Divisibility Tricks - Numberphile 

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.

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

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)

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.

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.

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

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 :)

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.

I actually learned this in school

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

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

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

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!

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

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.

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.

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.

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

27 mins of Numberphile. Yes. Yes. Yes!

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

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

9 is the largest single digit in Base 10, and 3 is a factor of 9, so that all makes sense. Given the proof for cross sums, It would appear that the same thing should work for other number bases. For example: that you can do a cross sum to check for divisibility of the number 5 in Base 6, or the number 12 in Base 13. And if it works for 12, it should also work for 2 or 3 or 4 in base 13. A quick check of my programming calculator, and the cross sums divisibility check works for 7 in octal, and 3 and 5 and 15 in hexadecimal.

I think you can actually generalize the tests for evenness, divisibility by 4, divisibility by 8, and so on. For any potential divisor 2^N, you only need to determine if the number formed by the last N digits is divisible by that divisor. So you'd need to look at 4 digits to test for divisibility by 16, 5 digits to test for divisibility by 32, etc.

I'm not great at maths, but I knew 464 was divisible by 8 bc it's 400 and 64 which are both easily recognisable as being divisible by 8.

There's a generalization you can do about divisibility by 2, 4 and 8 - number is divisible by 2^n if last n digits of a number are divisible by 2^n. Not quite useful on higher n values but still it is a valid rule that I think should've been mentioned. Also, second method of divisibility by 7 that Tony explains is appliable not to just 3 digit numbers but to any amount of digit numbers, and what's more interesting - it's recursive. Let's take that 6976984 as example: 6976984 -> 697698 - 2*4 = 697690, 697690 -> 69769 - 2*0 = 69769, 69769 -> 6976 - 2*9 = 6958, 6958 -> 695 - 2*8 = 679, 679 -> 67 - 2*9 = 49, 49 is divisible by 7, therefore 6976984 is divisible by 7. I find this one to be easier doing mental arithmetic than 'blocks by 3' one, because you only really need to keep track of one number and make simple operations of doubling and subtracting.

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.

Get Norrie on this channel or I RIOT!

With regard to that featured brilliant question, there are exactly six numbers less than 200 that have exactly three divisors. The solution for this one is simple: it is the set of the squares of prime numbers p such that p^2 < 200: 2^2 = 4 2x3 = 6 < Not in our set because 6 has 4 divisors, so we must ignore composite numbers that are not squares 3^2 = 9 4^2 = 16 < Not in our set because 16 has 5 divisors, so we must ignore squares of composite numbers 5^2 = 25 7^2 = 49 11^2 = 121 13^2 = 169 17^2 = 289 < Not in our set because 289 is obviously greater than 200, therefore it does not count And thus we have reached the end of the solution algorithm. Six numbers less than 200 with exactly three divisors. The rationale for this is that for a number n to have exactly three divisors, its prime factorization MUST be p^2 such that p is prime. Thus, the only three divisors of n will be 1, p, and n, and nothing else. And a quick proof by exhaustion: 1) If n is of the form q^2 for some composite number q, even if q is of a form p^2, then n will have more than one divisor, specifically: 1, p, q, pq (or p^3), and n. Five divisors, not three. 2) But if q is of a form ab such that a

There's a fairly straight forward way to derive these rules using remainders of dividing powers of ten. So 10^n mod 7 gives us {1, 3, 2, -1, -3, -2, 1, 3, 2, ...}. The alternating signs tell us that we can take these in blocks of 3 and do alternating sum. But further, for a 3-digit number, 100*a_2 + 10*a_1 + a_0, it's clear that it's divisible by 7 if a_0 + 3*a_1 + 2*a_2 is. Which, to me, is a simpler test, but your mileage may vary. For 11, same pattern of remainders is {1, -1, 1, -1...}, so we have a simple alternating sum. And for 3 and 9 it's {1, 1, 1, ...}, so that's the simplest case. In general, for any N not multiple of 2 or 5, this will produce a pattern at most (N-1) elements long, which might include some creative sign alterations, and can then be massaged into a practical rule. So if you want a harder one, lets do 13. The remainders are {1, -3, -4, -1, 3, 4, 1, -3, ...}. This one happens to repeat in just 6 elements instead of maximum possible 12, which is fortunate. Also, there is a pattern with signs creating groups of 3, but it's shifted by 1 from beginning. So taking a number like 847,559,388 the way you'd test it is by making groups of 3, but padding one's place with zeroes: 84 755 938 800. And do an alternating sum (right to left or left to right doesn't actually mater for any of these, since negative of a number has same divisors.) 800 - 938 + 755 - 84 = 533. So the whole number is divisible by 13 if 533 is. And again, {1, -3, -4} portion of the pattern can be used to test 3 digit numbers. 3 - 3*3 - 4*5 = -26, which is clearly divisible by 13. So the number was, in fact, divisible by 13. Overall, not any harder than test by 7, even if the rule itself looks stranger.

