Finding Cube Root – Vedic Maths Way
This is an amazing trick which was always appreciated by the audience I have addressed in various workshops. This awe inspiring technique helps you find out the cube root of a 4 or 5 or 6 digits number mentally.
Before going further on the method to find the cube root, please make a note of the following points –
1) Cube of a 2-digit number will have at max 6 digits (99^3 = 970,299). That implies if you are given with a 6 digit number, its cube root will have 2 digits.
2) This trick works only for perfect cubes, it will not work for any arbitrary 6-digit
3) It works only for integers
Fast Multiplication Tricks
Simple Fast Multiplication Tricks & Techniques
Fast Multiplication by 5: Multiply by 10 (just place 0 after the original number) and divide the result by 2.
Fast Multiplication by 6: Sometimes subsequent multiplication by 3 and then 2 is easy.
Fast Multiplication by 9: Multiply by 10 (just place 0 after the original number) and subtract the original number.
Multiply by 12: Multiply by 10 and add twice the original number.
Multiply by 13: Multiply by 3 and add 10 times original number.
Multiply by 14: Multiply by 7 and then multiply by 2
Multiply by 15: Multiply by 10 and add 5 times the original number, as above.
Multiply by 16: You can double four times, if you want to. Or you can multiply by 8 and then by 2.
Multiply by 17: Multiply by 7 and add 10 times original number.
Multiply by 18: Multiply by 20 and subtract twice the original number (which is obvious from the first step).
Multiply by 19: Multiply by 20 and subtract the original number.
Multiply by 24: Multiply by 8 and then multiply by 3.
Multiply by 27: Multiply by 30 and subtract 3 times the original number (which is obvious from the first step).
Multiply by 45: Multiply by 50 and subtract 5 times the original number (which is obvious from the first step).
Multiply by 90: Multiply by 9 (as above) and put a zero on the right.
Multiply by 98: Multiply by 100 (just place 00 after the original number)and subtract twice the original number.
Multiply by 99: Multiply by 100 (just place 00 after the original number)and subtract the original number.
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Vedic Multiplication of two numbers close to Hundred
Vedic Method of Multiplication: Base System of multiplication
Application: Multiplication of two numbers close to Hundred
Case 1: Both numbers greater than 100.
Rule: You will get the answer in two parts
First part, to get left hand side of the answer: Add the difference between 100 and either of the numbers to the other number
Second part, to get right hand side of the answer: multiply the difference from 100 of both the numbers
Example
103 x 104 = 10712
The answer is in two parts: 107 and 12,
107 is just 103 + 4 (or 104 + 3), and 12 is just 3 x 4.
Similarly 107 x 106 = 11342
107 + 6 = 113 and 7 x 6 = 42
123 x 103 = 12669
(123 + 3) | (23 x 3) = 126 | 69 =12669 .
If the multiplication of the offsets is more than 100 then this method won’t work. For example 123 x 105. Here offsets are 23 and 5.
Multiplication of 23 and 5 is 115 which are more than 100. So this method won’t work.
But it can still work with a little modification. Consider the following examples:
Example 1
122 x 123 = 15006
Step 1: 22 x 23 = 506 (as done earlier)
Step 2: 122 + 23 (as done earlier)
Step 3: Add the 5 (digit at 100s place) of 506 to step 2
Answer: (122 + 23 + 5) | (22 x 23) = 150 | 06 = 10506
Example 2
123 x 105 (Different representation but same method)
123 + 5 = 128
23 x 5 = 115
128 | 115
= 12915
In the next post I'll tell you about vedic multiplication, i.e., how to multiply two numbers lesser than the base (in this case 100).
Here's the promised post for you - http://www.quickermaths.com/base-method-of-multiplication/
If you liked this method of vedic multiplication included in ancient Vedic Maths, Please leave a comment to let us know.
Shortcut to find the Cube of a number
Very often we have to find the cube, i.e. third power of 2 digit numbers. Cubes of very large numbers are rarely used.
Cubes of all the single digits should be memorized. Find below the table of cubes of first ten natural numbers -
13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125,
63 = 216, 73 = 343, 83 = 512, 93 = 729, 103 = 1000
To find the cube of any 2 digit number, we have to take the following steps
First Step: The first thing we have to do is to put down the cube of the tens-digit in a row of 4 figures. The other three numbers in the row of answer should be written in a geometrical ratio in the exact proportion which is there between the digits of the given number.
Second Step: The second step is to put down, under the second and third numbers, just two times of second and third number. Then add up the two rows.
Finding the cube of 12
Or, 123 = ?
First Step: Digit in ten’s place is 1, so we write the cube of 1. And also as the ratio between 1 and 2 is 1:2, the next digits will be double the previous one. So, the first row is
1 2 4 8
Step II: In the above row our 2nd and 3rd digits (from right) are 4 and 2 respectively. So, we write down 8 and 4 below 4 and 2 respectively. Then add up the two rows.
Ex 2: 163 = ?
