Why 7 is Weird - Numberphile

2022 ж. 21 Там.
1 811 217 Рет қаралды

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This video features Dr James Grime on divisibility.
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Пікірлер
  • Divisible by 7 can be useful in figuring out if there is a whole number of weeks in a number of days.

    @andreroussel@andreroussel Жыл бұрын
    • I came here to make that exact point :)

      @amysteele2488@amysteele2488 Жыл бұрын
    • Could you demonstrate that please, thank you. :)

      @Reachermordacai@Reachermordacai Жыл бұрын
    • @@Reachermordacai "hey dude I see on sundays, I gotta work the next 2794 days!" "Damn so you stop work in the middle of a week?" "How did you figure that one out?"

      @Nate-bd8fg@Nate-bd8fg Жыл бұрын
    • Yeah but that isn't very useful at all unfortunately. We use days and months most of the time

      @YounesLayachi@YounesLayachi Жыл бұрын
    • @@YounesLayachi But we also use weekdays and weekends.

      @muizzsiddique@muizzsiddique Жыл бұрын
  • this feels like a very old-school numberphile video, love it

    @mazza420@mazza420 Жыл бұрын
    • Yeaah! It really does! fun!

      @shpensive@shpensive Жыл бұрын
    • When I saw this video on KZhead at a first glance, I thought I was looking at some old video from seven years ago or something, was surprised when I realized it was from today

      @jimi02468@jimi02468 Жыл бұрын
    • James Grimes is classic!

      @robinsonrom@robinsonrom Жыл бұрын
    • @@jimi02468 Same. It's quite funny

      @julianw10@julianw10 Жыл бұрын
    • What's changed

      @lucasng4712@lucasng4712 Жыл бұрын
  • 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!"

    @12Q46HPRN@12Q46HPRN Жыл бұрын
    • My rule was doubling the last digit and subtracting it from the rest of the number.

      @darpanjain4250@darpanjain4250 Жыл бұрын
    • yo man, will there be ant man 4?

      @BlueWhiteWiper@BlueWhiteWiper Жыл бұрын
    • my rule was imagine that number was base 3 and convert to denary so 343 3 + 4x3 + 3x9 = 42 divisible

      @dabama7483@dabama7483 Жыл бұрын
    • @@darpanjain4250 i used the same. If we try to prove the method, the process remains same for both methods.

      @phanibhushantholeti9446@phanibhushantholeti9446 Жыл бұрын
    • V0

      @simonkara4907@simonkara4907 Жыл бұрын
  • 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)

    @luketurner314@luketurner314 Жыл бұрын
    • For the ones with a y coefficient greater than 10, we could split the number like so: 100a + b; where b is the last 2 digits and a is the rest: 37: a +10b (k=10, j=999) 43: a - 3b (k=6, j=559; originally 6b - 2a which is even and backwards, so divide by -2) 47: a + 8b (k=8, j=799) 53: a - 9b (k=18, j=1749; originally 18b - 2a -> divide by -2) 67: a - 2b (k=12, j=1139; orig. 12b - 6a -> div. -6) Couldn't find a decent formula for 73, 83, and 97 in a timely manner. I might revisit this later

      @luketurner314@luketurner314 Жыл бұрын
    • Isn't iterability always necessarily given by the nature of these kinds of divisibility proofs?

      @leo848@leo848 Жыл бұрын
    • @@leo848 I suppose, but for 2, 5 and therefore 10, for example, you don't need to iterate since you're just looking at the last digit (is it even?, is it 5 or 0?, and is it 0? respectively). For the alternating sum for 11, I have not taken the time to reverse engineer its algebraic derivation; it too may have the same property because it may be a similar algorithm.

      @luketurner314@luketurner314 Жыл бұрын
    • @@leo848 not really. You can construct a finite state automaton (for instance, a 3-state automaton for divisibility by 3) that, after you run the digits through its arrows will straightaway give you a ‘yes’ or a ‘no’ without any need for iteration. I guess, we're just more used to addition and stuff, and devise these chevksum kind of tests, which fall into the trap of iteration.

      @nnaammuuss@nnaammuuss Жыл бұрын
    • @@nnaammuuss Of course. I meant that once you have a divisibility test that gives back a number that is divisible iff the original number is, one can always iterate.

      @leo848@leo848 Жыл бұрын
  • 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.

    @melaniehall5885@melaniehall5885 Жыл бұрын
    • Wow I’m in the same boat exactly. Can’t believe it’s been 10 years

      @juliusreiner5733@juliusreiner5733 Жыл бұрын
    • Exactly the same!! Finally getting myself to watch numberphile again after all these years. And it's just as brilliant.

      @Cannitbliss@Cannitbliss Жыл бұрын
    • I swear he hasn't aged a day in that time

      @matt69nice@matt69nice Жыл бұрын
    • It's awesome to watch him explain anything about math.

      @CafePorLaNoche@CafePorLaNoche Жыл бұрын
    • That's 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.

      @nicolestewart2843@nicolestewart2843 Жыл бұрын
  • 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.

    @3rdand105@3rdand105 Жыл бұрын
    • the test defeats the purpose of the test because it takes 5 times longer than just dividing by 7.

      @BeBopScraBoo@BeBopScraBoo Жыл бұрын
    • @@BeBopScraBoo But it doesn’t test your mental arithmetic so much (pardon the pun).

      @PC_Simo@PC_Simo Жыл бұрын
    • @@BeBopScraBoo this is exactly what i was thinking 😂😂

      @kkdpsudpsu@kkdpsudpsu Жыл бұрын
    • @@BeBopScraBoo Yes, and the video is also completely missing the stated topic. I wanted to know why non-roundly dividing by 7 always creates the same decimal pattern.

      @Dowlphin@Dowlphin Жыл бұрын
    • @@Dowlphin decimal system is based on ten, which factors to 2 and 5. any fraction with a denominator that factors with any other number will have a repeating pattern.

      @BeBopScraBoo@BeBopScraBoo Жыл бұрын
  • 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.

    @nicolestewart2843@nicolestewart2843 Жыл бұрын
  • Even Numberhile now recognizes the elegance of the number 7 Thala for a reason

    @mememan7682@mememan76823 ай бұрын
    • 7 is the GOAT of multiple things

      @user-yz2xl1tu6t@user-yz2xl1tu6t2 ай бұрын
  • 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.

    @frasco_5518@frasco_5518 Жыл бұрын
    • this is brilliant

      @kraken4354@kraken4354 Жыл бұрын
    • :o that's given in my book

      @arnoygayen1984@arnoygayen1984 Жыл бұрын
    • What about the magic 11s in pascal's triangle. Your algorithm reminded me of that.

      @ChrisSwaningAround@ChrisSwaningAround Жыл бұрын
    • I'd like to see a proof of this. This looks like coincidence so far to me.

      @michaelempeigne3519@michaelempeigne3519 Жыл бұрын
    • @@michaelempeigne3519 no idea what the proof is but it worked with every number i tried

      @frasco_5518@frasco_5518 Жыл бұрын
  • I use divisibility for finding primes and root approximations during tutoring. I lacked the 7 test so, thanks for that!

