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.  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.

22 Comments

Add a Comment

Your email address will not be published. Required fields are marked *