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!
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| Sample Sum for Left to Right Addition |
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| 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.
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| 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.
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| 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
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| 13240 is the dividend ; 5 is the divisor ; 2648 is the quotient ; 0 is the remainder |
2. 968 / 4
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| 968 is the dividend ; 4 is the divisor ; 242 is the quotient ; 0 is the remainder |
3. 143622936 / 6
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| 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
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.
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| 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.
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| 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.
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| Targeting the Ones Place |
iv. Conducting a step operation and solving the sum
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| 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.
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| 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.
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| 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.
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| 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.
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| 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.
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| Targeting the Ones Place |
vi. We us the step operator and solve the sum.
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| 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.
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| 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.
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| 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.
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| 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.
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| 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.
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