    @FHBStudio@FHBStudio Жыл бұрын
    • I usually find the closest number divisible by 70 then subtract/add from that. It’s not really a trick for 7 but it’ll always work.

      @dominicellis1867@dominicellis1867 Жыл бұрын
    • 7 has a lot of ways for divisibility determination

      @thethirdjegs@thethirdjegs Жыл бұрын
    • One trick that might help for finding primes is that except for 2 and 3 all primes follow the pattern of a multiple of 6 plus or minus 1. This is true because +2 would be divisible by 2, +3 by 3, +4 by 2, and +5 is also -1. Unfortunately I discovered this long after I needed to write Basic computer programs to find primes in school.

      @thomaswilliams2273@thomaswilliams2273 Жыл бұрын
    • Because of this video, I found out about the 1001 test, alternate sum of 3 digits: 123456788 is divisible by 7 because 123-456+788=455 is divisible by 7! Best part is that this test works with 11 and 13... 455 is also divisible by 13 so 123456788 is divisible by 13. This is golden when testing for primes!!

      @ifulea@ifulea Жыл бұрын
    • @@thomaswilliams2273 it will catch all the primes but you also have to filter out the multiples of 5’s.

      @dominicellis1867@dominicellis1867 Жыл бұрын
  • 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.

    @Glizzygobbler6969@Glizzygobbler6969 Жыл бұрын
    • 6468=6300+168 =900*7+24*7=924×7

      @goatgamer001@goatgamer0018 ай бұрын
    • the method in the video can be programmed into a computer

      @bornts8944@bornts89444 ай бұрын
    • ​@bornts8944 Use the modulo operator for computer and compare the result to zero. Works for any number.

      @itsmemailingyou4234@itsmemailingyou4234Ай бұрын
  • 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

    @dajaco81@dajaco81 Жыл бұрын
    • This sounds much easier than what was shown in the video.

      @loganroy3381@loganroy3381Ай бұрын
  • 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 ✌️

    @pietronardelli622@pietronardelli622 Жыл бұрын
    • Like using the missing fingers to do your nine times tables.

      @Celtic_Thylacine@Celtic_Thylacine Жыл бұрын
    • I love the fact that everyone keeps saying this video feels like it was from years ago but no one is commenting how Dr. Grimes didn't age that much. He's a vampire who knows all the secrets of numbers

      @ChrisM-qo1jc@ChrisM-qo1jc Жыл бұрын
    • It's the hidden beauty of numbers, and I love when I learn a new one.

      @izme1000@izme1000 Жыл бұрын
    • @@ChrisM-qo1jc The Count! 5 fingers, 4 fingers, bwahaha...! 💚✌️😎

      @erinmcdonald7781@erinmcdonald7781 Жыл бұрын
  • 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!

    @StevenStJohn-kj9eb@StevenStJohn-kj9eb Жыл бұрын
    • Mathematicians -hate- _love_ it!

      @AaronOfMpls@AaronOfMpls Жыл бұрын
    • The twinkle of mischief in his eyes was a nice touch too!

      @backwashjoe7864@backwashjoe7864 Жыл бұрын
    • If it appeared in the title, I'd have passed on it. Everyone knows that "This one weird Trick!" is web-speak for clickbait! 🙂

      @terryjwood@terryjwood Жыл бұрын
    • I kept looking for the CONTINUE button

      @richardsmith881@richardsmith881 Жыл бұрын
    • This weird trick will allow you to anger people in the comment section *Read more*

      @medexamtoolsdotcom@medexamtoolsdotcom Жыл бұрын
  • This is such a great video.. i’ve shared with my kids and the “presentation” with special british commentary and voice intonations is perfect. Useful and just enough entertainment!

    @chrisz6860@chrisz6860 Жыл бұрын
  • Wow. I love this. I really love when the concept is abstracted out. Fantastic.

    @theforeverpuddle8754@theforeverpuddle8754 Жыл бұрын
  • 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

    @JordynPi@JordynPi Жыл бұрын
    • Yeah, I think its easer way

      @Elnadrius@Elnadrius Жыл бұрын
    • The given method is pointless, much easier just to divide by 7.

      @christopherellis2663@christopherellis2663 Жыл бұрын
    • I just subtract 7 over and over again. To check whether the number 70,000 was divisible by 7, I had to subtract 7 10,000 times. Turns out it is!

      @EebstertheGreat@EebstertheGreat Жыл бұрын
    • @@EebstertheGreat So easy, Eebster10010100.

      @jamielondon6436@jamielondon6436 Жыл бұрын
    • @@christopherellis2663 it was joke :D

      @lydianlights@lydianlights Жыл бұрын
  • 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

    @brian5553@brian5553 Жыл бұрын
    • that's really odd!

      @duimaurisfootball8134@duimaurisfootball8134 Жыл бұрын
    • Cool! I've tried something like that as well, and it seems that powers of 7 all eventually reduce down to 49 as well: 7^2 = 49 --> 49 7^3 = 343 --> 49 7^4 = 2401 --> 245 --> 49 7^5 = 16807 --> 1715 --> 196 --> 49 And so on.

      @Merione@Merione Жыл бұрын
    • Loops like this remind me of Goldbach’s conjecture… nice result!

      @antonmiserez934@antonmiserez934 Жыл бұрын
    • One gets the next term in those sequences by multiplying by 5 modulo 49.

      @w.nickel2792@w.nickel2792 Жыл бұрын
    • @@antonmiserez934 I think you mean the Collatz-conjecture

      @dennismuller1141@dennismuller1141 Жыл бұрын
  • You have to love James Grime. Really! His passion, energy, and love for mathematics are just splashing on me from the monitor. Truly fantastic.👏 I wish I had such a passionate teacher/lecturer for my math lessons. I'm glad I could enjoy on YT for the least. 🙂

    @jirinovotny9704@jirinovotny9704Ай бұрын
  • This channel needs more James Grime 🥰

    @excentriqueesque@excentriqueesque Жыл бұрын
  • 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.

    @GregAlpar@GregAlpar Жыл бұрын
    • 100x + y -98x (multiple of 7) = 2x + y

      @alaskaoptimumvamps8127@alaskaoptimumvamps8127 Жыл бұрын
    • Nice, this was actually the answer i expected from the video. For smaller numbers (under 6 digits) i think your solution is easier. For larger numbers i think the videos solution is better as each step requires multiplying 1digit by 5 and reduces the test number by. 1 digit, the mod 100 version each step requires multiplying an n-2 digit number by 2, it potentially reduces the number of digits by 2 but 3/10 operations will only reduce it by 1. Regarding finding the remainder the video requies 1 extra step, each iteration of the algorithm multiplies the remainder by 5, if you keep track of the number of iterations mod 6 you can multiply your final remainder by (1,3,2,6,4,5) mod 7.

      @someknave@someknave Жыл бұрын
    • Cool! I think the method shown in the video also preserves mod.

      @joseville@joseville Жыл бұрын
    • @@joseville the method in the video multiplies the remainder by 5 for each iteration of the algorithm.

      @someknave@someknave Жыл бұрын
    • This is very cool! I "discovered" the same method while staring at the digital clock and splitting the number at the colon when I was little. Thought I was one of the great mathematicians of the century.

