# Reaching the finish line by finding Fraction of digits Ending with 9 IMAGE VIA GYAN SANCHAR

This blog post marks the completion of my learning project. Finally, I have successfully learned and implemented Vedic Techniques  in solving problems on Arithmetic GPY VIA GIPHY

operations . It was indeed fun and my journey was mesmerizing. Vedic Mathematics is a great way to make Mathematics simpler and joyful for everyone and yes, it is Magic math.

In my first blog post ,we learned what is Vedic Mathematics and how it  can make Mathematics simpler and joyful. I started this journey of learning and knowledge sharing with a great interest in the area of Vedic mathematics. My goal was to implemented Vedic Technic’s in solving problems on Arithmetic operations. My second blog post concentrated on implementing Vedic tricks in multiplication, we witnessed how easily one can multiply complex numbers in a minute and that too without using a calculator. My third and fourth blog post focused on finding Square and Square root of any numbers quickly and am sure most of us start looking for calculators if we want to find square or square root of numbers more than three digits and now we can do it mentally and that too within few seconds.

In my Fifth and Sixth blog post we learned how cool Vedic math is when it comes to finding cube and cube root of any numbers. This awe-inspiring technique helps us to easily find cube and cube root of any given digits mentally. Further Vedic Techniques made my seventh blog post more interesting by simplifying the division  process, now we can do more than just painlessly split a check. In my Ninth blog post Vedic mathematics offered a neat solution to calculate percentage  in our mind to overcome our day to day challenges. IMAGE VIA EDVIA

In this blog post  we are going to learn how to convert any fraction  ending with 9 and denominator into its decimal equivalent. Vedic mathematics helps us to shorten the complex fraction process into simpler way ,In Sanskrit  this technic is called  as “Ekadhikena Purvena
To solve  fractions ending with 9 there is a Sutra or formula i.e.
BY ONE MORE THAN THE ONE BEFORE ,by using this formula let me solve two examples

Below image is a simple  illustration of the difference between usual method and Vedic method 1)Find the fraction of 1/39
STEPS:
i)   In 1st step, i will find the divisor ,here  the number 9 is the one before, previous             .    number of is 3 and the number 3  will be increase by 1, so we will get
.                                                       3+1=4
ii)   In 2nd step, on every example, i will start with zero point,  and  i will divide 1 by 4 i.e. 1/4 = 0.25 ,i  will  write 0.25 as 025
i.e                                              1/39=0.025
iii)   In 3rd step, i will divide 25/4, i will get remainder as 1 and quotient as 6
i.e                                             1/39=0.0256
.                                                                          1                                           (REMAINDER)
iv)  In 4th step, i will combine remainder and quotient, it will become 16, then i will             .      divide again 16 by 4 which gives me 4
i.e.                                            1/39=0.02564
(REMAINDER)
v)  In 5th step, i am going to divide 4 by 4 which gives me the value as 1
i.e.                                          1/39=0.02564 1
.                                                                         1                                           (REMAINDER)
vi) In 6th step, last digit is 1 ,if we divide 1 by 4, i will get as 0.25 again i will  write it          .      as 025,it became same as in 1st step 25 divide it by 4 gives me 1 as remainder and     as  quotient
i.e.                                           1/39=0.02564 10256
.                                                                                        1                         (REMAINDER)
vii) In next step again we are going to combine remainder and quotient i.e.16, and i will        divide it by 4 which gives me 4 and again 4 by 4 gives me 1,i will stop up-to last           .      digit  1, if i go on solving the digit i will get recurring values ,we can stop up-to 5         .      decimal  values.
i.e.                                           1/39=0.02564 1025641
.                                                                          1                1                       (REMAINDER)

