The rule of divisibility for 13
A divisibility rule is a useful way to determine whether an integer is divisible by a fixed divisor without leaving a remainder. An integer a is divisible by 13 if there exists an integer b such that the product of b and 13 is equal to the integer a. The notation 13 | a means that a is divisible by 13. 4 divisibility rules are given in this article including rules, examples and their proofs.
Theorems about divisibility
In the course of derivation of divisibility of 13, it's always necessary to use some of basix theorems on divisibility.
Theorem 1: If a∣b, then a∣bc.
Theorem 2: If a∣b and a∣c, then a∣(b+c).
Theorem 3: If a∣b and b∣c, then a∣c.
Theorem 4: If a∣b and a∣c, then a∣mb+nc for all integers m and n.
4 divisibility rules for 13
Divisibility Rule 1:
Multiply the last digit of a number by 9, and subtract the product from the truncated number. If the difference is divisible by 13, then the numeral is divisible by 13.
For example, the number 494, the last digit is 4, the truncated number is 49.
which is divisible by 13. Therefore, the number 494 is divisible by 13.
How to prove the rule of divisibility of 13?
Let’s evaluate 3-digit number that is denoted as \overline{abc}, in which a, b, c are three digits at hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \overline{abc} is rewritten as 100a+10b+c
\dfrac{\overline{abc}}{13} = \dfrac{100a+10b+c}{13}
=\dfrac{90a+9b+10c+10a+b-9c }{13}
Let's denote
Z = \dfrac{10a+b-9c}{13}
Then
\dfrac{\overline{abc}}{13}=\dfrac{90a+9b+10c}{13}+Z
=\dfrac{91a-a+13b-4b+13c-3c}{13}+z
=7a+b+c+\dfrac{-a-4b-3c}{13}+ \dfrac{10a+b-9c}{13}
Next, we will show that the last two terms are dependent in terms of divisibility by 13. If \dfrac{10a+b-9c}{13} is an integer, \dfrac{-a-4b-3c}{13} is also an integer.
Let A = -a-4b-3c
On the other hand,
Z = \dfrac{10a+b-9c}{13}
=\dfrac{13a-3a+13b-12b-13c+4c}{13}
= a+b+\dfrac{-3a-12b+4c}{13}
Let B = -3a-12b+4c, then 13|B
-3A+B = -3( -a-4b-3c) + -3a-12b+4c =13C
which shows -3A+B is divisible by 13. If 13|B, then 13|A
Therefore, \overline{abc} is divisible by 13 if B is divisible by 13, that is, the difference of the truncated number and 9 times of the last digit is divisible by 13.
Divisibility Rule 2
Truncate the last digit of a number, multiply it by 4, and add the product to the rest of truncated number. If the result is divisible by 13, then the number is divisible by 13.
Let z =\dfrac{10a+b+4c }{13}
\overline{abc}=100a+10b+c
\dfrac{\overline{abc}}{13} = \dfrac{100a+10b+c}{13}
=\dfrac{90a+9b-3c +10a+b+4c }{13}
=\dfrac{91a-a+13b-4b-3c}{13}+z
=7a+b+\dfrac{-a-4b-3c}{13}+ \dfrac{10a+b+4c }{13}
Let A = -a-4b-3c and B = 10a+b+4c
-3A+B = -3(-a-4b-3c)+-a-4b-3c = 13a+13b+13c
Therefore
If 13|B, then 13|A
If \overline{abc} is divisible by 13, if and only if 10a+b+4c is divisible by 13.
Divisibility Rule 3
Subtract the number formed by the last 2 digits of the number from 4 times of the number formed by the rest of the number. If the difference is divisible by 13, the number is divisible by 13.