      @shanghandi-notrelatedtomah8534@shanghandi-notrelatedtomah8534 Жыл бұрын
  • Just seeing Dr. Grime already tells me this is going to be sweet. Loved it.

    @PieterDeStickere@PieterDeStickere Жыл бұрын
    • Dr Grime looks so much older than the old days of the channel. Now I feel old 💀

      @iwatchwithnoads7480@iwatchwithnoads7480 Жыл бұрын
    • @@iwatchwithnoads7480 And somehow he got even better at explaining

      @profbbfab6211@profbbfab6211 Жыл бұрын
  • It's much easier to find the nearest obvious number that can be divided by 7, e.g. 7000. Minus 6468. It's 532 and check if it's can be divided. Btw, you can do the same for 532 and 700. or 560 (7*80) - 532 = 28. Easy even doing in mind.

    @alexeyUK@alexeyUK Жыл бұрын
  • Disiion by 7 will be useful for following cases: 1) Hand calculation of Area of circle , volume of sphere etc ... involing pi=22/7 2) Days and weeks conversion.

    @crimsoncanvas51@crimsoncanvas51 Жыл бұрын
  • In a training material for level 2 math olympiads here in Brazil they outline this technique. It is lovely. It was very rewarding to learn this, I love the number 7.

    @lauramourasantos7379@lauramourasantos7379 Жыл бұрын
  • 1:09 I do it by head instead, decomposing the multiples. 434 -14 = 420 7x60 = 420 7x2= 14 434 = 7x62

    @Manu-Official@Manu-Official Жыл бұрын
  • Loved it, thanks so much

    @faithmargeuxcaubang7037@faithmargeuxcaubang7037 Жыл бұрын
  • I had learnt divisibility tests in school. But we hadn't learnt any for 7. I genuinely though it wasn't possible, (thinking they would've taught it if it was that way) But that 10x + y explanation completely blew me away. It's these very basic simplistic things in math which amaze you. Because it is so simple and 100% sure it was possible to come up with myself. Really encourages one to keep trying out stuff and working out things just for fun!

    @geetasharan5061@geetasharan5061 Жыл бұрын
    • if i was a teacher i'd refuse to teach the 'trick' because it's just friggin faster to divide by 7.

      @BeBopScraBoo@BeBopScraBoo Жыл бұрын
    • @@BeBopScraBoo You would be a crappy teacher. The WHOLE point of Mathematics is to recognize (and prove) beautiful patterns of set theory which can be inspiring for curiosity. I.e. The proof of Div by 7 is _Number Theory,_ specifically using *cyclic addition* which differs from *linear addition.* This _difference of perspective_ can inspire students to want to learn more.

      @MichaelPohoreski@MichaelPohoreski Жыл бұрын
    • Sadly I never learnt the Div by 7 mod trick in school either. When I was 30 I heard about the _subtract double the last digit from the remaining digits_ but didn’t know _why_ it worked so shortly after I set out to prove it. I independently derived some circular addition rules (modular arithmetic) and from that it was relatively straightforward: 10x + y ≡ 0 (mod 7) 50x + 5y ≡ 0*5 (mod 7) 49x + x + 5y ≡ 0 (mod 7) [49x ≡ 0 (mod 7)] + [x + 5y ≡ 0 (mod 7)] 0 + x + 5y ≡ 0 (mod 7) x + 5y - 7y ≡ 0 - 7y (mod 7) x - 2y ≡ 0 (mod 7) QED The sad part is kids do modular arithmetic (circular addition) when they learn how to tell analog time but no one every tells them they are doing Number Theory!

      @MichaelPohoreski@MichaelPohoreski Жыл бұрын
    • ​@@MichaelPohoreski That's not the point of school though. The purpose of school is to learn skills you can apply in real life. The point of divisibility tests is that you can do them in your head without writing anything down. If you have to write it down, it's pointless and you can just divide in the first place. Yes, it's interesting how numbers converge into patterns, but there's not point worrying about number theory and patterns in elementary school

      @Dragongaga@Dragongaga7 ай бұрын
  • Seven is weird because it eight nine

    @jifo360@jifo360 Жыл бұрын
    • 🤓👍🏻

      @systemoutprinthakim@systemoutprinthakim Жыл бұрын
    • Most underrated comment ive seen

      @tarasj6908@tarasj6908 Жыл бұрын
    • That's not weird, eating 3 square meals.

      @chriskuta6178@chriskuta6178 Жыл бұрын
    • Darn it. I came here to say that.

      @BodaciousWench@BodaciousWench Жыл бұрын
    • This is colder than 7 degrees Fahrenheit

      @goatgamer001@goatgamer001 Жыл бұрын
  • I completely forgot about this channel & how much math(s) makes my brain happy… Just subscribed.

    @AndTheBloodhound@AndTheBloodhound3 ай бұрын
  • 5:45 James’s reaction really goes to show that Maths-nerds aren’t big on sports 😅.

    @PC_Simo@PC_Simo Жыл бұрын
  • When you have Dr. Grime on, you know it's gonna be great.

    @LurkerPatrol5@LurkerPatrol5 Жыл бұрын
  • I remember I used a USB stick duplicator at work once, with 32 slots - 1 for the master USB and 31 for the clones. It also had a display that counted the total of successful duplicates, and checking if the number of successfully made duplicates was divisible by 31 was a quick way to tell if the machine was working properly. So, I used the similar divisibility test for 31, as it felt simpler once I got used to the method. "Subtract 3 times the last digit from the rest".

    @knooters@knooters Жыл бұрын
    • Hmmm I think I get your trick: for 31, you could do -3*(10x+y) = -30x-3y = -31x - (x - 3y). So indeed the original number (10x+y) it can be divided by 31 iff x-3y is divisible by 31. It's a bit of fiddling to find the right factor (in this case -3), or at least it's not obvious to me immediately, but it's not too hard to derive these tricks for other numbers actually! For example I found 13 too: 4(10x+y)=39x+(x+4y) so you can check x+4y (because 39=13*3).

      @ikbintom@ikbintom Жыл бұрын
    • alternative to thisss ssubtract 3 times method you indicate, you can also do " rest + 28 times the unit digit"

      @michaelempeigne3519@michaelempeigne3519 Жыл бұрын
    • @@michaelempeigne3519 that's true! I just thought 3 times a number was a bit easier than 28 times 😀

      @ikbintom@ikbintom Жыл бұрын
    • @@ikbintom Well, you thought wrong. It’s a lot easier to multiply by 28 then it is by 3, OBVIOUSLY.

      @chriswebster24@chriswebster24 Жыл бұрын
    • @@ikbintom Finding those fiddling factors comes in a couple ways if you have nice numbers: if you're looking for something divisible by XY, where X and Y are digits, then for a number ABC... you can multiply the last digit (ex: C) by X/Y, and subtract from the rest. Why? Every time you add a multiple of XY, you add Y to the last digit, and X/Y to the other digits. If you multiply Y by the ratio, and subtract that, you've essentially said 'ok, great, let's remove Y copies of XY from ABC, guaranteeing the last digit is now 0. Is the remainder left 10x a multiple we're aware of?. And if you repeat the process, you're just now doing the same operation, but with the 10s and 100s places. The second nice way is finding a convenient multiple near 100. For 7, 98 works great. Since that's 2 away from 100, we can just chop off all but the last 2 digits (would be 3 digits for a multiple near 1000), and add on 2 for every hundred we cut off - 343? 3*2+43 = 49. For 31, 93 is kinda close to 100, but 7 off. So let's try it : 568, 35+68=103, not a multiple. In fact, since it's 10 more than a multiple (103-10=93), you know 558 would be a multiple of 31, which it is (18x31).

