Why 7 is Weird - Numberphile

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This video features Dr James Grime on divisibility.

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[andreroussel]

Divisible by 7 can be useful in figuring out if there is a whole number of weeks in a number of days.

[mazza]

this feels like a very old-school numberphile video, love it

[luketurner]

The same algorithm (3:51) can be used to construct a formula for other primes:
11: x - y (k=10, j=99)
13: x + 4y or x - 9y (k=4, j=39)
17: x - 5y (k=12, j=119)
19: x + 2y (k=2, j=19)
23: x + 7y (k=7, j=69)
29: x + 3y (k=3, j=29)
31: x - 3y (k=28, j=279)
37: x - 11y (k=26, j=259)
41: x - 4y (k=37, j=369)
43: x + 13y or x - 30y (k=13, j=129)
47: x - 14y (k=33, j=329)
53: x + 16y (k=16, j=159)
59: x + 6y (k=6, j=59)
61: x - 6y (k=55, j=549)
67: x - 20y (k=47, j=469)
71: x - 7y (k=64, j=639)
73: x + 22y (k=22, j=219)
79: x + 8y (k=8, j=79)
83: x + 25y (k=25, j=249)
89: x + 9y (k=9, j=89)
97: x - 29y or x - 30x + x (k=68, j=679)

where k is the multiplier of 10x + y: 10kx + ky (4:11)
and j is a multiple of the prime that is subtracted: 10kx - jx + ky = x + ky (4:30)
then subtract the prime if it helps

And since the same algorithm is used, the results have the same properties, like iterability (1:24)

[melaniehall]

Can't believe that 10 years later, I am still loving watch James Grime on Numberphile. I watched him when I was a nerdy high schooler and now I'm a nerdy adult. Thank you so much for all the videos over the years.

[rdand]

When I was in the 7th grade, I was taught all the tests for divisibility except for 7. What we were told was "try dividing it by 7, " which completely defeats the purpose of a divisibility test. Thank you for filling in this particular gap in my education.

[Not-THAT-ChrisPratt]

For 40+ years I've been saying, "there is no rule for 7, " meaning no way to check for divisibility as with 3, 5, 9 etc. Thank you for this. I now have my "rule for 7!"

[frasco_]

Me and a friend of mine discovered a trick for 11 in middle school:
Take 121, u split it in 1+21 = 22 so if the result is divisible by 11 the original is too.
It works even for larger numbers like
35673 split in 3+56+73 = 132
132 split in 1+32 = 33
In case of even digits
1078 split in 10+78 = 88
Basically you split the number in sets of two digits starting from the end.

[pietronardelli]

This video took me back in time, when I was a middle-schooler amazed by these tricks and properties of numbers.
Even now in uni, I watch videos on this channel with the same flabbergasted look, enjoying every minute as an extraordinary discover.
I really appreciate your content, you're doing great ✌️

[nicolestewart]

One of the great things about youtube that is not talked about enough. There are so many creators I have watched for a decade or longer and they were so important to me and continue to be so.

In the media too much attention is paid to the drama and such. What about people like this? Or the green brothers? Vsauce? Rhett and link? I could go on. It's not an easy thing to do, to be able to take your fans with you as they pass into different eras of their lives. Most adults don't still watch what they watched as kids but so many of us do with creators like I mentioned above.

These people are my teachers and actually helped guide me to being a better adult. Me and so many others. It's a beautiful thing.

The creators, as well as fellow followers, feel like family and it's such a comfort.

[JordynPi]

another way to work out if something is divisible by 7 is to just do all your math in base 7 and see if it ends in 0

[Glizzygobbler]

I’d always do it by comparing to other numbers. 7000 is divisible by 7, 7000-6468=532. If 532 is divisible by 7 then 6468 was. 700-532=168. 168-140=28. 28 is divisible by 7 so 6468 is. This method is much easier for me to do in my head because it’s only addition or subtraction.

[FHBStudio]

I use divisibility for finding primes and root approximations during tutoring. I lacked the 7 test so, thanks for that!

[dajaco]

I always found multiples of 100, subtracted them and added on double the number of 100s you removed to what is remaining. Taking advantage of 98 being a multiple of 7. This works pretty well and is easy to do in your head

[brian]

I poked around in excel to find out what happens with that seven trick in how other numbers terminate. 49 is in a loop of 49's as we saw in the video and all numbers 1-48 that aren't divisible by 7 are in a loop together and all of the numbers divisible by seven are in their own loop.

1, 5, 25, 27, 37, 38, 43, 19, 46, 34, 23, 17, 36, 33, 18, 41, 9, 45, 29, 47, 39, 48, 44, 24, 22, 12, 11, 6, 30, 3, 15, 26, 32, 13, 16, 31, 8, 40, 4, 20, 2, 10
and
7, 35, 28, 42, 14, 21

[GregAlpar]

When I was 13, I found another method for the divisibility by 7. It's in a way even more useful than the one in the video. The trick is to cut the number at the second digit. For instance, for 343, it's 3 and 43. After that, you double the rest (in the example, from 3, you get 6). If the sum is divisible by 7, then the original number is divisible by 7: 6+43 = 49. So, 343 is divisible by 7.

What this method also provides is that it preserves modulo. So, if the number is not divisible by 7, the method also tells you how much is the remainder mod 7. For example, taking 716, you get 2×7 + 16 = 30, which is 28 + 2. So, 716 gives 2 as a remainder after dividing by 7.

[nathanenzo]

There is a more smart way to see if it is divisible by 4.
In three steps.
1: we look only in the last two digits.
2. It's even.
3: if the units are divisible by 4, which are 0, 4, 8 then the tens must be even, if it's not divisible by 4 which are 2, 6 then the tens must be odds.
So, for example with units as a 2, tens must be odds.
12, 32, 52, 72, 92

[mattv]

I just instantly see either 350+84 or 420+14 which makes it obviously divisible by 7. But I guess the point of the video is to showcase some more interesting methods. nice job.thanks.

[allonahoya]

The general divisibility trick for any prime p (other than 2 and 5) is as follows:
Let our dividend be k and express it as 10x + y (x, y are non-negative integers)
Find an integer z such that 10z when divided by p yields remainder 1.
Now if x + (z)y is divisible by p, then 10x + y is divisible by p. This yields the required algorithm.

Note: You can replace 10 with any natural number as long as it isn't a multiple of our chosen prime p.

[lamenwatch]

(1s x 3^0) + (10s x 3^1) + (100s x 3^2) + (1000s x 3^3) + ... + (10^ns x 3^n) should be divisible by 7 if the number itself is divisible by 7.

EX: 343
3x1 + 4x3 + 3×9
3 + 12 + 27
42
✅ Divisible ✅

Figured this out myself.

[StevenStJohn-kjeb]

The most amazing thing about this video? James said "this weird trick" in the video, but that phrase doesn't appear in the title. Well done!