There is a bit simpler rule (at least for mental calculations) for divisibility by 8. Start with the last 3 numbers just as shown if the first of those is odd, then add or or subtract (whichever is easier) 4 from the last 2 digits Now if the last 2 digits are divisible by 8, then the number is divisible by 8. I’ll give a few examples: Using the example in the video, last 3 digits are 464. The first digit is even, so just check if the last 2 are divisible by 8. Which they are, so it’s divisible by 8. Xxx,324. First digit of last 3 is odd, so add (or subtract) 4 from 24 to get 28 (or 20). 28 (or 20) is not divisible by 8 so the number is not divisible by 8. 1536, first digit of last 3 is odd, so add or subtract 4 to get 40 or 32. 40 (32) is divisible by 8 so the number is divisible by 8. It does not matter whether you add 4 or subtract 4, just whichever is easier for you using those numbers. For instance, if the last two digits are 28, it may be easier for you to subtract 4 than to add 4.

At 11:43 Isn't it easier if you just do 46 + 4/2 = 48 and check if 48 is a multiple of four? The way he does in the video implies a multiplication and a final check of divisibility by 8, which is much more demanding than a sum and check of divisibility by 4. You can simplify the whole thing just by dividing everything by two

I tend to like t - nq instead (where n = [Divisibility number] - m), so the number I have to divide by is smaller. Example: Instead of 103 + 7(5 464 + 13(4) = 516 which is 43 * 12. But you could just as easily use 4644 -> 464 - 30(4) = 344, which is 43 * 8. I can't really remember two-digit multiplications of two-digit numbers, but I could intuitively recognise that 344 is smaller than 430 by 86, so it's an advantage in my case. (It's also easier to multiply by 30 than it is by 13.)

22:58 nice

The version of the trick you showed for divisibility by 7, also works for 11 and for 13!! That is, the first step, where you alternately add/sub 3-digit groups. Because it boils down to the "1001" trick - which you sort of allude to later along. That 3-digit +&- (repeated enough to get down to a 3-digit result) ends with a 3-digit number that preserves divisibility by 7, 11 and 13, because it amounts to subtracting multiples of 1001 = 7·11·13. Fred

Check out Brilliant: brilliant.org/numberphile (sponsor) Extra footage and trick from this interview: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ZWeUNUCf2rI.html Card videos on Numberphile: bit.ly/Cards_Shuffling Numberphile playing cards (limited and occasional supply): bit.ly/NumberphileCards I Love Norrie stickers (these are designed to annoy Tony): teespring.com/en-GB/i-love-norrie-numberphile

The 11's trick doesn't need to reverse order (the only difference is a possible sign change), making it a little easier to do in your head. Also, the first part of the 7's trick also works for 11's and 13's. Also, when you said the two facts you need are that 1001=7*11*13, and the second fact is that 999,999 is divisible by 7, the second fact actually follows from the first since 999,999=1001*999.

I have never inverted the numbers when checking divisibility by 11 and it has always worked well.

For once, Numberphile goes over a concept I was already familiar with! :D

I have the same gift as Norrie except I can tell if something is a multiple of 2.

Thank you. What fun! Have been fascinated with these "divisibility rules" since I was young, yet I could never remember the 7 and 11 ones for long after I learned them.

The cross sum trick for 3 and 9 is one of the first things we learn about divisibility. You didn't think of it on the huge prime check because it was so simple.

For multiples of 8, can you take the last 3 digits, divide them by two, and then do the check for 4; if that number is divisible by 4, then original is divisible by 8?

Here's an easy trick I use on checking if a number is divisible by 8. If the hundreds place digit is even and the last two numbers are divisible by 8, then the number is divisible by 8. If the hundreds digit is odd, then if the last two digits plus 4 is divisible by 8, then the number is divisible by 8. Thanks for making awesome videos!

