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 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.
Multiply 2 numbers, sum of whose unit places is 10
Vedic Multiplication: This method of multiplication which is from Vedic Maths will make it very easy to multiply two numbers when sum of the last digits is 10 and previous parts are the same
You will get the answer in two parts.
First part, to get left hand side of the answer: multiply the left most digit(s) by its successor
Second part, to get right hand side of the answer: multiply the right most digits of both the numbers.
Example
First part: 4 x (4+1)
Second part: (4 x 6)
Combined effect: (4 x 5) | (4 x 6) = 2024
*| is just a separator. Left hand side denotes tens place, right hand side denotes units place
More Examples
37 x 33 = (3 x (3+1)) | (7 x 3) = (3 x 4) | (7 x 3) = 1221
11 x 19 = (1 x (1+1)) | (1 x 9) = (1 x 2) | (1 x 9) = 209
As you can see this method is corollary of "Squaring number ending in 5"
It can also be extended to three digit numbers like :
E.g. 1: 292 x 208.
Here 92 + 08 = 100, L.H.S portion is same i.e. 2
292 x 208 = (2 x 3) x 10 | 92 x 8 (Note: if 3 digit numbers are multiplied, L.H.S has to be multiplied by 10)
60 | 736 (for 100 raise the L.H.S. product by 0) = 60736.
E.g. 2: 848 X 852
Here 48 + 52 = 100,
L.H.S portion is 8 and its next number is 9.
848 x 852 = 8 x 9 x 10 | 48 x 52 (Note: For 48 x 52, use methods shown above)
720 | 2496
= 722496.
[L.H.S product is to be multiplied by 10 and 2 to be carried over because the base is 100].
Eg. 3: 693 x 607
693 x 607 = 6 x 7 x 10 | 93 x 7 = 420 / 651 = 420651.
Note: This Vedic Maths method can also be used to multiply any two different numbers, but it requires several more steps and is sometimes no faster than any other method. Thus try to use it where it is most effective
How do you like this Vedic Maths technique, please let us know. You can also share this with your friends.
Squaring number ending in five : Vedic Maths Trick
This is the most common, yet very interesting trick of Vedic Maths. Using this technique you can find the square of any number ending in 5 very easily. Given below is the step by step explanation of this Vedic Maths Method
Let us take a 2 digit number,
Say the number is a5 (=10a+5), where is the digit in ten's place
Square of a5= a x (a+1) | 25
For example,
452 = (40 + 5) 2, It is of the form (10a+b) 2 for a = 4
Giving the answer a (a+1) | 25 ( | stands for concatenation}
i.e. 4 x (4+1) | 25 = 4 x 5 | 25 = 2025
Similarly we can proceed for 3 digit numbers ending in 5
Few more examples:
952= 9 x 10 | 25 = 9025
1252 = 12 x 13 | 25 = 15625
5052 = 50 x 51 | 25 = 255025
Test yourself
Find out the square of 85, 245, 145, 35, 15, and 95?
Answer: 7225, 60025, 21025, 1225, 225, 9025
Please let us know if you like this Vedic Maths trick
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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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