# Divisibility by 7 and Its Proof

This is the 6th post in the Divisibility Rules Series.  In this post, we discuss divisibility by 7.

Simple steps are needed to check if a number is divisible by 7. First, multiply the rightmost (unit) digit by 2, and then subtract the product from the remaining digits. If the difference is divisible by 7, then the number is divisible by 7.

Example 1: Is 623 divisible by 7?

3 x 2 = 6
626 = 56
56 is divisible by 7, so 623 is divisible by 7.

If after the process above, the number is still large, and it is difficult if to know if it is divisible by 7, the steps can be repeated. We take the difference as the new number, we multiply the rightmost digit by 2, and then subtract from the remaining digits.

Example 2:  Is 3423 divisible by 7?

3 x 2 =
3426 = 336

We repeat the process for 336. We multiply 6 by 2 and then subtract it from 33

6 x 2 = 12
3312 = 21
21 is divisible by 7, so 3423 is divisible by 7.

Note that if the number is still large, this process can be repeated over and over again, until it is possible to determine if the remaining digits is divisible by 7.

The following portion are for students who have basic knowledge on proofs. In particular, we will be proving an  if and only if statement. A if and only if B requires to prove that A implies B and B implies A.

Let $N$ be the number that we want divide by 7.  Let $b$ be the unit’s digit and $a$ be the rest of the digit. Then  N = 10a + b.

Explanation: All whole numbers N can be expressed as the product of 10 and a  number added to its units digit. For example 983 = 10(98) + 3, 5896 = 10(598) + 6, and so on.

We assign the following statements to A and B.

A: a – 2b is divisible by 7.
B: N is divisible by 7.

As we have mentioned above, we have to show that (1) A implies B  and (2) B implies A.   This means that we have to show that if $a - 2b$ is divisible by $7$, then $N$ is divisible by $7$. The statement $a - 2b$ is the step where we multiplied the unit’s digit by 2, and then subtracted from the remaining digits $a$.

For (1) We have to show that A implies B. That is, we have to show that if $a - 2b$ is divisible by $7$, then $N$ is divisible by $7$.

Proof

If $a - 2b$ is divisible by $7$, then we can find a natural number $k$ such that $a - 2b = 7k$ (Can you see why?).

Multiply both sides by $10$, we have $10a - 20b = 70k$. Adding $b$ on both sides, we have $10a - 20b + b= 70k + b$. Now, $10a + b = 70k + 21b$. Notice that the left hand side of our equation is $N$ and the right hand side can be divided by $7$. Therefore, $10a + b = N$ is divisible by $7$. That proves our first statement that If $a - 2b$ is divisible by $7$, $N$ is divisible by $7$.#

For (2), we have to show that B implies A. That is, we have to show that if $N$ is divisible by 7, $a - 2b$ is divisible by $7$.

Proof

If $N$ is divisible by $7$, then $10a + b$ is divisible by $7$.  This means we can find a natural number $k$ such that $10a + b = 7k$. Subtracting $21b$ from both sides, we have $10a + b - 21b = 7k - 21b$. This means that $10a - 20b = 7k - 21b$. Factoring, we have $10(a-2b) = 7(k-3b)$

Now, since $10$ is not divisible by $7$, $a - 2b$ is divisible by $7$. This proves the second statement if $N$ is divisible by $7$, then $a - 2b$ is divisible by $7$#

From above, we have shown that A implies B and B implies A. We have shown that the process that we have done above will hold for all cases.