Divisibility Tricks - Numberphile

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Tony is a professor at the University of Nottingham.

NUMBERPHILE

Videos by Brady Haran

[SomeNerdOutThere]

What I love about these tricks is that they aren't actually properties of the numbers themselves, but of their relationship to the base of notation. For instance, 3 and 9 can be tested for by checking the sum of digits because they both have a multiple that is one less than 10. If we were working in hexadecimal, this would work for 3, 5 or F (fifteen in decimal), but not for 9. In octal, it could test for multiples of seven.

[Mswordx]

That 749 thing was funny and proves that the more advanced in your field you get, the more easy things seem hard

[EpicMathTime]

If you add up all the cards you get -1/12

[achyuthramachandran]

In the divisibility rule for 11, you don't need to reverse the number before taking the alternating sum. It works just fine with the given order of digits.

[comteharbour]

You can test if a number is divisible by 8 with a sligntly simpler method (it needs no multiplication)
1. you look at the 3 last digits
2.1 If the left digit is even, the 2 last digits must be a multiple of 8
2.2 If the left digit is odd, the 2 last digits must be a multiple of 4 but not a multiple of 8

sketch of proof:
200 is a multiple of 8, so every 3-digits multiple of 8 starting with an even digit can be written as 200n + 8m
100 is a multiple of 4 but not a multiple of 8. It means that it has an offset of 4 compared to a multiple of 8
So every 3-digits multiple of starting with an odd digit can be written as (200n + 100) + (8m + 4)

[SchutzmarkeGMBH]

19:48 Is 749 divisible by 7? That's a bit hard, beyond my mental arithmetic.
Nice to see fellow astro-particle physicists struggle as well.

[Rhyff]

Love the energy in this guy, you can tell he's loving his job!

[funkee]

I like how Tony was humble enough to admit that he lost sight of the "beauty of math". Sometimes we trip ourselves up when the answer is to just step back and go simple :)

[Codricmon]

Someone in the comments probably brought it up already, but there is an alternate test for 11: You remove the last digit and subtract it from the remaining number (similar to the second test for 7, but without doubling the last digit). If the result is divisible by 11, then the original number is, as well.
For example: 121 -> 12 - 1 = 11
If you encounter larger numbers, you can just continue this process multiple times:
72, 435 -> 7, 243 - 5 = 7, 238 -> 723 - 8 = 715 -> 71 - 5 = 66

There is yet another way, which is basically just a different method of taking the alternating cross sum shown in the video, but I find this method easier to do in my head:
You subtract the first digit from the second, then subtract the result from the next digit, and so on, and then check the result:
96, 410, 985 -> 6 - 9 = -3 -> 4 - (-3) = 7 -> 1 - 7 = -6 -> 0 - (-6) = 6 -> 9 - 6 = 3 -> 8 - 3 = 5 -> 5 - 5 = 0

This also works in reverse by subtracting the second to last digit from the last and continuing towards the left:
96, 410, 985 -> 8 - 5 = 3 -> 9 - 3 = 6 -> 0 - 6 = -6 -> 1 - (-6) = 7 -> 4 - 7 = -3 -> 6 - (-3) = 9 -> 9 - 9 = 0

With some practice, this can make checking whether large numbers are divisible by 11 almost trivial... as long as you can see the number, that is.

[JJ-kleq]

I learned all these tricks by reading H.G. Wells, The Divisible Man.

[SlipperyTeeth]

I first discovered the pattern in the 9's when I was in first grade, and I remember thinking that I had discovered something truly amazing. Later I learned that it was already known - by most people even.
I wouldn't again feel that bittersweet feeling of a major revelation that had already become commonplace until 9th grade.

[earthbjornnahkaimurrao]

so in base 13 then the cross-sum rule would apply with 2, 3, 4 and 6!

[HackSawSees]

This has proved to be one of those times where watching a video on something I already knew has never the less proved valuable. Somehow, in my math education, the idea that a proof tells you *how* a thing works never stuck, if it was ever mentioned at all. Thanks, folks!

[Bodyknock]

With 3 and 9 it’s not just divisibility but also the sum of the digits has the same remainder divided by 3 or 9 as the original number (i.e. the sum of the digits equals the original number modulo 3 or 9). So 527 for example has sum of digits 14. 14 has remainder 5 divided by 9 so 527 is also remainder 5 divided by 9. Likewise 14 divided by 3 and 527 divided by 3 both have remainder 2. Divisibility is simply the special case where the remainder is zero.

Also to simplify the remainder calculation you can cast out digits which sum to 3 or 9 for whichever test you are doing. Again, 527 has 7+2 in the digits so you can ignore those in the test since they have remainder 0 divided by 9 and 3. Thus 527 equals 5 mod 9 = 5 and it equals 5 mod 3 = 2

[percentSNAFU]

I love how Mo Salah popped up in his thought bubble when he was thinking about dividing by 11, because of course he wears the #11 shirt.

[RecursiveTriforce]

26:23 Primes squared: 4, 9, 25, 49, 121, 169

[quinnbartlett]

The divisible by 7 test can be heavily streamlined

The number given is: 872, 123

You can take out consecutive digits that are multiples of 7 and replace them with 0s

7 can be replaced with 0

> 802, 123

21 can be replaced with 0

>800, 023

Now the test has easier numbers.

the test would now be 23 - 800

Also when subtracting your numbers just swap so the bigger number to the left as of course most people find 800 - 23 a lot easier than 23 - 800

using this method makes it a lot easier to do this trick in your head and impress your fellow nerds

[itioticginger]

My way for checking most annoying primes is to subtract multiples of the prime until it is easy. For example, with the 7 case: 872123. I subtract 840000, which is 12x7x10000. Once left with 32123 I subtract 28000, then I have 4123, I subtract 4200, then I have 77 and that is obviously divisible by 7

[Kaepsele]

For divisibility by any number n, you can also simply check the sum
sum a_i * (10^i mod n)
for divisibility by n (or calculate it in mod n directly). The (10^i mod n) is independent of the number used, so you can derive the divisibility trick.
E.g. for divisibility by 7, you'd need to remember: 1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5... repeating. So to check 6976984 for divisibility by 7 you'd do:
6*1 + 8*3 + 9*2 + 6*6 + 7*4 + 9*5 + 4*1 mod 7
= 6 + 3 + 4 + 1 + 0 + 3 + 4 mod 7
= 0 mod 7

Not really useful, but it recreates the tricks for all the other numbers too.

[aeniln]

Divisibility by 37 :
Separate the number in blocks of 3 like for 7. On the left of each block, write the left digit one more time, then separate this block of now 4 in two blocks of 2.
Do the alternating sum of all blocks of 2.
If the final result is divisible by 37, so was the original number.
6, 203, 346
006 203 346
0006 2203 3346
00-06+22-03+33-46 = 0*37
=> 37 | 6, 203, 346.
Also note that if a 3k digit number is divisible by 37, all its circular permutations are also.
The more you know.