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

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## Comments

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TopNewestThere are several typos. As I get time I will mention them with corrections.

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Divisibility rule of 3

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You have to just add the digits if the sum is divisible by 3 than its divisible eg. 2034= 2+0+3+4= 9 which is divisible by 3 so 2034 is divisible by 3

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Divisibility rule of 11

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Please Provide proof For 47

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it's really helpful.

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Thank you. Its very useful for Maths students.

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Wow! This is useful!

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To prove this... Oh no.

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What do you mean? I did this all in my head. E.g. $47\mid 10a+b\iff 47\mid 140a+14b\iff 47\mid 14b-a$

This exact algorithm works for every one of the provided rules. Just multiply $10a+b$ by the number mentioned in the rule (in the case I've shown it's $14$; let's call it $t$) and subtract $da$ until $|k|$ is the smallest, where $d\mid ka+tb$ (here $d$ is the divisor for which the rule is proved).

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I meant to prove that this works might require some time and might require 23 cases of spamming... Doesn't seem like the best thing to prove.

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