2)     Let us solve one more example.
2/39
i)    In 1st step,i will find the divisor ,here  the number 9 is the one before, previous                   number 9 is 3  and the  number 3 we will be increase by 1,we will get.
.                                            3+1=4
ii)   In 2nd step,on every example, i will start with zero point  and  we will divide 2 by 4 i.e.  2/4 = 0.5, i will write 0.5 as 05.
i.e                                   2/39=0.05
iii)  In 3rd step, i will divide 5/4, we will get remainder as 1 and quotient as 1.
i.e                                2/39=0.051
(REMAINDER)
iv)  In 4th step, i will combine remainder and quotient, it will become 11, then i will             .      divide 11 by 4, which gives me remainder as 3 and quotient as.
i.e.                               2/39=0.0512
1 3                                                       (REMAINDER)
v)   In 5th step, if i combine remainder and quotient, it will become 32 .then i will divide  .      32 by 4 which gives me .
i.e.                               2/39=0.05128
1 3                                                        (REMAINDER)
vi)  In 6th step last digit is 8 ,i will  divide 8 by 4, i will get the value as 2.
i.e.                               2/39=0.051282
1 3                                                        (REMAINDER)
vii)
In 7th step last digit is 2, if i divide 2 by 4, i will get the values as  0.5, i will write 0.5   .       as  05  then if i divide 5 by 4 it gives me remainder 1 and quotient 1 .
i.e.                                2/39=0.051282051
1 3           1                                          (REMAINDER)
vii)
In next step again i am going to combine remainder and quotient i.e. 11 and i will             divide it by 4 which gives me 3 as remainder and 2 as quotient and again if i                  .       combine remainder and quotient and divide by   4 gives me 32/4=8,i will stop                    up- to last digit 8, if i go on solving the digit i will get recurring values .we can stop          up- to  5  decimal values.
i.e.                              2/39=0.05128 205128
1 3             1 3                                    (REMAINDER) GIF VIA GIPHY

# PERCENTAGE MADE EASY BY USING VEDIC MATHEMATICS IMAGE via Aksharshilp

Hi ECI831,

There are situations in our day to day life when we need to calculate percentage in our mind and these scenarios are quite common. When was the last time we talked to our friends, colleagues or students about percentage? probably in restaurants, parting things with friends or while discussing with students about their grades, paying rent, mortgage these real-life situations are common and what if we can do this quicker and in our mind easily. Vedic mathematics offers a neat solution to these problems.

In my previous blog ,we  learned division method when the divisor is 9 and also Nikhilam formula,today in this post we are going to learn how to find percentage of any number using vedic maths.

Below image is a simple  illustration of the difference between usual method and Vedic method To find the percentage we are going to apply the sutra or formula i.e. Vertical and cross wise multiplication .which we have learned in multiplication method .
By using above formula i am  going to solve some examples.
1)Lets us find the percentage of  36⸓  of   85
STEPS
i) Write the given values in two rows
6
5
ii)Now we will apply the technic of Vertical and cross wise multiplication , consider the 1st pair where cross multiplication is not possible ,we will  solve it by vertical multiplication.
3      6
↓                                 (3×8=24)
8      5
24
iii)Consider 2 pairs, here cross multiplication is possible so we will apply the same rule.
3              6
.       Χ
8              5                (1548=63)
6                             (CARRIED FORWARD)
(24
iv) Next consider the last pair there is only one pair left , so we will apply vertical multiplication   method.
3              6
↓                       (6×5=30)
8              5
6      3                (CARRIED FORWARD)
(24 3   0
6
5
6        3                (CARRIED FORWARD)
(24)   3   0
vi) Divide the Answer by 100 we will get the final answer
3060÷100=30.60
36⸓  of   85 = 30.60

2)Lets us find the percentage of  52⸓  of   640
STEPS
i) Write the given values in two rows
0       5      2
4      0
ii)Now we will  apply  the technic of Vertical and cross wise multiplication, consider the 1st pair,cross multiplication is not possible , we will solve it by vertical multiplication.
5      2
↓                                 (0×6=0)
4      0
0
iii)Consider 2 pairs,  here cross multiplication is possible so we will apply the same rule.
5         2
.     Χ                                           (0+30=30)
6         4         0
3                                          (CARRIED FORWARD)
0  0
iii)Consider 3 pairs, here cross multiplication is possible,Multiply extreme digits by cross multiplication and middle digit by vertical method.
5         2
Χ                                     (0+12+20=32)
6         4        0
3   3                                      (CARRIED FORWARD)
0   0   2
iv) Next consider the last 2 pairs, cross multiplication is possible so we will apply same rule.
0       5       2
.              Χ                                     (0+8=8)
6       4       0
3     3                           (CARRIED FORWARD)
0     0    2    8
v)  Next consider the last pair there is only one pair left , so we will apply vertical multiplication   method.
0           5          2
.                         ↓                         (0×2=0)
6          4           0
3         3                        (CARRIED FORWARD)
0        0    2   8  0
5        2
6        4        0
3        3                      (CARRIED FORWARD)
0        0     2  8   0
3        3     2  8   0      (ANSWER)

vi) Divide the Answer by 100 we will get the final answer
33280÷100=332.80
52⸓  of   640 = 332.80 GIF via GIPHY