Still waiting for a follow up to the Feigenbaum Constant video...

This is the best numberphile ever! I can make people think I can see factors of huge numbers now. Incredible. I needed this in calculus

4:54 There's only one explanation for missing something that obvious - he must've had a Parker thought XD

Divisibility rules ik of up to 50: 1: Any whole number 2: Ends in an even number 3: Cross sum divisible by 3 4: Last 2 digits divisible by 4 5: Ends in 5 or 0 6: Both divisibility rules for 2 and 3 apply 7: Double the last digit of the number and subtract it from the rest, repeat until you get to a number you know is divisible by 7 8: Last 3 digits divisible by 8 9: Cross sum divisible by 9 10: Last digit is 0 11: Subtract the last digit from the number and repeat until you have a number you know is divisible by 11 12: Rules for 3 and 4 apply 14: Rules for 2 and 7 apply 15: Rules for 3 and 5 apply 16: Last 4 digits divisible by 16 18: Rules for 2 and 9 apply 20: Rules for 4 and 5 apply 21: Rules for 3 and 7 apply 22: Rules for 2 and 11 apply 24: Rules for 3 and 8 apply 25: Ends in 00, 25, 50, or 75 28: Rules for 4 and 7 apply 30: Rules for 2, 3 and 5 apply 32: Last 5 Digits divisible by 32 33: Rules for 3 and 11 apply 35: Rules for 5 and 7 apply 36: Rules for 4 and 9 apply 40: Rules for 5 and 8 apply 42: Rules for 2, 3 and 7 apply 44: Rules for 4 and 11 apply 45: Rules for 5 and 9 apply 48: Rules for 3 and 16 apply 50: Ends in 00 or 50 d i s l i k e p l s x d d D &

Those tricks are cool, but divide by zero then I’ll be impressed

16:10 any number you put at the beginning of an odd-lengthed number can also be put at the end. since A-B+C-D+E-F = -(F-A+B-C+D-E). In fact, if the alternating sum ends up being a single-digit, positive number, you can always put that number on the end.

You shuffled the cards in the graphic. The cards displayed are 42135 and the graphic is 42315

a neat thing about the proof for the 7’s trick: not only is 142857 related to cyclic numbers, but it is also the repeated decimal expansion of 1/7, which is related to the proof that 0.999999… = 1

For divisibility by 8, I'd just halve the last 3 digits and check if that's divisible by 4 (check last 2 digits).

You can generalize the 9's proof to any base. Take your base number N (10 for the normal decimal system). Find the factors of N-1 (in our case 9, whose factors are 3 and 9). If the cross sum of a target number in your base is divisible by any of those factors, so is the original number. An example in hexadecimal (N = 16) N-1 = 15, whose factors or 3, 5, and 15. Our target number = fa927e3 (262744035 in decimal) The cross sum = f + a + 9 + 2 + 7 + e + 3 = 3c (60 in decimal) which is divisible by 15, so the target number must be as well.

I learned a bunch of divisibility tests in elementary school and they haven't been mentioned anywhere since. Crazy to think if not for then I wouldn't even know of any until now

This is magic ASMR which puts me to sleep. Not because it's boring, but because it's really interesting. My overworked brain shuts up so it can listen.

This is the first time I've seen the 7 trick. I have seen more than one book in which the author states there is no divisibility test for 7.

"I lost sight of the beauty of maths" one of the most elegant lines I've ever heard.

Sounds as though Norrie has synesthesia, an overlapping of senses

Dividing by 4 has another step if you do not recognize the last two digits as a multiple of 4. It is similar to the rule given for 8. Double the second last digit and then add the last digit. If the result is a multiple of 4, then the original number is a multiple of 4. For example, 465: the number is not even, so can not be a multiple of 2, let alone 4. 768: this is even, so could be divisible by 4, but 68 is not on my memorized 4-times list, so double the 6 and add the 8. 6*2=12+8=20. 20 is a multiple of 4, so 768 is also a multiple of 4. To simplify the divisibility rule for 11, you don't have to reverse the order before alternate cross-sums; just be prepared to handle negative numbers as multiples of 11. e.g. 42135 -> 4-2+1-3+5 = 5, so this is not a multiple of 11. 421630 = 4 - 2 + 1 - 6 + 3 - 0 = 0. Therefore 421630 is a multiple of 11. Another short-cut. In a 3-digit number, if the first and last digits add to the digit in the middle, it is a multiple of 11. E.g. 473 -> 4+3 = 7. Therefore divisible by 11. This can also be checked by alternating cross-sums. 4 - 7 + 3 = 0. This short-cut does not work for all 3-digit multiples of 11, but it the ones it works for are always multiples of 11. e.g. 759 7+9 = 16, so that fails, but 7-5+9 = 11, so this is a multiple of 11.

