Divisibility by 7

 Below is a short, purely “algebraic” argument (no explicit modular arithmetic) for the common test for divisibility by 7:

Take the last digit of the number, double it, and subtract that product from the remaining truncated part. If the result is divisible by 7, then the original number is also divisible by 7.

In a formula:

• Let $N$ have last digit $b$ and truncated part $a$.
  So $N = 10a + b$.

• The rule says:

  $$\text{“\(N\) is divisible by 7”} \;\Longleftrightarrow\; \text{“\(a - 2b\) is divisible by 7.”}$$

We prove this equivalence purely via algebraic manipulation and the notion of “multiple of 7.”

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Forward Direction

(If $N$ is divisible by 7, then $a - 2b$ is also divisible by 7.)

1. Assume $10a + b$ is a multiple of 7.
   That is, there exists some integer $k$ such that

   $$10a + b \;=\; 7k.$$

2. Rewrite “$10a$” as “$(7 + 3)a$.”
   Since $10 = 7 + 3$, we have:

   $$10a + b \;=\; (7 + 3)\,a + b \;=\; 7a + (\,3a + b\,).$$

   So

   $$7k \;=\; 7a + (\,3a + b\,).$$

   Subtract $7a$ from both sides:

   $$3a + b \;=\; 7\,(k - a).$$

   Hence $\,3a + b$ is itself a multiple of 7.
   Call that multiple $7m$, i.e.

   $$3a + b \;=\; 7m, \quad \text{where } m = k - a.$$

3. Express $b$ in terms of $a$.
   From $3a + b = 7m$, we get

   $$b = 7m \;-\; 3a.$$

4. Check $a - 2b$.
   Substitute $b = 7m - 3a$:

   $$a \;-\; 2b \;=\; a - 2\bigl(7m - 3a\bigr) \;=\; a - 14m + 6a \;=\; 7a \;-\; 14m.$$

   Factor out 7:

   $$7a - 14m \;=\; 7\,(\,a - 2m\,).$$

   Thus $a - 2b$ is a multiple of 7.

Therefore, whenever $10a + b$ is divisible by 7, $a - 2b$ must also be divisible by 7.

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Reverse Direction

(If $a - 2b$ is divisible by 7, then $N$ is divisible by 7.)

1. Assume $a - 2b$ is a multiple of 7.
   That is, there exists an integer $n$ such that

   $$a \;-\; 2b \;=\; 7n.$$

   Rearrange it to solve for $a$:

   $$a \;=\; 2b + 7n.$$

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

   $$10a + b \;=\; 10\bigl(2b + 7n\bigr) \;+\; b \;=\; 20b + 70n + b \;=\; 21b + 70n.$$

   Factor out 7:

   $$21b + 70n \;=\; 7\,\bigl(3b + 10n\bigr).$$

   Hence $10a + b$ is also a multiple of 7.

Therefore, $a - 2b$ is divisible by 7 if and only if $N$ is divisible by 7.

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Conclusion

Putting both parts together, we have shown purely by algebra that

$$10a + b\; \text{is a multiple of 7} \quad\Longleftrightarrow\quad a - 2b \;\text{is a multiple of 7}.$$

In everyday language:

A decimal number is divisible by 7 if and only if, when you double its last digit and subtract that from the rest of the number, the result is divisible by 7.