# DIVISION MADE EASIER BY VEDIC MATHEMATICS Image via Nikhi

Hello Everyone,
Vedic math is an awesome technique for solving arithmetic problems in quicker and simpler way. The cool thing about this technic is that every math checks out. Whether you want to painlessly split a check, to amaze your friends, or to learn a different way to quickly divide numbers, this easy method can be learned within no time.
In this week blog post let us learn how to find division  of 3 or more digits using Vedic mathematics.

Below image is a simple  illustration of the difference between usual method and Vedic method Finding the division of 3 or more digit number when the divisor is 9.
1)Let us solve example on  3 digit number
Let us consider the number 102÷9

102÷9
STEP 1:
Splits the digit 102 into 2 parts as a divisor has only one digit as 2 goes in                                remainder part and rest all digits will goes in Quotient part i.e.
Q       R
1 0 ǀ  2

STEP 2:
To find the 1st digit of a quotient put 1 as it is i.e.
Q       R
1 0 ǀ  2

1
STEP 3:
i)To find the 2nd digit of a quotient ,add quotient and divident  digit i.e.                                         (1+0=1)
ii)To find the remainder ,add remainder and quotient i.e.(2+1=3) i.e.
Q        R
1 0 ǀ  2

1 1 ǀ  3
Quotient     = 11
Remainder =   3

2) Let us solve examples of 4 digit number
Let us consider the 1232÷9
1232÷9
STEP 1:
Split the digit 1232 into 2 parts as a divisor has only one digit as 2 goes in                                 remainder part and rest all digits will goes in Quotient part i.e.
Q          R
123 ǀ 2
STEP 2:

Q           R
123 ǀ  2

1
STEP 3:

i)To find the 2nd digit of a quotient add quotient and the divident i.e.(1+2=3) ,                      ii) To  find the  3rd digit of a quotient add 2nd digit of quotient with the divident                        digit i.e. (3+3=6)
iii)To find the remainder add 3rd digit of quotient with  remainder i.e. (6+2=8) i.e.
Q            R
1 2 3  ǀ     2

1 3  6  ǀ    8
Quotient = 136
Remainder=8

Division method when divisor is closer and smaller than power of 10.
By understanding the  Concept of Nikhilam formula i.e.all from 9 & last from 10,(i.e. Subtract all digits from 9 & last digit from 10)

Below image is a simple  illustration of the difference between usual method and Vedic method Lets us consider the value
234÷87
STEP 1:
Splits the digit 234 into 2 parts as a divisor has two  digit as 34 goes in                                 remainder part and rest all digits will goes in Quotient part i.e.
Q          R
2 ǀ 34
STEP 2:

Apply Nikhilam formula on 87 and we will get the compliment 13 (i.e. 9-8=1, 10-7=3)
Q           R
2 ǀ  34
13
2
STEP 3:

i)Multiply 2 with individual digit of 13  i.e.
Q      R
2   ǀ  3 4
13               ↓   ǀ  2 6
2  ǀ  6 0
Quotient    =   2
Remainder= 60

Lets us consider the value
113401÷997
STEP 1:
Splits the digit 113401 into 2 parts as a divisor has three digit so remainder will also have 3 digit ,as 401 goes in remainder part and rest all digits will goes in Quotient part i.e.
Q          R
113 ǀ 401
STEP 2:

Apply Nikhilam formula on 997 and we will get the compliment 003
(i.e. 9-9=0, 9-9=0,10-7=3)
Q           R
113 ǀ  401
003
1
STEP 3:

i)Multiply 1 with individual digit of 003  i.e.
Q      R
1 1 3  ǀ  401
003         ↓ 0 0 ǀ  3
1 1     ǀ
ii)Multiply 2nd quotient with individual digit of 003 ,and add the third column,to get 3rd quotient
Q      R
1 1 3  ǀ  401
003         ↓ 0 0  ǀ   3
0  ǀ  0 3
1 1 3  ǀ
iii)
Multiply 3rd quotient with individual digit of 003 ,and add the 4th,5th and 6th column,to get the  remainder.
(Note: Have to repeat the function till we get the number in last column)
Q      R
1 1 3  ǀ  4  0   1
003         ↓ 0 0  ǀ  3
ǀ  0  3
ǀ  0  0    9
1 1 3 ǀ   7  3  (10)  (1 is carried forward)
1 1 3 ǀ   7  4   0
Quotient    = 113
Remainder= 740

A quick update on answering questions from my previous blog post.
One of my fellow classmate Kara Tylor had a question on one of my blog post on how to find cube of two numbers using vedic technic. She asked me how to find cube of more than 2 digit using Vedic math, below I have solved few examples based on the method and my video will further aid the understanding the concept.

Finding cube of 3 digit number using vedic maths.
Lets us consider the number
(153)3

STEPS:
i)Split the given digits into two part
15     3
ii)consider the 1st part i.e. 15 from left to right go on reducing power i.e
153    152    15
iii) Next consider the 2nd part i.e. 3 from right to left go on reducing the power i.e.
153    152.3   15.32   33
iv) Find the values and write in 2nd Row
153          152.3         15.32           33
3375        675           135             27
v) Multiply middle terms by 3 i.e.
153         152.        15.32         33
3375     675           135             27
×3            ×3
3375       2025       405            27
vi)To get the final answer   we are going to keep unit digit as it is rest other digit are carried forward
153           152.3         15.32          33
3375        675           135             27
×3            ×3
3375       2025          405             27
(206)        (40)             (2)                              (CARRIED FORWARD)
3581         5                   7                 7
(153)3
= 35,815,77

I hope you like this post and found this technic interesting . GIF Via GIPHY

# FINDING CUBE ROOT USING VEDIC METHOD Hi Eci831,
In My Previous blog , we learned how to find cube of any number using Vedic maths. In this Post we are going to study how to find cube root . of any perfect cube easily by using Vedic maths. This is an remarkable trick and is always quite easy and appreciated by many This awe inspiring technique helps us to easily find cube root of 5 or 6  digits number mentally

Before going ahead on this method to find cube root, please make a note and memorize the below table.

To calculate cube root of any Perfect Cube easily, we need to Memorize the cubes, cube of unit digit and its relations from 1 to 10.

 Number Cube Cube of unit digit Relations 13 1 1 1 ⇒1 23 8 8 8⇒2 33 27 7 7⇒3 43 64 4 4⇒4 53 125 5 5⇒5 63 216 6 6⇒6 73 343 3 3⇒7 83 512 2 2⇒8 93 729 9 9⇒9 103 1000 0 0⇒0

Mathematics is very interesting and easy to learn when we use tricks to make it simple,now let’s see how easily we can solve cube root of perfects cube by using  Vedic method and also by viewing the table above.

Example 1: Find cube root of 19683
Check the last digit number & make a groups of three, three digits
from right side i.e. 19683  Can be written as
19   ,    683
STEP 1

1. Take the last group which is 683, in this case the last digit of 683 is 3
2. Now compare the last digit number i.e. 3 in the above table in the  relation we can see that if last digit of perfect cube is 3 then last digit of cube root is 737
3. So we replace 3 by 7 at the right most digit of the cube root
19   ,    683

STEP  2
1. Now check the remaining digit which is left i.e. 19
2. We have to check the cube which is less then or equal to the digit 19
3.  In this case the number 23=8 which is less than 19
4. We will replace 19 by 2

i.e.   19  ,   683
2          7
Hence the answer for cube root of 19683 is 27

Example 2: Find cube root of 551368
Check the last digit number & make a groups of three, three digits from                                      right side i.e. 19683  Can be written as
551 ,     368