For the 8 test isn't it easier to check if half the number is divisible by 4 ?

One of my favourite tricks is to ask people to factor 899. For reference, it factors into 29 x 31. That is, of course, a ridiculous ask, unless you realise that 899 can be expressed as 900-1. And both 900 and 1 are perfect squares. Factor the difference of squares to get: 899 = 900 - 1 = (30 + 1)(30 - 1) = 29 * 31

An easier check for the 3 digit number for 8s would be if the hundreds digit is even just check the last 2, if the hundreds is odd, subtract 4 from the last 2 and then check it

In general, to test if any number in base b is divisible by d, multiply each digit by b^n (mod d), where n is 0 for the rightmost place, 1 for the place to its left, etc. (i.e. in base 10 multiply the digits from right to left times 1 mod d, 10 mod d, 100 mod d, etc. ) If the sum of these products is divisible by d, then the original number was well. For example, to see if the hexadecimal number 4EF2 is divisible by ten ("A"), we have (2*1)+15*(16 mod 10)+14*(256 mod 10)+4*(4096 mod 10)=2*1+15*6+14*6+4*6, or regrouping, 6(15+14+4)+2=198+2=200, so it is (4EF2 hex = 20210 decimal) (and we can see here, then, that the divisibility for ten (A) in hexadecimal is "drop the units digit, multiply the cross sum of the remaining digits by 6, and add back in the unit, see if this sum divisible by ten("A")") This is the underlying basis for all your divisibility tests shown here.

749 divisible by 7 - "it's hard, beyond my mental arithmetic" Me: bro it's just 700 and 49. That's cake lol

That divisibility rule for 3 and 9 ultimately stems from the fact that we use base-10 numbers. There is a corresponding "divisibility by base-1" rule for every system; e.g. divisibility by15 in the hexadecimal system or by 7 in the octal scheme.

Not sure if this is covered already in the Numberphile, but this is essentially the concept of congruence. 10 = 1 (mod 3) 10 = 1 (mod 9) 10 = 2 (mod 4) 10 = 2 (mod 8) 10 = -1 (mod 11) 10 = 3 (mod 7) or 1/10 = -2 (mod 7) & 1000 = -1 (mod 7*11*13) And 1/10 = 7 (mod 23)

Alternatives: A number is divisible by 4 if (2*tens+units) is divisible by 4. A number is divisible by 8 if (4*hundreds+2*tens+units) is divisible by 8. General case: a number is divisible by 2^n if (2^n*(a.n)+(2^(n-1))*(a.n-1)+...+2*a.1+a.0) is divisible by 2^n. A number is divisible by 7 if the digit sum, when digits are taken in groups of 6, is divisible by 7. A

One of the best videos in a while on this channel! That 749 was funny though.

Ahh, the moment when you downloaded the C. C. Briggs first 1,000 divisibility rules for primes two months before the actual Numberphile video and know exactly what they're talking about even before they begin

For divisibility test you don't need to reverse the digit just take sum of digits at odd places and subtract it from sum of digits at even places...if it is a multiple of 11 the number is multiple of 11.... There is no need of reversing 🙃

General case divvy-number (odd numbers that don't end in 5) for n- 10a+b is: 9a+1 for b=1, 3a+1 for b=3, 7a+5 for b=7, a+1 for b=9. By the way, the "blocks of 3" test works for 7, 11, *and* 13 because 1001 divides them all.