      @undine120@undine120 Жыл бұрын
  • I personally find 6300+140+28 easier to figure out in the head, works for arbitrary sized numbers quite quickly too if you know your x7 table well enough. Its a bit of doing the full division but in blocks of 2-3 digits... and in a way its related to this one...

    @PlasmaHH@PlasmaHH6 ай бұрын
  • I always felt the subtraction rule was a bit clumsy and I was faster at division (or finding the remainder, i should say) than doing that test. The addition rule is easier to do!

    @PunnamarajVinayakTejas@PunnamarajVinayakTejas Жыл бұрын
  • Man, I love that seven one, I feel like I must have learned it once but completely forgot it, as I knew all the rest. Awesome job!

    @magnus0017@magnus0017 Жыл бұрын
    • I'm really happy to learn the 7 one because it completes the set. Most of them I learned in elementary school, but 7 was always troublesome.

      @vigilantcosmicpenguin8721@vigilantcosmicpenguin8721 Жыл бұрын
  • Please dont let it be a 7-8-9 joke

    @fizikabi6358@fizikabi6358 Жыл бұрын
    • I'm angry and happy to see this comment came through.

      @laurenf.7922@laurenf.7922 Жыл бұрын
    • It turns out that 7 was a 6 offender

      @carltonleboss@carltonleboss Жыл бұрын
    • The answer is no! 7 !8 9

      @arse124@arse124 Жыл бұрын
  • Thank you ❤ it was fun with your passion 🎉 you reminded me how much I loved numbers when I was a kiddo 🥰

    @Anonnius@Anonnius4 ай бұрын
  • 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.

    @allonahoya316@allonahoya316 Жыл бұрын
    • I usually do it with a multiple of p that ends in 1. Call the multiple 10m+1. Then if x-my is divisible by p, then 10x+y is also divisible by p. For example, 17x3=51, so you can take off the last digit, multiply it by 5, and subtract that from the rest of the number. Repeat until you get a number that obviously is or isn't a multiple of 17. If it is, then the original number was divisible by 17, and if not, it wasn't.

      @PhilBagels@PhilBagels Жыл бұрын
    • Nice, yeah that yields the trick(s) for 11 of either adding 10N or subtracting N (for trailing digit N). And for 19 and 21 you get tricks involving ±2N.

      @maljamin@maljamin Жыл бұрын
  • I always love the James Grime videos!

    @swankitydankity297@swankitydankity297 Жыл бұрын
  • Enjoyed the video ... great stuff ... Cheers

    @algorithminc.8850@algorithminc.88502 ай бұрын
  • There's a maths based "latin square" puzzle that I love, and I'm usually trying to figure out if a given number is divisible by 7 somewhere in the puzzle. I'll try this method. Usually I'd break it up as 420 + 14, or 6300 + 140 + 28. For reference, the puzzle is called "Keen" and is on "Simon Tatham's Portable Puzzle Collection".

    @bg6b7bft@bg6b7bft Жыл бұрын
  • Here's a trick that only requires you to know up to 10x7: take your number (let's use his example of 6468), and find the multiple of 7 that ends in the same digit. So, 6468 ends in 8, and 28 ends in 8, so subtract 28 and you have 6440. Drop the last 0, so you have 644. Repeat. 644 ends in 4, 14 ends in 4, 644-14=630, drop the 0, and you have 63, which is divisible by 7, so it's all divisible by 7. It's similar, but you don't have to deal with the 'multiply by 5' step, so it might be a bit mentally easier.

    @joeyjohnson2828@joeyjohnson2828 Жыл бұрын
    • That is about the same as dividing by 7 normally, except you're going right-to-left instead of left-to-right. However, your method is what I would do to check divisibility by many larger prime numbers like 17 or 23, except that I may add or subtract and attack the number from both the left and right. For example, to test if 3893 is divisible by 17, I would find it easier to add 17 rather than subtract 153. 3893 + 17 = 3910 or 391 after dropping the 0. At that point, I would notice: 391 = 17 * 23 = (20 - 3)(20 + 3) = 20² - 3² However, I could have could have continued the process by subtracting 51 (or adding 119) to 391 to get 340 (or 510). As to how often this is necessary in day-to-day life, well...

      @deepowls@deepowls Жыл бұрын
    • I prefer this method as well, it's a 7 based multiplication + subtraction vs a 5 based multiplication + addition. So end of the day it's pretty much the same

      @Pussyguardian@Pussyguardian8 ай бұрын
    • I have an easier approach for this one. Take 6468 for example. Consider first two digits i.e. 64. 7*9=63. So 7*900= 6300. 6468-6300= 168. Now consider first two digits again i.e. 16. 7*2=14. 7*20=140. 168-140= 28. 28 is divisible by 7 so number is divisible by 7. Fun fact - this is nothing but doing elementary division by 7 in a more complicated way. The method given in the video and and in the comment section by people are are more complex than simply dividing by 7 and check. It's fun math but not practical at all

      @azeemuddinkhan923@azeemuddinkhan9235 ай бұрын
  • An alternative method (that I worked out for myself right now) you can do is *last 2 digits plus twice the rest* . Will shrink the number much faster, and the mental maths is barely harder. Proof is exactly the same as in the video. It's actually surprisingly easy to generate any number of these tricks.

    @QuantumHistorian@QuantumHistorian Жыл бұрын
    • I guess the hard part, for some people, would be to double the rest if its a big number, at least mentally.

      @leoirias3506@leoirias3506 Жыл бұрын
    • Nice find, way easier to remember than the one Grime suggested

      @sang1s160@sang1s160 Жыл бұрын
    • @@leoirias3506 true, but if you're dealing with 3-4 digit number, you've only got to double 1-2 digits. Not very hard to do mentally I'd say? If you've got a much larger number... You won't. Nobody needs to know if a 5+ digit number is divisible by 7 in their head.

      @QuantumHistorian@QuantumHistorian Жыл бұрын
    • @@QuantumHistorian For a larger number, you can split it into 2 digit numbers starting from the end, and then multiply by increasing powers of 2. Eg: 12936 -> 36+29*2+1*4 -> 98 -> divisible by 7 To make it simpler, you can do a mod 7 for each of the individual elements: 36 -> 35+1->1 29*2->(28+1)*2->2 1*4->4 1+2+4=>7

      @rohitraghunathan@rohitraghunathan Жыл бұрын
    • @@rohitraghunathanIngenious... thank you!!

      @ifulea@ifulea Жыл бұрын
  • Excellent information

    @kanankazimzada2500@kanankazimzada2500 Жыл бұрын
  • An alternate method is as follows. Using the example of 434 43 4, we double the last digit, getting 8. Subtract 8 from 43, leaving 35, which is divisible by 7.