STEP 1

1. Take the last group which is 368, in this case the last digit of 368 is 8
2. Now compare the last digit number i.e. 8 in the above table in the  relation we can see that if last digit of perfect cube is 8 then last digit of cube root is 2 i.e 82
3. So we replace 8 by 2 at the right most digit of the cube root
551   ,    368
2

STEP 2
1. Now check the remaining digit which is left i.e. 551
2. We have to check the cube which is less then or equal to the digit 551
3. In this case the number 83=512 which is less than 551
4. We will replace 551 by 8
i.e.  551  ,   368
8           2
Hence the answer for Cube root of 551,368 =82

Example 3: Find cube root of 148,877
Check the last digit number & make a groups of three, three digits from right                         side i.e. 148,877 Can be written as
148     ,     877
STEP 1

1. Take the last group which is 877, in this case the last digit of 877 is 7
2. Now compare the last digit number i.e. 7 in the above table in the  relation we can see that if last digit of perfect cube is 7 then last digit of cube root is 3 i.e 73
3. So we replace 7 by 3 at the right most digit of the cube root
148 ,    877
3

STEP 2

1. Now check the remaining digit which is left i.e. 148
2. We have to check the cube which is less then or equal to the digit 148
3. In this case the number 53=125 which is less than 148
4. We will replace 148 by 5
i.e       148   ,   877
5      3

Hence the answer for Cube root of 19683 =53
In the below video i have solved few more problems based on the vedic technics

I hope you liked the video and  found the Technic interesting and it simplified the process of finding cube root in a faster way, I appreciate your time for reading my blog post . GIF VIA GIPHY

# FINDING CUBE OF ANY NUMBER USING VEDIC MATH image via

Hello Everyone,
It was indeed fun and interesting journey in EC&I 831, I have learned a lot by reading my fellow classmates blog post, every week was so interesting and now we are almost halfway through the course. I have always been fascinated by mathematical studies and having a flair for the subject, there was never any doubt that I would choose  concepts in mathematics as my major learning project. I would call this as journey of learning, knowledge sharing. My project was focused on learning how to make mathematics simple, easy and fun for everyone and the tricks of Vedic mathematics GIF Via GIPHY

would be perfect to answer to it. In my first blog post we learned what is Vedic mathematics  and its sutra’s(Formula’s), the idea behind this was to introduce the concept of Vedic math to my classmates. In my second blog post I focused on learning how to implement this technic in various mathematical concepts, applying Vedic math in solving multiplication problems of complex numbers made multiplication so easy, we could solve 4-5 digit problems in less than a minute and that too without using calculator. My next post focused on learning how to find square of any number using Vedic trick, followed by another week blog post where we learned to find square root of any number using this trick.

I feel happy regarding my progress after reading the comments about my journey by my fellow classmates. Initially it was not so easy as I thought it would be, I had to understand the whole technic first and solve problems related to it. After searching a lot over internet I understood the best way to showcase this concept and learn with my classmates is by making a video of my own where I solve problems, which in-return help everyone understands the Vedic tricks and it applications. Going forward in the second half of our course we learn few more applications of Vedic technic in mathematics and by the end of our journey in EC&I 831 we will make Mathematics simpler and joyful using the tricks of Vedic Mathematics.

Now that we have learned how to find square root of any numbers using Vedic math , we will go ahead and use this technic to find cube of numbers. Finding cube of the number plays a vital role in Mathematics, so it is useful to find quick ways to find the cube of the number. We will learn the Vedic trick which will help us finding the cube of numbers in quick way. To find the cube of numbers we are going to learn four types Vedic Tricks, In the first we are going to find the cube of a number which starts with 1 i.e. 11,12,13……. In second type we are going to find the cube of a number which ends with 1 i.e.21,31,41……In the third type we are going to find the cube of a number having same digit i.e.22,33,44……and Finally in the fourth type we are going to find the cube of a number having different digit.

By knowing the values below will help us solve the problems easily.

