Checking of Calculations: Casting Out Nines
While doing arithmetic calculations, we should normally check our calculation. But the checking should not be as tedious as the original problem. To solve this problem I am explaining below a very frequently used method which is discussed in Vedic Mathematics as well as by many other mathematicians.
Vedic Sutra: Vedic Mathematics Technique
Beejank: The Sum of the digits of a number is called Beejank. If the addition is a two digit number, then these two digits are also to be added up to get a single digit.
Find the Beejank of 632174.
As above we have to follow
632174 --> 6 + 3 + 2 + 1 + 7 + 4 --> 23 --> 2 + 3 --> 5
But a quick look gives 6 & 3 ; 2 & 7 are to be ignored because 6+3=9,2+7=9.
Hence remaining 1 + 4 --> 5 is the beejank of 632174.
Checking of Addition
Thumb Rule: Whatever we do to the number, we also do to their digit sum: then the result we get from the digit sum of the number must be equal to the digit sum of the answer.
For example: The number: 12+45+96+75+25 =253
The digit sum = 3+9+6+3+7 =28=10=1
Answer’s digit sum: 2+5+3 =10=1 (verified)
Another example: 3.5+23.4+17.5 = 44.4
The digit sum: 8+9+13=8+9+4=21=3
Answer’s digit sum: 12=3 (verified)
Casting Out Nines
This method is also known as "casting-out-nines". The method involves converting each number into its "casting-out-nines" equivalent, and then redoing the arithmetic. The casting-out-nines answer should equal the casting-out-nines version of the original answer. Below are examples for using casting out nines to check addition.
We get the casting-out-nines equivalent of a number by adding up its digits, and then adding up those digits, until you get a one digit number. If our answer is 9, then that becomes 0. As a short cut, we don't have to add in any of the 9's in our work, as these are the equivalent of 0. We can just "cast out" those 9's. For example, 19 becomes 1, without even adding 1 and 9 and getting 10, and then adding 1 and 0 and getting 1. As a further short cut, we can group numbers together which add up to 9, and replace them with 0. 2974 becomes 4, because we can cast out the 9 and the 2+7 (which is also 9 or 0). Well, let's try an arithmetic problem:
137892 3
+ 92743 + 7
------ --
230635 1
3+7=10, casting out 9 we get 1.
This rule is also applicable to subtraction, multiplication and up to some extent to division also
In the next post I will explain the use of this method for all of them.
Fast Multiplication by 5
This fast calculation trick or vedic maths trick will teach you how to multiply any number by 5. The concept can be divided in two parts as shown-
MULTIPLYING 5 TIMES AN EVEN NUMBER
Memory Trick: Halve the number you are multiplying by and place a zero after the number.
Example:
i. 5 × 136, half of 136 is 68, add a zero for an answer of 680.
ii. 5 × 874, half of 874 is 437; add a zero for an answer of 4370.
MULTIPLYING 5 TIMES AN ODD NUMBER: subtract one from the number
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
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)
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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