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
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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Speed Multiplication by 111 : Vedic Maths
Multiplication of a number consisting of only ones with another number becomes very easy using Vedic Maths techniques. You must see the earlier post on shortcut for multiplying a number by 11
MULTIPLYING A NUMBER BY 111
To multiply a two-digit number by 111, add the two digits and if the sum is a single digit, write this digit TWO TIMES in between the original digits of the number. Some examples:
36x111= 3996
54x111= 5994
The same idea works if the sum of the two digits is not a single digit, but you should write down the last digit of the sum twice, but remember to carry if needed. So
57x111= 6327
because 5+7=12, but then you have to carry the one twice.
For 3 digit numbers
Carry if any of these sums is more than one digit.
Thus 123x111 = 1 | 3 (=1+2) | 6 (=1+2+3) | 5 (=2+3) | 3
Similarly,
241x111 = 26751
For an example where carrying is needed
Say, 352x111=3 | 8 (=3+5) | 10 (=3+5+2)| 7 (=5+2)| 2
= 3 | 8 | 10 | 7 | 2 = 3 | 9 | 0 | 7 | 2
= 39072
(Because of the carries, it may be easier to do the sums and write the answer down from right to left.)
Let me know, if you liked this Vedic Maths trick.
Vedic Multiplication by 11
Speed Vedic Multiplication Trick
Vedic Multiplication by 11
Step 1.
Assume that there are two invisible 0 (zeroes), one in front and one behind the number to be multiplied with 11
say if the number is 234, assume it to be 0 2 3 4 0
Step 2.
Start from the right, add the two adjacent digits and keep on moving left
02340
Add the last zero to the digit in the ones column (4), and write the answer below the ones column. Then add 4 with digit on the left i.e. 3. Next add 3 with 2. Next 2 with 0.
0+4 = 4
4+3 = 7
3+2 = 5
2+0 = 2
So answer is 2574
Similarly,
36 x 11 = 0+3 | 3+6 | 6+0 = 396
74 x 11 =0+ 7 | 7+4 | 4+0 = 7 | 11 | 4 = 814 (1 of 11 is carried over and added to next digit, so 7+1 = 8 )
6349 x 11 = (0+6) | (6+3) | (3+4) | (4+9) | 9+0 = 69839
This method works for all the number, no matter how long or short, times 11. Just try it yourself and get amazed at the simplicity of the concept.
In the next post will learn Vedic Multilplication by 111, 1111, 11111, and so on.
Vedic Maths Workshop
This is to announce that Quickermaths.com is conducting a workshop on VEDIC MATHEMATICS on this Sunday and Monday, 19th and 20th July, 2009.
The workshop will be on VEDIC MATHEMATICS [quicker calculation techniques that would help you to increase your calculation speed tremedously sharpen your intellect. The seminar will by me.
To know about me - click here.
Seminar Details
Days & Dates - Sunday and Monday, 19th and 20th July, 2009
Time 10.30 am to 1.30 pm on both days
Venue BG-109, Sector - II, Salt Lake, Kolkata -700091
Fees: INR. 750/ Student (inclusive of charges for study material)
You can pay by Cash, Cheque or Demand Draft, payable at Kolkata (in favour of "Preksha Consultancy Private Limited")
The fee needs to be deposited on or before 17th July, 2009 (Friday) at BG-109, Sector - II, Salt Lake, Kolkata -700091.
The feedback from students from eminent colleges, specially students preparing for competitive examinations like CAT, GMAT, GRE, MAT, etc. is very encouraging and I am happy to see such acceptance of Vedic Maths: a great gift from the sages of India.
For any clarification or further information, please feel free to mail me at vineetpatawari [at] gmail [dot] com.
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+91-98302-64971
Vedic Maths Subtraction
Learn Amazingly Fast Vedic Mathematics Subtraction
Very often we have to deduct a number from numbers like 1000, 10000, 100000 and so on.
This Vedic Maths Subtraction method found as sutra in ancient vedas, is given below is very useful for such subtractions.
Memory Trick: ALL FROM 9 AND THE LAST FROM 10
Use the formula all from 9 and the last from 10, to perform instant subtractions.
For example 1000 - 357 = ? (subtraction from 1000)
We simply take each figure in 357 from 9 and the last figure from 10.
Step 1. 9-3 = 6
Step 2. 9-5 = 4
Step 3. 10-7 = 3
So the answer is 1000 - 357 = 643
And that's all there is to it!
This always works for subtractions from numbers consisting of a 1 followed by noughts: 100; 1000; 10,000 etc.
Similarly 10,000 - 1049 = 8951 (subtraction from 10000)
9-1 = 8
9-0 = 9
9-4 = 5
10-9 = 1
So answer is 8951,
For 1000 - 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083.
So 1000 - 83 becomes 1000 - 083 = 917
Corollary: If last term is 0, keep that last term as 0 and subtract the last non Zero term from 10 .
Illustration: 10000 - 920 = 10000 - 0920 = (9-0) (9-9) (10-2) 0 =9080
Illustration: 100000 - 78010 = (9-7) (9 - 8 ) (9- 0) (10 - 1) 0 = 21990
If you like this vedic maths subtraction, please leave a comment.
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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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