    @claudeabraham2347@claudeabraham2347 Жыл бұрын
  • James discussing Number Theory, a perfect start to my day. The idea that we see here as a ‘trick’ can be further generalised to derive an algorithm to check whether a number is divisible by a prime number that ends with 1,3,7,9 (i.e. all primes greater than 5). Very useful indeed.

    @marksman_561@marksman_561 Жыл бұрын
  • This method seems more complicated than just using classic division tricks to check. For example, with 686, you could just divide 68 by 7 with a remainder, then you take that remainder, apply it to the remaining digit to get 56 and 56 is evenly divisible by 7. Plus, doing it this way also gives you the quotient, not just the yes/no on divisibility. And, just like the trick you showed, you can repeat the steps for chains longer than 3, _and_ it works for divisibility with _any_ number, not just 7.

    @RabblesTheBinx@RabblesTheBinx Жыл бұрын
    • You are the smartest person in the comments section.

      @azeemuddinkhan923@azeemuddinkhan9235 ай бұрын
    • One of the points of this division trick is to avoid division. I teach both methods, and people seem to prefer the one highlighted in the video for whatever reason. For larger numbers, say 675081 you have to keep dividing by 7 and adding the remainder, OR you could multiply and add, and just check the last number for divisibility. If you noticed, all of these tricks are just to find divisibility, not the quotient. But I agree getting the quotient is a nice bonus =) I personally do not see one more complicated than the other, but different people work differently, and have their preferences!

      @henrygreen2096@henrygreen20964 ай бұрын
    • isn't this just a long division though

      @allengrove1864@allengrove18643 ай бұрын
    • So to figure out if a number is divisible by seven you have to figure out if it's divisible by seven? That's not very helpful

      @blu3260@blu32602 ай бұрын
  • These techniques are precisely elaborated in Sulva Sutras from ancient times in India and we are practicing these for questions asked by government recruitment examinations.

    @JamuiVlogs@JamuiVlogs Жыл бұрын
  • The general rule for any number relatively prime with 10, call it p, is multiply the last digit by the multiplicative modular inverse of 10 mod p, and add the rest, and check the sum.

    @marcusscience23@marcusscience23 Жыл бұрын
  • I own a funnel cake stand and I used to sell them for $7 each (not anymore... inflation!). This would have been handy when counting up the money at the end of an event to make sure the drawer was right!

    @ryanallen2001@ryanallen2001 Жыл бұрын
    • And if the drawer wasn't right, what next? I'm not sure I'd want to count.

      @charliebell5073@charliebell5073 Жыл бұрын
  • 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.

    @mattv2099@mattv2099 Жыл бұрын
    • I saw 700 - 14 for 686 :D

      @jonasb.236@jonasb.236 Жыл бұрын
    • I think is more for students who struggle to immediately see that and how they can be supported :)

      @Matthew-bu7fg@Matthew-bu7fg Жыл бұрын
    • @@Matthew-bu7fg I do have to say that this method is indeed great, the examples could have been chosen better imo. But nevertheless great method.

      @jonasb.236@jonasb.236 Жыл бұрын
    • There's another similar method which I use if the divisibility test is just too hard: Keep adding or subtracting the number which you want to know divisibility by to your number until it's a multiple of 10. If, say, your number which you want to know divisibility by is even, and your number is odd, there's no way of getting to a multiple of 10, so it isn't divisible. Divide your 10 multiple by 10 and try to see if that's divisible, if not, repeat the process until you see something you recognize. Example: divisibility by 13 for 832 832(i picked this number completely at random no joke)+13=845 845+(13*5)=845+65=910 910/10=91 91=13*7, but lets say i'm still unsure 91+13=104 104+13=117 117+13=130 No doubt about it, 130 is divisible by 13. 832 must be divisible by 13

      @cheeseburgermonkey7104@cheeseburgermonkey7104 Жыл бұрын
  • I loved the summary 😘

    @enissay9950@enissay9950 Жыл бұрын
  • This is gonna be reccomended 4 years later😊

    @weiyanwu3942@weiyanwu3942 Жыл бұрын
  • Man, what a nice guy professor Grime seems. I've known him for so long on this platform, I feel he's part of my family now.

    @naedolor@naedolor Жыл бұрын
    • parasocial relationship moment

      @ParasocialCatgirl@ParasocialCatgirl4 ай бұрын
  • I seriously love watching Numberphile videos and you are my absolute favourite, James Grime Sir! I adore you James Grime Sir to a great extent. (Your smile is literally the best!) Thanks for making me love mathematics more than ever.

    @saarthaksinghal.@saarthaksinghal. Жыл бұрын
  • Really Good Video ...! Thanks Numberphile !!!

    @SamadhanSalunke7@SamadhanSalunke72 ай бұрын
  • dr. James

    @ralitsa-ost@ralitsa-ost2 ай бұрын
  • I'm delighted! This filled in the one remaining gap in my knowledge of divisibility by single digit numbers! I will use it often, because I often feel curious about whether a number is prime, and I like to see what I can work out before simply searching the web for the answer.

    @Priapos93@Priapos93 Жыл бұрын
  • Another interesting thing my friend and I figured out- if you divide a number which is not divisible by 7, by 7; you get repeating patterns after the decimal place- which is always 142857… in a cyclic order.

    @m.islamnafees5770@m.islamnafees5770 Жыл бұрын
    • Yeah this one is completely nuts. But it's the best known cyclic number. Multiply 142857 by anything you get an anagram or 9's repeated if it's a power of 7.

      @ivanski28@ivanski28 Жыл бұрын
    • basically, what i do is i add up all of the digits in the number, then i add 7 to the original number and add up the digits of that number. if the total of (original +7) is 2 less than the original sum, it’s a multiple of 7. it works *a lot* of the time but it isn’t foolproof because occasionally, the sun resets to a larger number e.g 7(7), 14(5), 21(3) 28(*10*) wait, so maybe it works if the next number is 2 less or 7 more? idek

      @osaruguex@osaruguex8 ай бұрын
    • @@aaronmarchand999that is typology, i think you meant anagram

      @xaryop7950@xaryop79505 ай бұрын
    • @@xaryop7950 No look it up, comes from Gurdjieff, unfortunately has been used for typology in recent years but that's not the original meaning

      @aaronmarchand999@aaronmarchand9995 ай бұрын
    • True, but the decimal remainder starts with a different number in the sequence which is in numeric order in respect to the remainder. example: put the cycle in numeric order - 1,2,4,5,7,8 which correlates with remainders 1,2,3,4,5,6 If you divide 37 by 7 you get 5 with a remainder 2. So the decimal remainder is .285714 repeated. For 38/7, it would be 5.428571... (3 is the remainder. The third number in the sequence is 4.)

      @danlayne9436@danlayne94364 ай бұрын
  • I appreciate you showing the reasoning behind why this works, tricks like these are far more useful (to me at least) when I understand the underlying math

    @Defectivania@Defectivania3 ай бұрын
  • My trick for seven was take the first 2 digits of the number, find the highest multiple below those digits, subtract that from the number and repeat. So for 6468, take 64, the highest multiple below that is 63, so subtract 6300 it leaves you with 168. Take 16, 14 is just below, subtract 140, you have 28 and that is easily recognizable as divisible by 7.