The 7's second divisibility rule can be made for all primeslarger than 3 : take a prime (say 23), multiply with some number to get the last digit to be 1 (23*7=161). The rule is to subtract the number of this product without the last digit (16) times the last digit of the number under consideration from the number formed by the rest of the digits of that number. This process can be continued indefinitely: if the number obtained is divisible by the prime, so is the original one. As an example: 2323->232-16*3=184->18-64=-46 and 23|-46 so 23|2323. Actually, sometimes it is easier to use another very similar rule: instead of subtracting 16*last digit, one can add (23-16)=7 (that is -16(mod 23)) times the last digit and this rule works as well. As one more example, let's generate a rule for dividing by 37. 37*3=111 so -11 is one fitting number and 37-11=26 is the other. Check 3959=107*37. With -9: 3959->296->-37. With 26: 3959->629->296 and 37|-37 and 37|296.

You could also do an easier test for 11, The difference between the sum of the odd and the even digits must be divisible by 11 (including 0) Examples: 8316 =(8+1)-(3+6)=0 11027016 =(1+0+7+1)-(1+2+0+6)=0 93163582512 =(9+1+3+8+5+2)-(3+6+5+2+1)=28-17=11

Let's call multiples of 3 "threeven" from now on.

The 9s (and 3s) are easy to do in your head even if you have a large number, remove multiples of 9 (or 3). If you are left with 0, the number is divisible by 9 (or 3). This works because you're basically doing the modulus function. For 945136872 : remove 9, 4+5, 3+6, and 7+2 leaving 1+8 which you can also remove, leaving 0.

Why is there calendar for 2012 on the wall?

Great video! I have an bit of a nicer proof for the 3 and 9 rule. 1. The cross sum of any number N will be 1 more than the cross sum of N -1 or will be 1 2. With this, you can picture a circle of numbers from 1 to 9, and each consecutive number takes on step around the circle. 3. Since there are 9 numbers on the circle, taking 9 steps around brings you right back to where you started, and taking 3 steps at a time will have you always land on the same 3 spots. 4. This means this trick works for different numbers in different bases! In base B, this trick works for B-1. Base 8 would mean that any number whose cross sum is 7 is divisible by 7. In base 13, this trick would work for 2,3,4,6, and 12!

I do /3 on licence plates while driving all the time

Possibly easier way of working out if a number is divisible by 11: Subtract the last digit (i.e. units) from the 10s digit ('borrowing' one from the hundreds if necessary. Then subtract what you have left from the thousands digit etc. If, when you subtract from the leading digit, you are left with zero, the number divides by 11. e.g. 1573 7 - 3 = 4 5 - 4 = 1 1 - 1 = 0 Therefore divides by 11.

I was shown these tricks in elementary school. Are these not common knowledge elsewhere or have you dang kids just been using calculators so much that no one uses them anymore?

Fun carnival tricks, but you can imagine that these methods may have been pretty important before we all walked around with small supercomputers in our pockets.

749 = 700+49 (both divisible by 7, thus the whole number is also a multiple of 7).

This is why I love math. I swear sometimes I see more when I learn things like this. I feel like I acquire a diluted version Norrie's skill. I need more. I said before I need to watch all the numberphile videos from first video to recent video, I hope I follow through sometime soon.

what's the rule for finding numbers divisible by 0 tho

For 7, you can also just go digit by digit starting from the left. Each time, you'll multiply what you have by 3 and add the next digit (casting out 7s / working mod 7 the whole time). As an example: 456739 (mod 7) 4*3+5=17=3 (mod 7) 3*3+6=15=1 (mod 7) 1*3+7=10=3 (mod 7) 3*3+3=12=5 (mod 7) 5*3+9=24=3 (mod 7) so then 456739 = 3 (mod 7) not 0 (mod 7) so 456739 is not a multiple of 7. [In particular 456739 = 65248 * 7 + 3, where it's clear that the remainder is 3 after dividing by 7.] With this method, you are essentially replacing the starting number with a new shorter number that'll have the same remainder after dividing by 7. For this example those numbers are: 456739 036739 001739 000339 000059 000003 This makes it more clear that this method is extremely similar to regular "short division" but you only care about the remainders, not the quotients. If instead of 1 digit at a time, we want to do 2 digits at a time, we break up our number into pairs of digits and we'll change the multiplier to 2. 456739 45*2+67=3*2+4=10=3 0339 03*2+39=6+4=10=3 (mod 7) so 456739 = 3 (mod 7) as before [as a note, in the modular arithmetic parts, those = signs should be triple-line congruence signs instead, but I don't know where those are in this keyboard layout.]

when teaching my students to reduce fractions I show them that handmade questions overwhelmingly only require you to check for common factors of 2, 3 and 5, since 7 behaves strangely (mostly due to how it interacts with base-10) and 11 and up are generally too large for test makers to focus on (when looking at fraction to decimal conversions, though, 9 and 11 are related, since n/9 yields a repeating decimal of n*11, and n/11 yields a repeating decimal of n*9, but 7 becomes even more bothersome). 2 is easy, since both just need to be even, 5 is easy since they just need to end in 0 or 5, and 3 is easy b/c the sum of the digits will be a multiple of 3, which is a process you can keep iterating, or which you can simulate by just crossing off sets of digits that add up to 3, 6 or 9.

