GENIUS Test Divisibility By 7 - Graph Visualization

This graph will tell you if a number is divisible by 7. Start at the circle YES. Move a number of black arrows equal to the first (leftmost) digit of the number. Then move one green arrow to get to the next digit. Repeat this for each digit (except don't move a green arrow for the last digit). If you end up at YES, then the number is divisible by 7.

Sources
Tanya Khovanova's Math Blog, Guest Blogger David Wilson

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[NotQuiteFirst]

Thanks this is really useful. It's going to save me loads of time every day at work (my job is to go through all the numbers to see which ones are divisible by 7).

[ffggddss]

Ah, I see.
The way it works is, it's counting mod 7.

You start at 0. This is the initial sum.

Each trip of n black arrows is addition of a digit to the sum.
Then each green arrow multiples the current sum by 10 ≡ 3 (mod 7). The green arrows are arranged to do that.

So this can be adapted to divisibility by any integer, n > 1.
It just takes a bit of figuring to get the diagram right for n.

[Pika]

The green arrows on the graph (other than the one to "yes") represent the multiplicative group modulo 7, as the powers of the cyclic generator 3 (which 10 is congruent to mod 7), so a green arrow would indicate "multiply by 10" (or 3 mod 7), with the exception of the very last digit (the ones digit), which is why the graph works as intended.

[AngryArmadillo]

Aha each green arrow maps a number n to n*10 (mod 7). Very clever

[eternalfizzer]

More than the instructions on how to use this algorithm, I'd really like to learn who came up with this and how it works.

[iamstickfigure]

Seems like finite automata.
We had pretty much exactly the same problems in my finite automata class, but with binary.

[HenryFabianGT]

Holy Jesus! *That was freaking awesome!!* :D

[tamirerez]

Another method is as follow:
We want to check the number 1456.
1. We separate the unit digit: 145 | 6
We calculate 145 minus 2 times 6.
133=145-12

We repeat this operation again with the new number 133.

2. Separate the unit digit:
3 | 13
And now 13- (2x3)= 7
Since we know 7 is divisible by 7 (7 or 0)
we found that 1456 divisible by 7 too.
Another example:
Check for 2895
279=289-10
9=27-18
We get 9, so 2895 not divisible by 7.

One last:
Check for 12566
1244=1256-12
116=124-8
1-=11-12
So
12566 not divisible by 7

I sended a mail to mr. Presh to ask him WHY IS IT WORKING??
I mean what is the logic behind this method?
mr. Presh probably didn't have time to answer, or maybe I didn't explain clear.
Amyway, I didn't get an answer.
So PLEASE anybody:
If you read this, can you send a question to this channel and ask?
Or maybe you can amswer me yourself?
I will be happy to see a new video of mr. Presh explaining this method and WHY IT'S WORKING
Thank you.

[johnnypoker]

A graph like this can be constructed that works for any base and any divisor provided they are relatively prime to each other.

Each green arrow points a number n to n*base (mod divisor)

(as others have essentially pointed out already)

[TheHappybilmore]

2:50 are you not supposed to move a green arrow on the last digit? not that it changes the result.

[bpark]

There is a much simpler way, and it can be applied to any divisor. Add to units digit, 3 times 10's, 2 times 100's, -1 times 1000's, -3 times 10000's, -2 times (multipliers repeat). Result has same modulo as original number. Modulo can be taken during this process to keep numbers small. You can easily calculate the multipliers on-the-spot for any divisor.

[paulobouhid]

Very, very clever!!! How about this one:

Is 5, 555^2, 222 + 2, 222^5, 555 divisible by 7? ("^" means exponentiation)

[ThisNameisalreadytaken.]

Wohoo cool!! Better if you can also explain how it works, I mean the logic behind it.

[MumboJ]

Since the only green arrow that ends at "YES" also begins at "YES", it is unnecessary to skip the final green arrow, since it only ever changes the final position from a non-yes to a different non-yes.

[BarnibusMaximusMusic]

Is it possible to make these graphs for all numbers or is it a property of 7 that allows it? :)

[williamstier]

Another trick you can use is take every digit except the last one and multiply all of it by 3. Then add the last digit in. Keep repeating this until you reach a single digit number. If that single number isn't a result of 7 then, the number also isn't divisible by 7. e.g 14. Take 1 multiply it by 3 which is 3, then add 4 and you get 7. 49 multiply 4 by 3 to get 12. Add the last digit 9 to get 21. Finally multiply 2 by 3 and then add 1 and again you wind up at 7.

[rob]

It's quicker to just recalculate the decimal number in modulo 7 arithmetic.
e.g.
21364 = (((2*10 + 1)*10 + 3)*10 + 6)*10 + 4
do the calculation in modulo 7 arithmetic
2*10 + 1 == 0 (modulo 7)
0*10 + 3 == 3 (modulo 7)
3*10 + 6 == 1 (modulo 7)
1*10 + 4 == 0 (modulo 7)
nice curiosity, though and it had me for a while.

[sighmaniacrotmg]

Or take the number, take off the last digit and subtract it twice. If it's divisible by 7, so is the original number.

Example:
133
13 - (2*3) = 13 - 6 = 7
133 is divisible by 7.

[rob]

Nice an easy to convert to fast working code to determine the modulo 7 equivalence of a number. What about other numbers besides 7. How would you determine the graph for them?

[johnz.]

Another way is to do
Let’s say 126 is you number
You divide the number in two parts: the 12 and the 6.
You do 12-2*6=0
0 dividable by 7. Then 126 dividable by 7.
Same for 139
13-2*9=-5 not dividable by 7
And you’ll find 5 is the remainder of the division as well