Divisibility Rule of 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
You might have seen divisibility rules of various numbers. But most of them very conveniently skip the ones which are very difficult and a divisibility rule for which is very much required. Today I’ll show you the divisibility rule for 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.
|
Number |
Method |
Example |
|
7 |
Subtract 2 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 7, the original number is also divisible by 7 | Check for 945: : 94-(2*5)=84. Since 84 is divisible by 7, the original no. 945 is also divisible |
|
13 |
Add 4 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 13, the original number is also divisible by 13 | Check for 3146:: 314+ (4*6) = 338:: 33+(4*8) = 65. Since 65 is divisible by 7, the original no. 3146 is also divisible |
|
17 |
Subtract 5 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 17, the original number is also divisible by 17 | Check for 2278:: 227-(5*8)=187. Since 187 is divisible by 17, the original number 2278 is also divisible. |
|
19 |
Add 2 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 19, the original number is also divisible by 19 | Check for 11343:: 1134+(2*3)= 1140. (Ignore the 0):: 11+(2*4) = 19. Since 19 is divisible by 19, original no. 11343 is also divisible |
|
23 |
Add 7 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 23, the original number is also divisible by 23 | Check for 53935:: 5393+(7*5) = 5428 :: 542+(7*8)= 598:: 59+ (7*8)=115, which is 5 times 23. Hence 53935 is divisible by 23 |
|
29 |
Add 3 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 29, the original number is also divisible by 29 | Check for 12528:: 1252+(3*8)= 1276 :: 127+(3*6)= 145:: 14+ (3*5)=29, which is divisible by 29. So 12528 is divisible by 23 |
|
31 |
Subtract 3 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 31, the original number is also divisible by 31 | Check for 49507:: 4950-(3*7)=4929. Since 492-(3*9) is divisible by 465:: 46-(3*5)=31. Hence 49507 is divisible by 31 |
|
37 |
Subtract 11 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 37, the original number is also divisible by 37 | Check for 11026:: 1102 - (11*6) =1036. Since 103 - (11*6) =37 is divisible by 37. Hence 11026 is divisible by 31 |
|
41 |
Subtract 4 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 41, the original number is also divisible by 41 | Check for 14145:: 1414 - (4*5) =1394. Since 139 - (4*4) =123 is divisible by 41. Hence 14145 is divisible by 41 |
|
43 |
Add 13 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 43, the original number is also divisible by 43.
*This process becomes difficult for most of the people because of multiplication with 13. |
Check for 11739:: 1173+(13*9)= 1290:: 129 is divisible by 43. 0 is ignored. So 11739 is divisible by 43 |
|
47 |
Subtract 14 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 47, the original number is also divisible by 47. This too is difficult to operate for people who are not comfortable with table of 14. | Check for 45026:: 4502 - (14*6) =4418. Since 441 - (14*8) =329, which is 7 times 47. Hence 45026 is divisible by 47 |
Notes: In applying the above methods, stop repeating the step of adding or subtracting (as the case may be) from the remaining truncated number as soon as you realize that the truncated number is divisible by the given number.
Popularity: 14%
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
Popularity: 23%
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"
Popularity: 9%
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)
Popularity: 17%
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 |
Popularity: 14%
Speed Multiplication by 111 : Vedic Maths
Multiplication of a number consisting of only ones with another number becomes very easy using Vedic Maths techniques.
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.)
If you like this Vedic Maths Trick, please let us know.
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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.
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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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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
44 x 46 = (4 x (4+1)) | (4 x 6) = (4 x 5) | (4 x 6) = 2024
*| is just a separator. Left hand side denotes tens place, right hand side denotes units place
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"
Also it can be extended such as
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 then multiplying by hand. 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.
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