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. This post includes the divisibility rule for some such numbers like 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

While reading this you have to be little patient. Read this carefully and try to apply it practically. If you master divisibility rules or tests explained below, I am sure these will come very handy in various examinations including competitive ones.

### Divisibility Rules with Examples

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 13, 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 29 |

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 :: 492-(3*9) :: 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 37 |

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.

While carrying the above process if during addition/subtraction any time the last digit is 0, that has to be ignored.

Also check out the **divisibility rule** of some other numbers.

Please give some trick about divisibility by 7

For all this everyone must know How To find a ‘ key digit ‘

For divisibility of 7 :

Key digit is

Write multiple of 7 till then it reaches to (multiple of 10 -1) or (multiple of 10+1)

Here starts

7

14

21

Stops here because 21 is ( 2*10+1)

Obtain 2 and from + sign obtain other – sign.

Hence key digit is here -2

Key digit also called osculator.

With this -2 means

Take any number like 24562 check this take last number 2 multiply it with osculator -2 means 2*-2= -4

Now eliminate last 2, means 2456-4=2452

Again take last 2, multiply it with -2= -4

Removing last 2,

245-4=241

Again last 1*-2=-2

Now 24-2=22

22 is not divisible by 7,

Hence 24562 is not divisible by 7.

You are great mathematician

THanx!!!!!

but what about 87661012 divisibility by 17 ?

It will show yes but it isn’t.

the most simplest way to find out the divisibilty of a number by 3 is:

Add all the digits of that number and if the sum is divisible by 3 then the number is olso divisible by 3.

example : if we take number 43692..

then add all the digits.

(4 + 3 + 6 + 9 + 2) = 24

24 is divisible by 3 so the number 43692 is also divisible by 3!

thank you very much . it is awwwwwwsome.

thanx,u all r great mathematicians,heads off…………………………………….

hey boddies ur are great mathematicians..

.

….

…

.

.i want to be like u..

Previously, I did some research in school with regards to this rule and found out that, the divisibility rule for prime numbers, or perhaps even all of the numbers that DO NOT have the factors 2 or 5, can be expressed as such:

1. To make things simple, let’s call the dividend “w” and the divisor “x”.

2. Find the first multiple of x that ends with a “1”. For example, the corresponding number for 3 and 7 would be 21, for 17 would be 51, etc.

3. Remove the last digit “1” from that multiple. Call the resulting number y.

4. Remove the last digit of w, multiply the digit by y, then subtract the total away from what is left of w. Call this difference w’.

5. Check if w’ is divisible by x. If it is, then w is also divisible by x, and vice versa. 6. If any of the differences turn out to be negative, just remove the negative sign.

7. If you are not sure whether w’ is divisible by x, then repeat steps 4 and 5.

Of course, it will be easier for dividends with “9” as the last digit of their divisor to use directly the y value that is 1/10th of (x+1) and ADD the product to the “cut-off” number as mentioned in the post, since the y value used will be the smallest, but that’s a different approach.

Example 1: Is 4653 divisible by 3?

3. The first multiple of 3 that ends with a “1” is 21. So y=2.

4. Multiply 3 (from 4653) by 2 and subtract it from 465: 465-(2×3) =459.

Multiply 9 (from 459) by 2 and subtract it from 45: 45-(2×9) = 27.

5. 27 is divisible by 3. So is 4653.

Example 2: Is 1756 divisible by 11?

3. The first multiple of 11 that ends with a “1” is 11. So y=1.

4. Multiply 6 by 1 and subtract it from 175: 175-6 =169.

Multiply 9 by 1 and subtract it from 16: 16-9 = 7.

5. 7 is not divisible by 11, so 1756 is also not divisible by 11.

Example 3: Is 1612 divisible by 13 or 91?

3. The first multiple of 13 (or 91) that ends with a “1” is 91. So y=9.

4. Multiply 2 by 9 and subtract it from 161: 161-18 =143.

Multiply 3 by 9 and subtract it from 14: 14-27 = -13.

6. Remove the negative sign: 13.

5. 13 is divisible by 13, but not by 91. Hence 1612 is divisible by 13, but not 91.

Example 4: Is 2923 divisible by 79?

3. The first multiple of 79 that ends with a “1” is 711. So y=71.

4. Multiply 3 by 71 and subtract it from 292: 292-213 =79.

5. 79 is divisible by 79. So is 2923.

Now the proof.

Using the same definitions for w, x and y above, the proof is as follows:

4. Express the dividend as the “sum” of individual digits, as in Am(10^m) + Am-5[10^(m-1)] + Am-2[10^(m-2)] +…+ A2(10^2) + A1(10^1) + A0 (where the j in Aj is in subscript).

5. Hence, the first digit is Am(10^m), the second is Am-1[10^(m-1)] and so on.

6. We shall use the example of dividing by 7 here. Hence, x=7 and y=2.

7. When we “take away the last digit” from the dividend, the resulting number becomes Am[10^(m-1)] + Am-1[10^(m-2)] +…+ A2(10) + A1 – 2A0. Let’s assume that this number is a multiple of 7, and we shall call it 7R.

8. Then we multiply the number above by 10. We get: Am(10^m) + Am-1[10^(m-1)] +…+ A2(100) + A1(10) – 20*A0 = 70R.

9. When we add 21*A0 to 70R, it becomes the original number w: Am(10^m) +…+ A1(10) + A0(-20+20+1) = Am(10^m) +…+ A1(10) + A0.

