# Divisibility by 7 question

• December 7th 2008, 07:39 PM
Rafael Almeida
Divisibility by 7 question
Hi there! I'm having some big trouble with this:

Let there be $x, y \in \mathcal{Z}$. Show that $10x + y$ is a multiple of $7$ if, and only if $x - 2y$ is a multiple of 7. Using the equivalence above, find a divisibility criteria for $7$.

So, I'd love to show you where I'm stuck, if it was not in the very beginning. Tips?
• December 7th 2008, 08:03 PM
o_O
Split into the two statements:

(1) If $x - 2y$ is a multiple of 7, then $10x + y$ is a multiple of 7 as well.

Since $x - 2y$ is a multiple of 7, then $x - 2y = 7k_1 \ \Leftrightarrow \ x = 7k_1 + 2y$ for some $k_1 \in \mathbb{Z}$. Now let's plug this in: $10(7k_1 + 2y) + y = 70k_1 + 21y = 7 (10k_1 + 3y)$

Obviously, this is a multiple of 7 so we're done.

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(2) If $10x + y$ is a multiple of 7, then $x - 2y$ is a multiple of 7 as well.

Since $10x + y$ is a multiple of 7, then $10x + y = 7k_2 \ \Leftrightarrow \ y = 7k_2 - 10x$ for some $k_2 \in \mathbb{Z}$

Plug this into $x - 2y$ and the procedure is the same as above.
• December 7th 2008, 08:51 PM
Soroban
Hello, Rafael!

I'll do part of it for you . . .

Quote:

Let there be $x, y \in \mathcal{Z}$.
Show that $10x + y$ is a multiple of $7$ if and only if $x - 2y$ is a multiple of 7.

We are given: . $x - 2y \:=\:7a\,\text{ for some integer }a.$

Multiply both sides by 10: . $10x - 20y \:=\:70a$

$\text{Add }21y\text{ to both sides: }\;10x + y \:=\:70a + 21y\:=\:7(10a + 3y)$

Therefore, $10x+y$ is a multiple of 7.

• December 8th 2008, 12:02 AM
Opalg
Quote:

Originally Posted by Rafael Almeida
$10x + y$ is a multiple of $7$ if, and only if $x - 2y$ is a multiple of 7. Using the equivalence above, find a divisibility criteria for $7$.

Suppose for example that you have a 5-digit number abcde. This is equal to 10*abcd + e. The equivalence shows that this is divisible by 7 if and only if abcd – 2*e is divisible by 7. But this number is equal to 10*abc + (d–2*e). So it is divisible by 7 if and only if abc – 2(d–2*e) is divisible by 7. But this number is equal to 10*ab + (c–2*d+4*e). So it is divisible by 7 if and only if ab – 2(c–2*d+4*e) is divisible by 7.But this number is equal to 10*abc + (d–2*e). So it is divisible by 7 if and only if abc – 2(d–2*e) is divisible by 7. But this number is equal to 10*a + (b–2*c+4*d–8*e). So it is divisible by 7 if and only if a – 2(b–2*c+4*d–8*e) = a–2*b+4*c–8*d+16*e is divisible by 7.

As an example, look at the number 42931. The criterion says that this should be divisible by 7 if and only if 4–2*2+4*9–8*3+16*1 is divisible by 7. But 4–2*2+4*9–8*3+16*1 = 28, which is divisible by 7. Therefore so is 42931.

You can extend this criterion from 5-digit numbers to numbers with any number of digits in an obvious way.