 Number Square of a number Cube of a number 1 12 = 1 13 = 1 2 22 = 4 23 = 8 3 32 = 9 33 = 27 4 42 = 16 43 = 64 5 52= 25 53 = 125 6 62=36 63 = 216 7 72=49 73 = 343 8 82=64 83 = 512 9 92=81 93 = 729 10 102=100 103 = 1000

Type 1: Numbers starting with 1
(16)3
1)   We consider 1 as 1st term and 6 as 2nd term, we will write the given term as it is  i.e.                               6
2)  Square the 2nd term i.e. 62=36 and also cube the 2nd term i.e.=63=216 and write the         values in 1st row  i.e.
1     6      36       216
3)  In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
1      6      36       216
12     72                       (Doubled the value)
4)  Add them vertically in column. carry forward the 10th places digit to next column
1      6       36       216
+         12       72
3   12       21                       (Carried forward)
(16)3=4,096

Type 2: Numbers Ending with 1
(61)3
1)  We consider 6 as 1st term and 1 as 2nd term, we will write the given term as it is but           in reverse order i.e.        6     1
2)  Square the 1st term i.e. 62=36 and also cube the 1st term i.e.=63=216 and write the
values in 1st row i.e.
216    36       6     1
3)  In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
216     36       6      1
72     12                                (Doubled the value)
4)  Add them vertically in column. carry forward the 10th places digit to next column
216      36            1
+                72     12
10        1                                 (Carried forward)
(61)3=226,981

Type 3: Numbers having same digit
(55)3
1)  We consider 5 as 1st term and 5 as 2nd term, here both the digit is same so we take             any one digit .Cube of 5 =125 and write 4 times
i.e.              125   125   125   125
2)  In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
125    125   125   125
250    250                 (Doubled the value)
4)  Add them vertically in column. carry forward the 10th places digit to next column
125    125    125    125
+               250    250
41        38      12                (Carried forward)
(55)3=166,375

Type 4: Numbers having different digit
(32)3
1)  We consider 3 as 1st term and 2 as 2nd term, Cube the 1st term and 2nd term
i.e.       (3)3 =27     and         (2)3=8
27                                  8
2)  Square the 1st   term i.e. 32=9 then multiply by 2nd term i.e.9×2=18
27         18                   8
3)  Square the 2nd term i.e. 22 =4 then multiply by 1st term i.e.4×3=12
27          18      12       8
4)  In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
27          18       12      8
36       24               (Doubled the value)
5)  Add them vertically in column. carry forward the 10th places digit to next column
27         18       12      8
+                     36       24
5           3                             (Carried forward)
(32)3=32,768

I hope you found the technic interesting and it simplified the process of finding cube       of a number in a faster way, I appreciate your comments and suggestions. GIF Via GIPHY

# FINDING SQUARE ROOT OF A NUMBER USING VEDIC MATH Hello Eci831,

In my previous blog post we learned how to square any number using Vedic math, I hope you found the technic interesting and easy. In this post we will learn how to find a square root of a number. Finding the square root of a number is the inverse operation of squaring that number. Remember, the square of a number is that number times itself.

For Example: Square of n=n2, so square of 10= 10×10= 102= 100.

Finding square root by Vedic methods:

The square root of any number means getting a number multiplying by itself gives the given number in the regular method. To find the square root, the divisor continues always bigger at each step. This increases the calculation time as well as the complexity of the problem.  Now we shall see the quick trick of Vedic math to find the square root of perfect square, before going ahead to find the square root look at the image below Steps to find Square root of 3, 4, & 5 digits’ number.

1) Finding Square root of 3 digit

Consider a number

√729

1)Splits the number keeping last 2 digits aside i.e.

7 ǀ 29

2) Check the square digit of number 9 & its unit Square digit is 3& 7

3) Next check Square 7 it lies between 2 & 3 Square (22=4,32=9)

4)Choose the smaller number i.e.2

5)Now there is two possibilities for answer i.e. 23 or 27.

6) The smaller number is 2, consider the next number of 2 is 3 multiply those two   number 2×3=6 ,6 is smaller as we compare to given digit i.e. 7 (6˂ 7).

7)We have to consider the greater number i.e.27 so the final answer will be 27.

√729=27

2 )Finding Square root of 4-digit number

Consider the number

1024

1)Splits the number keeping last 2 digits aside i.e.