For example, the number 4524, the number formed by the last two digit is 24, 4 times of the rest of digits is 4\times 45 = 180. The difference is 156, which is divisible by 13. Therefore, 4524 is divisible by 13, that is
To prove the rule of divisibility of 13, let's evaluate a 4-digit number \overline{abcd}, which can be represented to
\overline{abcd}=1000a+100b+10c+d
Let
Z= \dfrac{40a+4b-10c-d}{13}
\dfrac{\overline{abcd}}{13} =\dfrac{1000a+100b+10c+d}{13}
=\dfrac{1000a+100b+10c+d - 40a-4b+10c+ d+40a+4b-10c-d}{13}
=\dfrac{960a+96b+20c+2d}{13} + Z
=\dfrac{74\cdot 13a-2a+91b+5b+2\cdot 13c-6c+2d}{13} +Z
=74a +7b+2c+ \dfrac{-2a+5b-6c+2d}{13} + \dfrac{40a+4b-10c-d}{13}
Let A = -2a+5b-6c+2d and B = 40a+4b-10c-d
7A+B = 7(-2a+5b-6c+2d)+-2a+5b-6c+2d
= 26a+39b-52c+13d
which is divisible by 13.
Therefore,
If 13|B, then 13|A, which shows the \overline{abcd} is divisible by 13 if and only if B is divisible by 13.
Divisibility Rule 4
Subtract the sum of numbers formed by alternate 3 digits starting from even places from the sum of numbers formed by alternate 3 digits numbers starting from odd places. If the difference is divisible by 13, then the number is divisible by 13.
For example, the number 9,345,674, the sum of alternate 3 digits starting from odd places is 674+9=683, the sum of alternate 3 digits starting from even places is 345. Their difference is 345, which is divisible by 13. Therefore,
From this rule, it’s obvious that the number has to be large enough to apply this rule. So let’s evaluate a 7-digit number, which is denoted as \overline{abcdefg}
which is rewritten as,
\overline{abcdefg}
=1000000a+100000b+10000c+1000d+100e+10f+g
Let
Z = \dfrac{100e+10f+g - (100b+10c+d )+a}{13}
Before proofing the rule, let’s have a review of some of interesting multiples of 13. For numbers of more than 4 places, the numbers are divisible by 13 in the following patterns alternating every 3 places.
\begin{array}{|c|c|c|} \hline \text{place}&\text{number} & n/13 \\ \hline 1&1& \\ \hline 2&10& \\ \hline 3&100& \\ \hline 4&1001&77 \\ \hline 5&10010&770 \\ \hline 6&100100&7700 \\ \hline 7&999999&76923 \\ \hline 8&9999990&769230 \\ \hline 9&99999900&7692300 \\ \hline 10&1000000001&76923077 \\ \hline 11&10000000010&769230770 \\ \hline 12&100000000100&7692307700 \\ \hline 13&999999999999&76923076923 \\ \hline 14&9999999999990&769230769230 \\ \hline 15&99999999999900&7692307692300 \\ \hline \end{array}
Let’s divide the number by 13,
\dfrac{\overline{abcdefg} }{13} = \dfrac{1000000a+100000b+10000c+1000d+100e+10f+g }{13}
=\dfrac{999999a+100100b+10010c+ 1001d+(100e+10f+g ) - (100b+10c+d )+a }{13}
=76923+7700b+770c +77+ Z
Therefore, the number \overline{abcdefg} is divisible only if Z is an integer.
What are the numbers within 1000 that are divisible by 13
There are 76 numbers that are multiples of 13 within 1000.
\begin{array}{|c|c|c|c|c|c|c|} \hline 13&156&299&442&585&728&871 \\ \hline26&169&312&455&598&741&884 \\ \hline39&182&325&468&611&754&897 \\ \hline52&195&338&481&624&767&910 \\ \hline65&208&351&494&637&780&923 \\ \hline78&221&364&507&650&793&936 \\ \hline91&234&377&520&663&806&949 \\ \hline104&247&390&533&676&819&962 \\ \hline117&260&403&546&689&832&975 \\ \hline130&273&416&559&702&845&988 \\ \hline143&286&429&572&715&858& \\ \hline \end{array}
Steven Zheng
posted 4 months ago