    @colink3728@colink3728 Жыл бұрын
  • Knowing divisibility checks is really useful for prime factorizations too...7 and 11 in particular are terrific for this since they're on the bigger end of the early primes.

    @thane9@thane9 Жыл бұрын
  • Isn't it amazing that the sponsor is placed at the _end_ ? I can't believe Numberphile is the only channel I've seen that doesn't intrude on its own content!

    @0ia@0ia Жыл бұрын
    • shitty advertisers (like Raid Shadow Legends and other mobile games, most VPN, etc.) force you to have the ad at the starts + link in pinned comment + link in description, while Brilliant, Kiwico, etc. don't.

      @minirop@minirop Жыл бұрын
    • Yet another reason to love Numberphile

      @itsreeah2663@itsreeah2663 Жыл бұрын
    • Most educational channels (all of the SciShow channels, PBS channels, Stand-up Maths, etc) put the ad at the end if they have one. If you can see the viewership graph on the seek bar, viewer retention usually flatlines on those videos as soon as the ad starts.

      @acoupleofschoes@acoupleofschoes Жыл бұрын
    • I know old Jacksfilms videos and Oddheader do that as well

      @bingbonghafu@bingbonghafu Жыл бұрын
    • Probably because Numberphile has more leverage in the situation than your average independent youtuber who may be more in need of the sponsorship money (and therefore much more able to say no to a sponsor demanding a more intrusive segment).

      @ParasocialCatgirl@ParasocialCatgirl4 ай бұрын
  • Just an addition to the old trick which is take the number's last digit double it and subtract it from the remaining number. For eg: Let abc be the number then if 7| ab -2×c then 7|abc . It's just that we add 7×c to ab-2×c which will be ab+5c. So just an addition to that.

    @loveshchitte@loveshchitte Жыл бұрын
  • Even if Argam Numerals were to extend up to digits in Base-720 (the factorial of 6), 7 would still be a somewhat interesting number, as it'd be the smallest prime number to not be divisible from the Base-720 sub-digit points, or in Dozenal's definition: Base-500.

    @SebastianGMSFB@SebastianGMSFB Жыл бұрын
  • Just wanted to say that I'm especially excited when I see Dr. James Grime on Numberphile! Thank you for sharing your love for Math and the numbers! ❤️

    @autumn_skies@autumn_skies Жыл бұрын
  • I've been waiting for this for 49 years actually. How fitting LOL. I use it for checking numbers for primailty in my head. Brilliant! Thank you! ( I did know all of the others already actually).

    @koeielul112@koeielul112 Жыл бұрын
  • Never heard of that "7" test, it's genius!

    @technik-lexikon@technik-lexikon5 ай бұрын
  • Thanks that was a really useful summary of all the divisibility "tricks" there a the end

    @quatarsr6217@quatarsr6217 Жыл бұрын
  • Very useful video. Lately I’ve been wanting to check Primality in my head, and this helps make it a lot easier. I knew the tricks for 2,3,&5, and memorized that 49, 77, and 91 were multiples of 7 to be able to do it for all two digit numbers. Knowing the tricks for 7 and 11 let me check all the way up to 169, so that’s pretty cool.

    @MitchBurns@MitchBurns Жыл бұрын
  • The method I found for determining divisibility by 7 was 3 times the ten's place plus the one's place. So 105 would be 10x3+5 or 35. 3x3+5 or 14. 1x3+4 or 7. For divisibility by 4, if the 10's place is odd the 1's place must be 2 or 6, or even but not divisible by 4. If the 10's place is even then the 1's place must be 0, 4, or 8, or divisible by 4.

    @thomaswilliams2273@thomaswilliams2273 Жыл бұрын
    • For 434 I did 3X43 and added the 4. This is 129 + 4 =133, or 7 less than 140, so, a 7-mer. Or, using 133 as a rest stop, do it as 3 X 13, + 3, which is 39 + 3 = 42. I think our way beats his, although I've not analyzed just why it works.

      @davidcovington901@davidcovington901 Жыл бұрын
    • @@davidcovington901 I believe it works because 1 is 1 more than a multiple of 7 while 10 is 3 away, so to compensate for dividing by 10 you multiply by 3. My Geometry teacher in high school told us that some POW had figured out a way but didn't tell us what it was. I guess I subconsciously kept working on the problem until I figured it, or rather apparently one, way.

      @thomaswilliams2273@thomaswilliams2273 Жыл бұрын
    • @@thomaswilliams2273 Thanks!!!

      @davidcovington901@davidcovington901 Жыл бұрын
    • @@thomaswilliams2273 Just divide the number by 7 ffs. Why would you do all these other steps as a “test”? Just do the division.🤦‍♂️🤡

      @spanqueluv9er@spanqueluv9er Жыл бұрын
    • This only works if the number has 3 digits. Consider 1105

      @batuozer7179@batuozer7179 Жыл бұрын
  • When I was like 10yo I figured out by myself a very similar method to verify if a number is divisible by 7 and I was so proud of it: In case of 434, multiply by 3 every digit except for the last one and add it to the result (43x3)+4=133; Then again (13x3)+3=42 that is divisible by 7 so 343 is divisible by 7

    @eldorado7mo@eldorado7mo Жыл бұрын
  • I like the pattern of multiple 2^n (2, 4, 8, 16, etc…). You truncate the last, 2 last, 3 last, 4 last digits, etc… It make sense since 10^n = 2^n * 5^n.

    @Christian_Martel@Christian_Martel Жыл бұрын
    • Precisely. And understanding that, you can also develop a pattern for powers of 5 :)

      @MuffinsAPlenty@MuffinsAPlenty Жыл бұрын
  • Well, it certainly makes sense that adding 5x and subtracting 2x would work the same when what you care about is divisibility by 7; when counting modulo 7, 5 and -2 are the same number.

    @markjreed@markjreed Жыл бұрын
    • Ooo you're right. If the number is 10x+y, then the first method is about checking if x+5y is divisible by 7, and the second method is about checking if x-2y is divisible by 7. (I like to call them the x+5y & the x-2y methods). But modulo 7, they're the same thing! (For people not familiar with modular arithmetic, if x+5y is divisible by 7, so is x-2y. Because it is smaller by exactly 7y.) In fact I can make an infinite number of 7 divisibilty tests like these, like the 8x-9y divisibility test & the 12y-6x divisibilty tests by adding or subtracting 7x's & 7y's. (If you frame it as a number is divisible by 7, if 12 times the last digit minus 6 times the first digit is divisible by 7, you can perhaps use it as a cool party trick ;) )

      @adarshmohapatra5058@adarshmohapatra5058 Жыл бұрын
    • Hey -2 is my method! I was wondering why that worked! Thanks!

      @witchcraft2264@witchcraft2264 Жыл бұрын
    • What’s neat is that the proof for x+5y mod 7 is contained in the proof for x-2y mod 7: 10x + y ≡ 0 (mod 7) 50x + 5y ≡ 0*5 (mod 7) 49x + x + 5y ≡ 0 (mod 7) [49x ≡ 0 (mod 7)] + [x + 5y ≡ 0 (mod 7)] 0 + x + 5y ≡ 0 (mod 7) x + 5y - 7y ≡ 0 - 7y (mod 7) [x - 2y ≡ 0 (mod 7)] - [7y ≡ 0 (mod 7)] x - 2y ≡ 0 (mod 7) + 0 x - 2y ≡ 0 (mod 7) QED

      @MichaelPohoreski@MichaelPohoreski Жыл бұрын
  • I love seeing someone so excited about divisibility by 7!

