Divisibility by 13

Divisibility “rule” for 13 (subtract 9 times the last digit)

A known test for divisibility by $13$ works by taking the last digit, multiplying it by $9$, and then subtracting that product from the remaining truncated part of the number. If the result is divisible by $13$, then the original number is divisible by $13$. Here’s how it looks for $533$:

1. Split off the last digit ($3$) and the rest of the number ($53$).

2. Multiply the last digit by $9$: $3 \times 9 = 27$.

3. Subtract that from the truncated part: $53 - 27 = 26$.

4. Since $26$ is divisible by $13$ (i.e., $26 = 13 \times 2$), $533$ is also divisible by $13$.

Either approach quickly confirms that $533 = 13 \times 41$.

Proof:

Below is a short, purely “algebraic” proof of the rule

A decimal number $N$ is divisible by 13 if and only if “take its last digit $b$, multiply it by 9, and subtract that from the truncated part $a$.” The resulting number, $a - 9b$, must also be divisible by $13$.

We never explicitly mention “modular arithmetic” or “congruences”; we only use the idea that “$x$ is divisible by $13$” means “$x$ is a multiple of $13$.”

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Setting Up the Problem

Let $N$ be a decimal number with last digit $b$ and “truncated part” $a$.
In other words,

$$N = 10a + b,$$

where $0 \le b \le 9$ and $a$ is the integer we get by removing the final digit of $N$.

We want to show:

$$13 \;\text{divides}\; N = 10a + b \quad\Longleftrightarrow\quad 13 \;\text{divides}\; (a \;-\; 9b).$$

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Forward Direction (if $13$ divides $N$, it divides $a - 9b$)

1. Assume $10a + b$ is a multiple of $13$.
   That is, there exists an integer $c$ such that

   $$10a + b \;=\; 13\,c.$$

2. Rewrite “10$a$” as “(13 - 3)$a$.”
   Since $10 = 13 - 3$, we have

   $$10a + b \;=\; (13 - 3)\,a + b \;=\; 13a \;-\; 3a \;+\; b.$$

   Hence

   $$13\,c \;=\; 13a \;+\; \bigl(b - 3a\bigr).$$

   Rearrange to get

   $$b - 3a \;=\; 13\,(c - a).$$

   So $b - 3a$ is also a multiple of $13$.
   Write $b$ as

   $$b \;=\; 3a + 13\,m,\quad \text{where }m \,=\, c-a.$$

3. Check $a - 9b$.
   Substituting $b = 3a + 13m$, we get

   $$a \;-\; 9b \;=\; a - 9\bigl(3a + 13m\bigr) \;=\; a - 27a \;-\; 117m \;=\; -26a \;-\; 117m.$$

   Factor out $13$:

   $$-26a \;-\; 117m \;=\; -13\,\bigl(2a + 9m\bigr).$$

   Thus $a - 9b$ is a multiple of $13$.

Hence, if $13$ divides $N=10a+b$, then it must also divide $\,a - 9b.$

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Reverse Direction (if $13$ divides $a - 9b$, it divides $N$)

1. Assume $a - 9b$ is a multiple of $13$.
   That is, there exists an integer $d$ such that

   $$a \;-\; 9b \;=\; 13\,d.$$

   Rearrange to get

   $$a \;=\; 9b + 13\,d.$$

2. Substitute into $N = 10a + b$.

   $$10a + b \;=\; 10\bigl(9b + 13d\bigr) \;+\; b \;=\; 90b + 130d + b \;=\; 91b + 130d.$$

   Factor out $13$:

   $$91b + 130d \;=\; 13\,\bigl(7b + 10d\bigr).$$

   So $10a + b$ is a multiple of $13$.

Thus, if $13$ divides $a - 9b$, then it must also divide $N$.

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Conclusion

Putting both directions together shows

$$\text{“\(10a + b\) is divisible by 13”} \;\;\Longleftrightarrow\;\; \text{“\(a - 9b\) is divisible by 13.”}$$

In words: A number $N$ is divisible by 13 exactly if, after you remove its last digit $b$, multiply that digit by 9, and subtract from the remaining part $a$, you get a multiple of 13.