Last night, I was exploring the Internet, when I stumbled across a very cool diagram. It look like this:

With this graph, you can see what numbers are divisible by 7. Lets take the number 1234567. You would first start at the white node at the bottom. First move 1 black arrow, then 1 white arrow, then 2 black arrows, then 1 white arrow, then 3 black arrows, the 1 white arrows, and so on and so forth. For each digit number of black arrows you move, you move 1 white arrow. If you end up at the start, then it is divisible by 7. However number of black arrows away from the start you are is the remainder. Can someone prove this?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestI think i have figured out the proof.

So, starting from the starting point mark the junctions as \(0, 1, 2, 3, 4, 5, 6\). Note that every junction represents the number \([mod 7] \).

Now choose any number and start the procedure. Let your first digit is \(a\). Follow the black arrow \(a\) times. You will currently be at a juction that represents \(a [mod 7]\).

Now, if you have one more digit then you are asked to move in the direction of the correspondong white arrow.

These white arrows are so arranged that the corresponding white arrow takes you to the place \(a \times 10 [mod 7]\) (You can check it for all \(7\) cases.).

It means now you are at \(\overline{a0} [mod 7]\).

Now, if your next digit is \(b\), you will , after moving \(b\) black arrows will reach at \(\overline{ab} mod[7]\) position.

This will continue and at any step you will be at the position that represents your current number \([mod 7]\).

By this, the working of this diagram is explained and using the same method I have made diagrams for divisibilty tests of \(6, 8\) & \(11\).

Log in to reply

Another cool thing about multiples of 7 is that if you multiply the last digit by 5, and add it to the remaining number, it will always be a multiple of 7. For example:

84: 8 + \(4 \times 5\) = 28 = \(4 \times 7\)

161: 16 + \(1 \times 5\) = 21 = \(7 \times 3\)

And this does not only apply to multiples of 7. Take multiples of 13, for example. When the last digit is multiplied by 4, the sum of that and the remaining number will also be a multiple of 13. For example:

26: 2 + \(6 \times 4\) = 26 (this kind of becomes a loop, but you get the point.

I have documented how this works for many numbers… (From what I have seen, it only occurs with some numbers)

3: Multiply last digit by

1(3 * 3 =10 − 1)7: Multiply last digit by

5(7 * 7 =50 − 19: Multiply last digit by

1(9 * 1 =10 − 1)11: Multiply last digit by

10(11 * 9 =100 − 1)13: Multiply last digit by

4(13 * 3 =40 − 1)17: Multiply last digit by

12(17 * 7 =120 − 1)19: Multiply last digit by

2(19 * 1 =20 − 1)The notes in parentheses are observations I have made. But can someone explain why?

Log in to reply

That's a nice observation. I will prove this for \(7\).

Let the given number be \(n=\overline {a_p a_{p-1} ..........a_0}\). This number is also given to be a multiple of \(7\).As the 'trick' involves only the last digit and the remaining number, let's reduce this whole number to \(\overline {ab}\) where \(b=a_0\) and \(a= \overline{a_p a_{p-1}.........a_1} (mod 7)\).

The algorithm is multipying \(b\) by \(5\) and adding it to \(a\).

So, we will be left with the number

\(m= a+ 5\cdot b\) and we are required to prove that \(m=0 (mod 7)\)

Now note that

\(10\cdot m (mod7) = 10\cdot a+ 50\cdot b (mod7)\)

\( =10\cdot a+ b (mod 7)\) [because, \(50=1 (mod 7)\)]

\( = \overline {ab} (mod7)=0(mod 7)\)

\(\Rightarrow 10\cdot m=0 (mod 7)\)

\(\Rightarrow m=0 (mod 7) \).

Hence proved.

On similar lines, you can also prove it for other numbers.

Log in to reply

882is divisible by7, or if912is divisible by19, I can say yes in an instant!!!Log in to reply

this

TryLog in to reply

Awesome! Upvoted!

Log in to reply

Can any of you guys create a diagram like this for testing multiples of 13 or 17 or 19 or any prime numbers greater than 5? It would be really cool!

Log in to reply

Here's one for 13. (I have used single arrow in place of black and double in place of white)

Log in to reply

And question: How do you come up with these pictures? Is there some sort of trick to it?

Log in to reply

Log in to reply

Log in to reply

Comment deleted Jun 04, 2015

Log in to reply

## Hi , if you want to create this effect , try using 6

`#`

's right at the start of the text that you are about to type .Log in to reply

Thanks!

Log in to reply

Can you give me the link to it ? I might be able to help you out

Log in to reply

I think he meant this problem.

Log in to reply

Wow! It's a great thing you even figured the whole trick . If I would have seen it somewhere on the net , I'd have thought it was something related to Physics and would have closed the tab :P

Log in to reply

Where did you learn about this? It is very neat

Log in to reply