    @volodask@volodask Жыл бұрын
  • Same can be done by diving the number in different 2 parts Eg 119 - 1 and 19 1x2 + 19 = 21 -> divisible by 7.

    @mangeshpuranik31@mangeshpuranik31 Жыл бұрын
  • I always come away smarter after just a few moments spent here Cheers

    @w.colonialboy9144@w.colonialboy91443 ай бұрын
  • A method which I find easier to do in my head is to split the number up into groups and look at each group individually. For instance, 434 becomes 43 4 becomes 42 14 and both groups are divisible by 7, so the original number is also. As a bonus, the quotient is immediately given as 62. Thus 434 / 7 = 62. Do the same with 6468: 6468 becomes 63 168 becomes 63 14 28, so it is divisible by 7 and the quotient is 924.

    @ib9rt@ib9rt Жыл бұрын
    • I agree. I would guess that this is not more work than the method presented in the video, in count of number of arithmetic operations ...

      @mmneander1316@mmneander1316 Жыл бұрын
    • Yes, this is exactly what I do. And really nothing to remember as the method in the video describes. And, for me at least, it’s faster!

      @tenapier@tenapier Жыл бұрын
    • Yeah I immediately noticed 434 was divisible by 7 this way

      @ravneiv@ravneiv Жыл бұрын
    • That's just long division with fewer steps.

      @badrunnaimal-faraby309@badrunnaimal-faraby309 Жыл бұрын
    • @@badrunnaimal-faraby309 Exactly. :-)

      @mmneander1316@mmneander1316 Жыл бұрын
  • I love your channel! An improvement for 8 is once you divided the last 3 digits by 2, you could use the divide by 4 rule and drop the 100’s digit-A lowly 62 y/o industrial engineer.

    @jimf2525@jimf2525 Жыл бұрын
    • oooo, I like it! definitely more efficient - a lowly 22 y/o college student

      @Defectivania@Defectivania3 ай бұрын
  • Awesome! For 4 and 8 divisibility for a number with the last three digits of ABC we can also check 4|C+2B or 8|C+2B+4A.

    @HS-fw2ed@HS-fw2ed3 ай бұрын
  • Take the first three prime numbers: 1, 3, 5. Duplicate each number in series: 11, 33, 55. Join them all together (113355) then divide into halves at the center: 113 and 355. Divide 355 by 113. You get Pi accurate to 6, arguably 7 digits when rounded off.

    @CatholicDefender-bp7my@CatholicDefender-bp7my5 ай бұрын
  • There's another way to check divisibility by 7. Take a number (91 in this case). Subtract its base-3 representation (91 in base 3 = 28 in base 10). 91 - 28 = 63, which is divisible by 7, therefore 91 is divisible by 7. This works for any base. In general, if N_b - N_t is divisible by b-t, then the number is divisible by b-t as well.

    @jozbornn@jozbornn Жыл бұрын
    • Yes, but your 91 in Base 3 would actually be 1001. However, that number is also divisible by 7 - or 21, in base 3 - and if you divide (B3) 1001 by (B3) 21, you would have (B3) 11. Quite a coincidence, too, because (B10) 3 × (B10) 7 = (B10)... (you guessed it) ...21 !

      @robertanthonyfairweather3416@robertanthonyfairweather3416 Жыл бұрын
  • I sent a video to you guys a long time ago where I explore the significance of dividing by 7, and thought this may have been your response 😅 but my video was pointing out the 142857 loop of decimal points. I love all of these neat patterns that lie in mathematics !

    @mattcoulter7@mattcoulter7 Жыл бұрын
    • I was hoping the video would be about that! I discovered that pattern when in school - how doubling 142857 shifts to 285714 and doubling again shifts to 571428.

      @thomasreichert2804@thomasreichert2804 Жыл бұрын
    • @@thomasreichert2804 yess such a cool pattern! I was trying to find this pattern with other numbers too, but it only seemed to exist if the divisor was a multiple of 7, such as 1/14 or 1/28. I looked at the pattern as treating the numerator as the index of the decimal number from low to high, then the pattern continues on. For example 3/7, the 3rd smallest number from the 142857 pattern is 4, hence 3/7 = 0.428571 and so on.

      @mattcoulter7@mattcoulter7 Жыл бұрын
    • @@mattcoulter7 yeh, it's great, you can work out any integer divided by 7 to as many decimal places as you want by just remembering that simple sequence

      @gameclips5734@gameclips5734 Жыл бұрын
  • A similar process (as I’m sure others worked out) is to divide the number by 50, then add the whole number portion of the quotient to the remainder. So for 434 it would be 8 + 34 = 42, which is divisible by 7. This works because 50 - 1 = 49, as related to his explanation of how his method works. For very large numbers, you can repeat this process as needed 😸

    @edielungreen@edielungreen10 ай бұрын
    • That's quick and easy!

      @dielaughing73@dielaughing734 ай бұрын
  • Multiply any prime number other than the prime numbers that have 1 and 9 in its ones place with 3. You will get a number that is either 1 more or 1 less than the nearest multiple of 10. For example, Let's see if 49 is divisible by 7. We multiply 7 with 3 to get 21, which is 1 more than 20. Here, 20 is (we get the number 2 from 20, as 20 the second positive multiple of 10. We multiply the ones place digit with 2. In case of 49, we multiply 9 with 2 and get 18 as the result. Lets suppose, 18 is called "x" here.) the nearest multiple of 10 in case of 21. As we got 1 more than the nearest multiple of 10, we need to deduct "x" from the number that is created from the rest of the digits. In case of 49, we deduct 18 from 4 and get -14 as the result, which is clearly divisible by 7. If we get a number ( after multiplying it with 3) which is 1 less than the nearest multiple of 10, we simply use addition in case of deduction (that is shown in the previous example). Sorry if I failed to explain the whole procedure clearly. Because, I discovered this method for myself and I simply follow these steps to find the divisibility rule of any prime number. I haven't tried to present this as a formula, but it works for me every time.

    @manah2882@manah2882 Жыл бұрын
  • I think the easier and more straight forward method would just be subtracting multiples of seven since it cannot change the remainder. This method is solid and works for any number, but for some small numbers there are better ways. 6468 - 6300 168 - 140 28 ✓ A nice bonus is that we get can easily perform the division right now, since 6468 = 6300 + 140 + 28 = 7 (900+20+4) = 7*924

    @norbi275275@norbi275275 Жыл бұрын
    • 434 = 420 + 14, 6468 = 6300 + 140 + 28. This is the skill I got from taking math tests without a calculator. This will always be faster

      @seventhtenth@seventhtenth Жыл бұрын
    • Literally just division by hand as we learned in primary school, but with 0's instead of spaces. I wish we'd have learned why that worked back then. I think it might have helped quite a fair few to better understand, or might at least somewhat dispel the myth that maths is just magic.