Soln:
Explanations: 13 (from 16) = 1. So, 1 is our first digit in the first row. Digits of 16 are in the ratio 1:6, hence our other digits should be 1×6 = 6, 6×6 = 36, 36×6 = 216. In the second row, double the 2nd and 3rd number is written. In the third row, we have to write down only one digit below each column (except under the last column which may have more than one digit). So, after putting down the unit-digit, we carry over the rest to add up with the left-hand column. Here,
i) Write down 6 of 216 and carry over 21.
ii) 36 + 72 + 21 (carried) = 129, write down 9 and carry over 12.
iii) 6 + 12 + 12 (carried) = 30, write down 0 and carry over 3.
iv) 1 + 3 (carried) = 4, write down 4.
Vedic Multiplication by 9, 99, 999 and so on
When any number has to be multiplied by a series of 9s, like 9, 99, 999, 9999 and so on than we can apply this very simple vedic maths technique to increase your speed of calculation.
Multiplication with 9/ 99 / 999 and so on.
we know, 789 × 999 = 788,211
You will get the answers in two parts,
- The left hand side of the answer: subtract 1 from 789, which is 788
- The right hand side of the answer subtract 789 from 1000 = 1000-789= 211
Thus, 999 x 789 = 789-1 | 1000-789 = 788, 211 (answer)
{for the right hand side of the answer, 789 should be subtracted from (999+1)}
or, 99999 x 78 = 78-1 | 100000 - 78
= 7799922
{78 should be subtracted from (99999+1)}
Another example:
1203579 × 9999999 = 1203579-1 | 10000000- 1203579
=120357887964 21
Number in red is 1 less than 1203579. Number in blue is (10000000-1203579). Hence the answer.
This method has to be altered a little bit when number of 9s are lessers than the number of digit in the divisor.
1432 x 9 = 1432 (10 – 1) = 14320 – 1432 = 12888
So for multiplication with 9, put a zero after that number and subtract the number itself from that.
Likewise for 99 put two zeroes after that number .
3256 x 99 = 325600 – 3256 = 322344
How to Find the Average Speed?
Lot of students gets confused while finding out the average speed, when various distances are travelled with different speed. Say for example, trip to Agra from Delhi is made at an average speed of 40 km/hr and the trip back at an average speed of 60 km/hr. Find their average speed for the entire trip. (Hint: It's Not 50 Km/hr)
Rule: If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.
Or, If a person travels half the distance at a speed of x km/hr and the other half at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.
So answer to the above question:
(2*60*40)/(60+40) Â = 48 Km/hr
If a person travels three equal distances at a speed of x km/hr, y km/hr and z km/hr respectively, then the average speed during the whole journey is 3xyz/xy + yz + xz km/hr.
This problem can also be dealt with assuming some hypothetical distance.
You can find thousands of such amazing faster calculation tricks in the "MAGICAL BOOK on QUICKER MATHS - by M. Tyra"
Zeller’s Rule: Day on any date in the calendar
Zeller's Rule : With this technique named after its founder Zeller, you can solve any 'Dates and Calendars' problems.
Zellers rule can be used to find the day on any particular date in the calendar in the history. All you have to know is the formula given below and how to use it.
Zeller's Rule Formula:
F = K + [(13xM - 1)/5] + D + [D/4] + [C/4] - 2C
K = Date => for 25/3/2009, we take 25
In Zellers rule months start from march.
M = Month no. => Starts from March.
March = 1, April = 2, May = 3
Nov. = 9, Dec = 10, Jan = 11
Feb. = 12
D = Last two digits of the year => for 2009 = 09
C = The first two digits of century => for 2009 = 20
Example: 25/03/2009
F = 25 + [{(13 x1)- 1}/5] + 09 + 09/4 + 20/4 - (2 x 20)
= 25 + 12/5 + 09 + 09/4 + 20/4 - 2x20
=25+2+09+2+5-40 [ We will just consider the integral value and ignore the value after decimal]
= 43 - 40 =
Replace the number with the day using the information given below.
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
7 = Sunday
So it's Wednesday on 25th march, 2009.
If the number is more than 7, divide the no. by 7. The remainder will give you the day.
I shall be very grateful if anyone of you can provide me the java script for this formula, so that I can post it here for everyone's convenience.
Thanks in advance. I hope you will find the above method very useful.
you may find another post named, cyclicity very interesting.
Vineet Patawari - PGDM, ACA, B.COM (H)
Cyclicity
Cyclicity
In CAT and other MBA entrance examinations you get questions based on cyclicity every year. Here we are explaining in details the concept of cyclicity and how it should be used for solving problems.
To understand cyclicity let us take a simple example.
Take any two numbers say 43 and 97.
If they are multiplied, the answer is 4171. The last digit of the product is same as the last digit of 3 x 7.
Hence, it is 1.
This concept could be extended to a host of situations. An interesting pattern emerges when we look at the exponents of the numbers. We would find conclusions as given below.
The last digits of the exponents of all numbers have cyclicity i.e. every Nth power of the base shall have the same last digit, if N is the cyclicity of the number. All numbers ending with 2, 3, 7, 8 have a cyclicity of 4.