10. Because 70R + 21*A0 = w, it turns out that w = 7 (10R+3*A0), which renders w to be divisible by 7.

11. Conversely, if the number in step 7 above is not a multiple of 7, we can call it 7R+p, where p is a variable integer between 1 and 6.

12. When we multiply 7R+p by 10, we get 70R+10p. 10p cannot be a multiple of 7 as well, for the multiplication by 10 does not give it a factor of 7.

13. The sum thus becomes w = 70R + 10p + 21*A0 = 7 (10R+3*A0) + 10p. Hence, w is not divisible by 7.

The proof can be similarly extended to other numbers:

14. From step 7 above, the “cut-away” number becomes [10^(m-1)] + Am-1[10^(m-2)] +…+ A2(10) + A1 – yA0. Let’s assume that this number is a multiple of x, and we shall call it Rx.

15. Then we multiply the number above by 10. We get: Am(10^m) + Am-[10^(m-1)] +…+ A2(100) + A1(10) – 10y*A0 = 10Rx.

16. When we add (10y+1)*A0 to 10Rx, it becomes the original number w.

17. Now w = 10Rx + (10y+1)*A0.

18. Let w = Qx. Then Q = 10R + [(10y+1)*A0/x]. Q is only an integer (that is, when w is divisible by x) when 10y+1 is divisible by x.

This proves that the relation between x and y in this subtraction function only works when 10y+1 is divisible by x, and that 10y+1 ends in the digit “1”. The condition is that x cannot be any multiple of 2 or 5, since those multiples will not end in “1”; the other numbers are all applicable.

Using this method, we can use the y value of 30 for an x value of 43 (for 43*7 = 301) to do subtraction, and it will be faster than to add a multiple of 13. Using the examples given:

4. Multiply 9 by 30 and subtract it from 1173: 1173-270=903.

Multiply 3 by 30 and subtract it from 90: 90-90=0.

5. 0 is (supposedly) divisible by 43. So 11739 is also divisible by 43.

tell me why divisibility test of 7 last digit multiply 2 then substract…

Using the multiplicative inverse of 10 mod 23.Write Matlab code to check whether is divisible by 23.

How to do that??Can you help me please!! 🙂

what is an examle of a number divisible by 13?

377910

hello sir if you submit other shortcut like mixture time and work then it will be good for our sake

Your all tricks are very good. Thanks for the help.

Your this page has one mistake.

In the trick of divisibility by 29, you have solve correctly but in example, by mistake you have write divisibility by 23.

Kindly correct it very soon…

Corrected! Sorry for the mistake.

Also look at the interesting comments by Nandeesh and Deep!

We r evaluating a method which will help us to reduce calculation.

I appreciate the methods suggested above but they are just increasing calculation time which is contradictory to our ultimate purpose.

Comments are personal.

same mistake in 37 also.

thank u DEEP

U R GENIOUS!!!

Sir, I it is very tough to remember all the numbers that are to be substacted or added in a same process to find wheather a prime number is divisible to the number concerned or not. Thinking of those number I found a easier process to form these kind of rules. & it really works.

Since the last digits of all prime numbers should have to be 1, 3, 7, 9 (except 2 & 5), the first step is to multiply the numbers in such a way that the last digit becomes 1.,i.e, for number ending with 1 is to be multiplied with 1, 3 with 7, 7 with 3 & 9 with 9. And what ever be the remaing digit, that will be substracted from 0 or the prime no. itself. The answer will be the number that is to be add in the above process.

Ex. For 7

7*3=21. Now, 2+k=0, or, k=-2.

Else, 2+k=7, or,k=5.

Therefore, we have to multiply the last digit with -2 or 5 and add it with the next digit & keep this process continue untill the we get our answer.

In case of 161 we do it with -16 or 145.

Actually these are a bit time consuming but funny also.

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Since there is apparently a separate divisibility rule for each prime number, I do not think it is practical to remember all of them. Instead, learning “Long division reduced to one line shortcut” will be more useful and certainly more general.

An example for Long Division reduced to one-line shortcut (divide 45026 by 47) is given below:

(I am not able to draw boxes around the following numbers. Assume the digits are inside boxes to facilitate explanation. Also in the divisor 47, 7 is the flagged digit which is to be superscripted for easy understanding.)

4 9 7 5

47 4 5 0 2 6

0 9 5 8 0

1. 4 ÷ 4 = 0 remainder 4. Put the quotient 0, the first digit of the solution, in the first box of the bottom row and carry over the remainder 4

2. The product of the flagged number (7) and the previous quotient (0) must be subtracted from the next number (45) before the division can proceed. 45 – 7 x 0 = 45

45 ÷ 4 = 9 remainder 9. Put down the 9 and carry over the 9.

3. Again subtract the product of the flagged number (7) and the previous quotient (9), 90 – 7 x 9 = 27

27 ÷ 4 = 5 remainder 7. Put down the 5 and carry over the 7.

4. 72 – 7 x 5 = 37

37 ÷ 4 = 8 remainder 5. Put down the 8 and carry over the 5.

5. 56 – 7 x 8 = 0

There is nothing left to divide, so this cleanly completes the division.

Hi Nandeesh,

Thanks again for your wonderful contribution or rather improvement of the above post.

It will be very kind of you if you can mail me your idea posted as comment in a word doc (with proper formatting). I will try to include it in this post.

Regards

Vineet