      @PwnEveryBody@PwnEveryBody Жыл бұрын
  • To check for a multiple of 8 i have an easier method (for me at least): I look at the last 3 digits, if the first is even then the last 2 must form a multiple of 8. if the first is odd, then the last 2 can't form a multiple of 8 but must form a multiple of 4. For example, 264 is a multiple of 8 because 2 is even and 64 is a multiple of 8. Another example, 128 is a multiple of 8 because 1 is odd and 28 is a multiple of 4 (but not of 8) I find this much easier than halving 3 times in a row. Checking whether the last 2 digits are a multiple of 8 is easy as long as you know your 8s table, checking whether they are multiple of 4 is a bit trickier, but you only have to do it half of the times anyways.

    @3snoW_@3snoW_ Жыл бұрын
    • For 2-digit multiples of 4 there is a similar "trick": if the 10s digit is even, then the 1s digit should be one of 0,4,8 if the 10s digit is odd, then the 1s digit should be one of 2,6

      @Demki@Demki Жыл бұрын
    • It does work everytime? So cool!

      @Hybban@Hybban Жыл бұрын
    • The same trick "scaled down" works to find multiples of 4 too. If the 2nd digit is even, the last one must be 0, 4, or 8. If the 2nd digit is odd, the last one must be 2 or 6.

      @QuantumHistorian@QuantumHistorian Жыл бұрын
    • Another way to do 8's is take the first 2 digits, multiply by 2 and add the 3rd digit. So for 264, 26*2 = 52+4 = 56 = 7*8 for 128, 12*2=24+8=32=4*8

      @nekogod@nekogod Жыл бұрын
    • @@nekogod Another similar one is: take the last three digits multiply first by 4, second by 2, third by 1 and sum, check if divisible by 8, i.e. 264 = 2*4 + 6*2 + 4 = 24 The advantage is that it gives you a smaller number.

      @the_real_glabnurb@the_real_glabnurb Жыл бұрын
  • My favourite way to tackle divisibility by 7 is to exploit the fact that 1001 = 7 x 11 x 13. It's relatively easy to cast out lots of 1001 (though you can occasionally get some nasty carries) and it's guaranteed to bring you down to a number below 1001. Even though this might require a bit more memorisation (or mental arithmetic; just doing long division with numbers that small is surprisingly manageable, especially with a bit of practice/experience) from this point onwards than most of the methods discussed here, it does have the huge advantage of being able to attack 7, 11 and 13 simultaneously!

    @joonasjurgenkisel5480@joonasjurgenkisel5480 Жыл бұрын
    • its guaranteed to get a 3 digit number, as 1-001 is 0 and 1-000 is 1.

      @goatgamer001@goatgamer001 Жыл бұрын
  • Alternatively a relatively easier approach that we were taught in school. Take the twice of the units digit and subtract it from remaining. If the remaining is divisible by 7, then the number is divisible by 7. Example: 784, units digit*2 = 8, subtract it from 78 gives 70 which is divisible by seven

    @shashankasarikonda7813@shashankasarikonda78133 ай бұрын
  • Simplification for 8 is to calculate Last + 2*SecondFromLast + 4*ThirdFromLast and check if the result is divisible by 8, which is slightly faster than dividing by 2 thrice and matches with the other digitsum methods

    @orangeninjaMR@orangeninjaMR Жыл бұрын
  • 1:25 For numbers with 6 digits, the easy thing to do is to split it in the middle, so you get two numbers with three digits each. If the difference of those numbers is divisible by 7, then so is the original numbers. This works, because the method corresponds to subtracting multiples of 1001, which is a multiple of 7 (and of 11 and 13, so you can use this to check those divisibilities too). Using this method on given example of (00)6468, the first step gives 468-6=462, which is notably divisible by 7, because 462 equals 420 + 42 😎

    @bass2564@bass2564 Жыл бұрын
  • If you're interested in the proofs for these divisibility tricks, look up "divisibility and modular arithmetic"

    @vandel_@vandel_ Жыл бұрын
  • I use a method that rounds to the nearest 10, 100, 1000... subtract the highest number that is divisible by 7 multiplied by whatever the rounding factor is, if that result added to the remainder of the first step is divisible by 7 the begining number was divisible by 7. Lets start with 2744, subtract 2100 leaving 644, subtract 63, if 14 is divisible by 7 then 2744 is also.

    @donerickson1954@donerickson19544 ай бұрын
  • There's a divisibility by eleven trick for three-digit numbers where if the sum of the hundreds and ones digits equal the tens digit, it is divisible by 11 (11×11 = 121, 1+1 = 2), and it can be applied to factoring quadratic equations, too. (1x^2 + 2x + 1 = (1x +1)^2) But the factorability trick can be used with any coefficients.

    @seanwilkinson7431@seanwilkinson7431 Жыл бұрын
  • I worked out my own test for 7: split off the RH digit, multiply the rest by 3 and add the RH digit. The reason is as follows: 10x +y can be re-written as 7x+ ( 3x +y) . Now as 7 divides 7x, to divide 10x+y, it needs to divide 3x +y. A similar trick can be done for 13, but this time the test is (RH digit ) MINUS ( 3 times rest) ( ignore the minus sign if the result is negative). Edit: I've just worked it out for 17: similar to 13, but the test is (RH digit) minus (7 times rest). Didn't bother with 14, 15 or 16 as they are simply combinations of 2 tests ( respectively 2 and 7, 3 and 5, 2 and 8 ( or 4 twice)).

    @alanclarke4646@alanclarke4646 Жыл бұрын
  • I've always known the rule about taking the last digit, multiplying it by 2, then subtracting the smaller number from the larger number. ... the x5 rule is new to me.

    @chrisrj9871@chrisrj9871 Жыл бұрын
    • Same. The one you said is my go-to 7 divisibility rule

      @bingbonghafu@bingbonghafu Жыл бұрын
    • It's essentially the same because +5 and -2 are the same modulo 7

      @aniruddhvasishta8334@aniruddhvasishta8334 Жыл бұрын
    • Same!

      @atornblad@atornblad Жыл бұрын
  • Spectacular. MAGIC for sure!!!

    @junkmail4613@junkmail46136 ай бұрын
  • Ya it was really fascinating tricks, and I want to share my experience on this. I already know divisible by 7 tricks when teachers told me in middle school. And recently, I just found out that you can do other numbers as well. For example, I noticed that 13 = 1 + 3 * (4). So if you take the rest added by 4 * last digit it works. Also works with 19 = 1 + 9 * (2). I noticed that on my bed, and I couldn’t sleep. 😂

    @silencesls7967@silencesls7967 Жыл бұрын
  • A game I've been playing since I was a kid is breaking down numbers I come across, like street addresses or telephone numbers, to find it's largest prime. I "win" if a number's largest divisor is less than 12. I've always known every other trick mentioned in this video but seven has ALWAYS been the bane of this game as I didn't have an easy method to quickly divide by it. Or at least, not until now. So glad to have learned this, this trick is going to speed up my games considerably!

    @AvsJoe@AvsJoe Жыл бұрын
    • I love this. Thank you for sharing with us. :)

      @nicolestewart2843@nicolestewart2843 Жыл бұрын
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