Divisibility Tricks in 5 Levels of Difficulty 

I didn't know there were more divisibility tricks than just adding up the digits. Thank you Dr. Sean!

Alternative trick for 7: If you can get good at finding the remainder mod 7 of 3 digit numbers you can find the remainder mod 7 of any sized number. And finding the remainder mod 7 of 3 digit numbers is easy. 1) double 100s digit 2) add result to the last two digits 3) result is congruent to the original number Example: 463 2x4 = 8 8+63 = 71 71 = 70 + 1 Therefore, remainder mod 7 of 463 is 1. Now say you have a large number, like 901,885,519. Starting on the right, group the number into blocks of 3 digits and do the alternating sum (similar to the trick for 11 described in the video): 519 - 885 + 901 The result of this calculation will be congruent mod 7 to the original number. But since that's a hard calculation to do in your head you can simplify it by converting each 3 digit block to its remainder mod 7 using the method described above: 519 -> 2x5 = 10, 10+19 = 29, 29 = 28 + 1 885 -> 2x8 = 16, 16+85 = 101, 101 = 98 + 3 901 -> 9x2 = 18, 18+01 = 19, 19 = 14 + 5 Now the alternating sum calculation can be simplified to: 1 - 3 + 5 = 3 Therefore original number is not divisible by 7, and in fact has a remainder of 3 mod 7. Takes a little bit of practice, but not as much as you might think. There are ways to streamline things further too. Not hard to do large numbers in seconds.

Next video suggestion: why 1 is not a prime.

There is one for 13: Take your number, divide it with D the number of tens, and N the digit of units (for instance, for 1782, D = 178, N = 2) and consider D + 4N, and repeat until you get a number smaller than 50. Your original number is divisible by 13 if and only if your final one is (13, 26 or 39)

A trick that only works sometimes, or might make other tricks simpler: Replacing known multiples of a number anywhere in the digits with zeros. This is a subtraction in a flimsy disguise. e.g. Is 463 divisible by 7? We might know that 63 = 7×9, so we can replace "63" with zeros and we get 400, which is clearly not divisible by 7, so neither was 463. Another handy trick is knowing that 1001 = 7×11×13. Combining these we can ask if, say, 4494 is divisible by 7. Zap the 49 (7×7) from the middle for 4004 and that's obviously 4×1001 which we know is divisible by seven, so 4494 must have been as well.

One of my favorite proofs I did was in number theory, it was for the final exam, the professor wanted us to prepare a one on one session, where we would go over a proof from class expanding on it if we wanted to. I did division by 9, then expanded that proof to all bases. The conclusion was for a base N if the sum of a numbers digits is divisible by N-1, then the number itself is divisible by N-1.

You finally made a video about division🎉🎉🎉What a legend!

That you for including the “ interbase “ divisibility level 5. I hadn’t thought about stuff like that before!

I've never seen those tricks before. Awesome! Thanks for the content.

Level 0: Divisibility rule for dividing by 1

The first vid of yours I watched was good, this is also good. Subbed.

A number is divisible by n if the digit sum of the number is divisible by n. That is always true, you just have to use base n+1.

Long ago, really long ago, my teacher, after having telling us about how to know whether or not a number is divisble by three, wrote some huge, pre-selected, numbers on the board. Summing all the digits would still give some 3 or 4 digit number and we were at loss then because we didn't know whether or not that number was divisible by three. So he said to us that given a number, we had to add the digits and if this sum of the digits produced a two digit number or greater, repeat the procedure each time repeatedly untill it terminated on either 3, 6 or 9. I remember he's smile bc we, 6 and 7 years old, were happy with this strange knowledge and kept literally demanding for him to write more numbers so we could apply this.

I really enjoyed this video thanks for making it.

Thanks

🔥🔥

There’s a general right-to-left trick for numbers n, with gcd(n, base)=1. Add or subtract a multiple of n to get a multiple of base, remove the 0 at the end, and repeat until the verdict is obvious. This trick will return a yes for any number, but only a no for numbers whose last digit has no common factors with the base.

If 10's place is odd, and it ends with 2 or 6, divisible by 4 If 10's plaxe is even, and it ends with 0 4 of 8, divisible by 4

To check divisibility by 13 multiply the last digit by 4 and add it to remaining number. For example 104 becomes 10+4*4=26

if x is divisble by y then you can check whether any number is divisble by y^n with only the first n digits if the number is written in base x

Here even before 10k

What about divisibility by 1

That’s the first time I found out why the divisibility reals work. I never saw this seven method

Level 0: divisibility by 1

My divisibility check works for all integers of reasonable size and is very fast: I pick up my mobile phone and divide the numbers. If there is a decimal point and something after it, the numbers are not divisible.

Divisibility trick of 25 and 125 comes under level 2 and they are arguably easier than 4 and 8