in the ten thousands place by multiplying the digits in the hundreds place. Hundreds × Hundreds = Ten Thousands.

Targeting the Ten Thousands Place

ii. Then we put a step operator and calculate the cross products of the digits in hundreds place with the digits in tens place and sum them up. Hundreds × Tens = Thousands.

Targeting the Thousands Place

iii. Then we put a step operator and calculate the cross products of the digits in hundreds place with the digits in ones place, the product of the digits in the tens place and sum them up. Hundreds × Ones = Hundreds ; Tens × Tens = Hundreds.

Targeting the Hundreds Place

iv. Then we put a step operator and calculate the cross products of the digits in tens place with the digits in ones place and sum them up to find the value in the tens place. Tens × Ones = Tens.

Targeting the Tens Place

v. Then we again put a step operator and calculate the value in the ones place by multiplying the digits in the ones place. Ones × Ones = Ones.

Targeting the Ones Place

vi. We us the step operator and solve the sum.

Performing step operation and ensuring that there is a two-digit number on both sides

1. 126 × 457

(1 × 4) └┐ (1 × 5) + (2 × 4) └┐ (1 × 7) + (2 × 5) + (6 × 4) └┐ (2 × 7) + (6 × 5) └┐ (6 × 7)

= 04 └┐ 05 + 08 └┐ 07 + 10 + 24 └┐ 14 + 30 └┐ 42

= 04 └┐ 13 └┐ 41 └┐ 44 └┐ 42

= 53 └┐ 41 └┐ 44 └┐ 42

= 571 └┐ 44 └┐ 42

= 5754 └┐ 42

= 57582

III. 2 Digit × 2 Digit × 2 Digit

Steps :

i. First we calculate the value in the thousands place by multiplying the digits in the tens place. Tens × Tens × Tens = Thousands.

Targeting the Thousands Place by calculating the product of the circled numbers

ii. Then we add a step operator and calculate the value in the hundreds place by multiplying the two tens digits with one ones digit and adding all such combinations Tens × Tens × Ones = Hundreds.

Targeting the Hundreds Place by calculating the product of the circled numbers in each box and adding all such combinations

iii. Then we add a step operator and calculate the value in the tens place by multiplying the one tens digits with two ones digit and adding all such combinations Tens × Ones × Ones = Tens.

Targeting the Tens Place by calculating the product of the circled numbers in each box and adding all such combinations

iv. Then we add a step operator and calculate the value in the ones place by multiplying the digits in the ones place. Ones × Ones × Ones = Ones.

Targeting the Ones Place by calculating the product of the circled numbers

v. We use the step operator and solve the sum.

1. 24 × 15 × 31

(2 × 1 × 3) └┐ (2 × 1 × 1) + (2 × 5 × 3)  + (4 × 1 × 3) └┐ (4 × 5 × 3) + (4 × 1 × 1)  + (2 × 5 × 1) └┐ (4 × 5 × 1)

= 06 └┐ 2 + 30 + 12 └┐ 60 + 4 + 10 └┐ 20

= 06 └┐ 44 └┐ 74 └┐ 20

= 104 └┐ 74 └┐ 20

= 1114 └┐ 20

= 11160

IV. Shortcuts for calculating squares and cubes using step operators

1. (az)² = a² └┐ 2az └┐ z²

Example : 64 × 64 = (6)² └┐ 2 (6 × 4) └┐ (4)² = 36 └┐ 48 └┐ 16 = 4096

2. (ijk)² = i² | 2ij | 2ik + j² | 2jk | k²

Example : 124 × 124 = (1)² └┐ 2 (1 × 2) └┐ 2 (1 × 4) + (2)² └┐ 2 (2 × 4) └┐ (4)² = 01 └┐ 04 └┐ 12 └┐ 16 └┐ 16 = 15376

3. (az)³ = a³ └┐ 3a²z └┐ 3az² └┐ z³

Example : 15 × 15 × 15 = (1)³ └┐ 3 ((1)² × 5) └┐ 3 (1 × (5)²) └┐ (5)³ = 01 └┐ 15 └┐ 75 └┐ 125 = 3375

10. Squaring a number ending with 5

Example :

1. 65 × 65

i. Multiply the portion of the number before 5 (i.e. 6) with its subsequent number (i.e. 7) : 6 × 7 = 42

ii. Append it with 25.

65 × 65 = 4225

Let's cross verify using cross multiplication :

(6 × 6) └┐ (6 × 5) + (6 × 5) └┐ (5 × 5)

= 36 └┐ 30 + 30 └┐ 25 

= 36 └┐ 60 └┐ 25 

= 420 └┐ 25

= 4225

2. 35 × 35

i. Multiply the portion of the number before 5 (i.e. 3) with its subsequent number (i.e. 4) : 3 × 4 = 12

ii. Append it with 25.

35 × 35 = 1225

3. 125 × 125

i. Multiply the portion of the number before 12 (i.e. 12) with its subsequent number (i.e. 13) : 12 × 13 = 01 └┐ 02 + 03 └┐ 06 = 156

ii. Append it with 25.

125 × 125 = 15625

11. Multiplication Shortcuts

I. Multiplying a number with 11

Example :

1. 42 × 11

i. Write the first digit = 4

ii. Then write the sum of the digits 4 and 2 = 6

iii. Then write the last digit = 2

42 × 11 = 462 

For verification : (4 × 1) └┐ (2 × 1) + (4 × 1) └┐ (2 × 1) = 04 └┐ 06 └┐ 02 = 462

2. 98 × 11

i. Write the first digit = 9

ii. Then write the sum of the digits 9 and 8 = 17 (since it is a two-digit number, 1 will be added to the 9)

iii. Then write the last digit = 9

98 × 11 = 1078

3. 76 × 11

i. Write the first digit = 7

ii. Then write the sum of the digits 7 and 6 = 13 (since it is a two-digit number, 1 will be added to the 7)

iii. Then write the last digit = 6

76 × 11 = 836

II. Multiplying a number with 99

Example :

1. 99 × 73

First : 73 - 1 = 72

Then : 100 - 73 = 27

99 × 73 = 7227

2. 99 × 64

First : 64 -1 = 63

Then : 100 - 64 = 36

99 × 64 = 6336

II. Multiplying a number with 999

Example :

1. 999 × 357

First : 357 - 1 = 356

Then : 1000 - 357 = 643

999 × 357 = 356643

2. 999 × 421

First : 421 - 1 = 420

Then : 1000 - 421 = 579

999 × 421 = 420579

For verification : (1000 - 1) × 421 = 421000 - 421 = 420579

Conclusion 

Finally I just want to say that it is very important that Indians are familiarized with these concepts at an early age. These techniques should also be implemented in schools so that parents don't lose their money in fraudulent EdTech companies charging exorbitant fees and not refunding them when the parents are dissatisfied with the services.

Thanks and Regards,

The Aadyanagha Foundation.