For instance,
2^1 ends with 2
2^2 ends with 4
2^3 ends with 8
2^4 ends with 6
2^5 end with 2 again.
The same set of the last digits shall be repeated for the subsequent powers. So, if we want to find the last digit of (say) 2^45, divide 45 by 4.
The remainder is 1
So the last digit would be the same as last digit of 2^1, which is 2
Let us take a CAT level example
(3) The digit in the unit place of the number represented by (795 – 358) is
A. 7
B. 0
C. 6
D. 4
Answer: D (4)
Cycle of 7 is
7 1=7
7 2=49
7 3= 343
7 4= 2401
If we divide 95 by 4, the remainder will be 3.
So the last digit of (7)95 is equals to the last digit of (7)3 i.e. 3.
Cycle of 7 is
31 =3
32 =9
33= 27
34= 81
35= 243
If we divide 58 by 4, the remainder will be 2.
The content in this post is a part of FireUp's Free E-Book on Number System. Click on Online CAT Preparation to download
Working out similarly for all other digits we get
CYCLICITY TABLE
| 1 | 1 |
| 2 | 4 |
| 3 | 4 |
| 4 | 2 |
| 5 | 1 |
| 6 | 1 |
| 7 | 4 |
| 8 | 4 |
| 9 | 2 |
| 10 | 1 |
Divisibility Rules including 7 and 13
Divisibility by 2: If its unit’s digit is any of 0,2,4,6,8.
Ex : 100 is divisible by 2 while 101 is not.
Divisibility by 3: If the sum of its digits is divisible by 3.
Ex: 309 is divisible by 3, since sum of its digits = (3+0+9) = 12 , which is divisible by 3.
Divisibility by 4: If the number formed by the last two digits is divisible by 4
Ex: 2648 is divisible by 4, since the number formed by the last two digits is 48 which is divisible by 4.
Divisibility by 5: If its units digit is either 0 or 5.
Ex: 20825 and 50545 are divisible by 5.
Divisibility by 6: If it is divisible by both 2 & 3.
Ex: 53256 is divisible by 6 because it is divisible by 2 as well as 3.
Divisibility by 7: If after subtraction of a number consisting of the last three digits from a number consisting of the rest of its digits the result is a number that can be divided by 7 evenly
Ex.: 414141 is divisible 7 as 414-141= 273 is divisible by 7
Many different ways to test divisibility by seven have been devised. Some are long and complex, a few involve rewriting the digits, and one even consists of a grid-like box. We have chosen one of the more simplistic versions even though in almost every case it is quicker to merely perform long division.
Divisibility by 8: If the last three digits of the number are divisible by 8.
Ex: 3652736 is divisible by 8 because last three digits (736) is divisible by 8.
Divisibility by 9: If the sum of its digit is divisible by 9.
Ex: 672381 is divisible by 9, since sum of digits = (6+7+2+3+8+1) = 27 is divisible by 9.
Divisibility by 10: If the digit at unit’s place is 0 it is divisible by10.
Ex: 69410, 10840 is divisible by 10.
Divisibility by 11: If the difference of the sum of its digits at odd places and sum of its digits at even places, is either 0 or a number divisible by 11.
Ex: 4832718 is divisible by 11, since:
(Sum of digits at odd places) – (sum of digits at even places)
= (8+7+3+4)-(1+2+8) = 11
Divisibility by 12: A number is divisible by 12 if it is divisible by both 4 and 3.
Ex: 34632
(i) The number formed by last two digits is 32, which is divisible by 4
(ii) Sum of digits = (3+4+6+2) = 18, which is divisible by 3.
Divisibility by 13:Â Remove the last digit of a number. Multiply it by 4 and add it to the remaining truncated number. Â Continue doing these steps until you reach a 2 digit number. If the result is divisible by 13, then so was the first number.
Example: 113945-->11394+20=11414-->1141+16=1157-->115+28=143 (since this number is divisible by 13, you can say 113945 is also divisible by 13)
You can go a step forward
14 + 12 = 26 is 2*13, so 113945 is divisible by 13.
Divisibility by 14: If a number is divisible by 2 as well as 7.
Divisibility by 15: If a number is divisible by both 3 & 5.
Divisibility by 16: If the number formed by the last 4 digits is divisible by 16.
Ex: 7957536 is divisible by 16, since the number formed by the last four digits is 7536, which is divisible by 16.
Divisibility by 24: If a number is divisible by both 3 & 8.
Divisibility by 40: If it is divisible by both 5 & 8.
Divisibility by 80: If a number is divisible by both 5 & 16.
Next in this series, based on your responses, we will share the divisibility rules of 17, 19, 23, 29, 31, 37, 41, 43, 47
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Who is behind QuickerMaths?
I am Vineet Patawari - PGDM (IIM Indore), ACA, B.Com(H). My passion for Mathematics, specially Vedic Maths encouraged me to start QuickerMaths
I believe that if trained properly using powerful tools like Vedic Maths, the immense intellect of human mind can be ignited instantly - find out more
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