10 ǀ 24

2) Check the square digit of number 4 & its unit Square digit is 2& 8

3) Next check Square 10 it lies between 3 & 4 Square (32=9,42=16)

4) Choose the smaller number i.e.3

5) Now there is two possibilities for answer i.e. 32 or 38

6) The smaller number is 3, consider the next number of 3 is 4 multiply these twos’

number 3×4=12 ,12 is greater as we compare to given digit i.e.10 (12> 7)

7) We have to consider the smaller number i.e.32 so the final answer will be 32.

1024=32

3) Finding square of 5-digit number

76176

1) Splits the number keeping last 2 digits aside i.e.

7 61 ǀ  76

2) Check the square digit of number 6 & its unit Square digit is 4& 6

3) Next check Square 716 it lies between 27&28 Square (272=729,282=784)

4) Choose the smaller number i.e.27

5) Now there is two possibilities for answer i.e. 274 or 276.

6) The smaller number is 27, consider the next number of 27 is 28, multiply those

two number 27×28=756, is smaller as we compare to given digit i.e.761 (756˂ 761).

7) We have to consider the greater number i.e.276 so the final answer will be 276

76176=276

In the video below ,I have solved few more problems based on  Vedic method which will help in understanding the technic, , and article by TEXTBOOK will further aid the process .

# SQUARE ANY NUMBER USING VEDIC MATH image via: Exam hot spot

Hello Everyone,

I hope you are having fun with math using Vedic mathematics technique, in my previous post we learned multiplication tricks and witnessed that how easily we can solve complex multiplication problems using Vedic math. Knowledge to execute fast mental mathematical calculation will help us enormously regardless of which field of life our deal with. The significance of these mental math tricks is that it will give us positive edge over others irrespective of whether we are a student or a teacher.

In this blog post, we will learn how to find a square of any number using Vedic math technique.

Now we will see the how we can implement Vedic technic and find the square of numbers ending with unit place digit as 5 (up to 95) in specific method.

Type 1: Square of any number with 5

252

1)2 is in ten’s place 5 is in unit’s place

2) Splits the digit

3) Tens place is 2 the next number of 2 is 3 multiply it i.e.2×3=6

4) Square of 5 is 25

5) Step 3 & step 4 together gives the final answer i.e.625

Type 2: Square of any number with 25

4252

1) Split two digits, hundred’s place 4 & ten’s and unit’s place is 25

2) Square hundred’s place i.e.42 = 16

3) Square of 25 i.e.252 = 625

4) Then divide 4/2=2

6) Multiply 18 by 10 we will i.e. 18×10=180

7) Step 3 & step 6 together gives final answer i.e. 180,625

Type 3: Square of any number lies between (30-80)

532

1)5 is ten place (Left side) & 3 is in unit place (Right side)

2) Consider the base, here 50 is nearest which can be taken as a  Power of 10

3) Deviate the Number i.e. 53= (50+3) (Base + n)

4) Take the half of the base and add the digit of deviation ‘n’ i.e. 25+3=28

5) Square unit place (Right side) i.e. 32=9, When we take square of right side, if we get  1-digit value then we have to consider the value as 0 & the digit hence the value will be   09

6) Step 4 & Step 5 together gives the answer i.e. = 2,809

Type 4: Square of any number to Base 100

982

1) Consider the base, here 100 is nearest base which can be taken as a power of 10

2) Deviate the number i.e. 98= (1002) (Base-n)

3) Then 982=96

4) (n 2) i.e. 22=4, if we get 1-digit value then we have to consider the value as 0 & the digit then the value will be 04

5) Step 3 & step 4 together gives the answer i.e. 9604

Type 5: Square of any number near to Base n×100

896 2

1) Consider the base, here 900 is nearest base which can be taken as a power of 10

2) Find the deviation for the given digits i.e. 8962= (9004)2

3) We will split Base as (9×1004) (n×100-n 1)

4) Multiply the given digit with n i.e. 896×9=8064

5) n×(-n 1) =9× (-4) =-36

6) Step 4-Step 3 i.e.806436=8028

7) Square n 1 i.e.42=16

8) Step 6 & Step 7 together gives the answer i.e. 802816 image via :Momscribe

Some of us might find it difficult when we just see these steps, with just a little bit of practice we can easily master these tricks and perform these simple math tricks in blink of an eye. In the below video I have solved few problems based on the above stated steps and I hope after seeing the video you find this technic simple and easy.

Thanks for stopping by.