I prefer doubling the last digit then subtracting from the rest of the number for divisibility by 7. Rinse and repeat until you get down to 2-3 digits

I always had a fascination about numbers. When I was 10-11, there was a weekly math competition organized by the teacher to help us learn and gain speed, especially for multiplication and division. That's how I figured out the divisibility tricks up to 10 and their compounds, except for 7

Well I knew the first trick when I was 6.

To test for divisibility by 7, chop the last digit off the number, double that last digit, then subtract from the rest of the number. Repeat until you get a multiple of 7 (or not). Example 1: 336 -> 33 - 12 = 21, which is a multiple of 7 (42 x 7) Example 2: 2401 -> 240 - 2 = 238 238 -> 23 - 16 = 7....so 2401 is also a multiple of 7 (343 x 7) Example 3: 48167 -> 4816 - 14 = 4802 4802 -> 480 - 4 = 476 476 -> 47 - 12 = 35.... so 48167 is a multiple of 7 as well (6881 x 7)

I learned it when I was 9 years old, 3rd grade.

18:22 I do like these methods but there is another general method that is mathematically boring but a quicker for me. All you need to do is know the first 10 terms in the times table and repeatedly subtract multiples of 10 of the largest one that doesn't make the number negative. e.g. is 6976984 divisble by 7?: 6976984 -> 676984 -> 46984 -> 4984 -> 84. In the first step I subtracted 6300000, then 630000, then 42000, then 4900. I end with 84 which is divisble by 7. While it seems like this has more steps, I find it a lot quicker than the 7s, 11s, or general method in this video. It'll work fine for 13, 17, 23, etc. but I would struggle with it for three digit factors and need some paper. At that point this video's general method may be easier.

7:54 (Pontifically): “Let it henceforth be known that the term ‘whole bunch of’ shall be defined as exactly, with no more and no less, equal to n-1 ! Well... Whatever n is...”

8 divisibility can be checked easier by adding another step in the middle: 1,752,904 is divisible if 904 is divisible if (here I take and remove 200's as much as possible since it's divisible by 8) 104 is divisible if 10 * 2 + 4 is divisible which is 24 and it is We don't have to calculate 2 * 90 + 4 which is 184 and then probably do it again.

When you have even length palindrome numbers, they're divisible by 11.

To test for divisibility by 7 on one hand: 1. Clear the hand (all fingers up). 2. Multiply by 3. 3. Add the next digit. 4. If three fingers in a row are down, or the thumb, middle finger, and pinky are down, raise them. 5. If the pinky is down but the index is up, put the index down and the pinky up. If the ring is down but the thumb is up, put the thumb down and the ring up. 6. If any digits remain, go to step 2. You can occasionally overflow the hand. How can you handle this?

What a divisive video

Another way to determine if a number is divisible by 7: Convert it to base 7 notation and see if the last digit is 0. (A reminder of how to convert a decimal number to base 7:) Divide the number by 7 and note the remainder. Divide the new number by 7 and note the remainder. Repeat as often as necessary. Write the noted remainders in reverse order to give your 'base 7' version of the original number. Example. 1000/7 = 142 rem 6; 142/7 = 20 rem 2; 20/7 = 2 rem 6; 2/7 = 0 rem 2. Hence 1000 (decimal) = 2626 (base 7) And it is easy to see that 2626 (base 7) is not divisible by 7 because it does not end in 0.

There is a simpler way to show that 999,999 is divisible by 7: it's 1001×999 and you just did 1001

Many years ago I was taught the 11 divisibility rule slightly diferent: 1) Sum the figures in odd positions. Let's call it a. 2) Sum the figures in even positions. Let's call it b. 3) If a - b is 0 or 11 or a multiple of 11, then the number is divisible by 11. And no need to reverse the number at the beginning.