Why it works: a b a + b a + 2b 2a + 3b 3a + 5b 5a + 8b 8a + 13b 13a + 21b 21a + 34b ---------------- 55a + 88b = 11(5a + 8b) ...which is 11 times the 4th number from the bottom.
a = first number (F1) b = second number (F2) F3 = a + b F4 = a + 2b ... F7 = 5a + 8b ... Sum F1 through F10 = 55a + 88b F7*11 = F10 Symmetric if you write them in the other order to start.
I used to add the fibonacci numbers to get to sleep. numbers and patterns have always brought me peace. Thank you for this trick! It's fun and great knowledge
Well this induction proof is astoundingly simple. You don't even need a piece of paper. Just pause a video and do it in your head. Beautiful piece of mathematics James!
It is - I was considering doing it for the video, but decided people who know induction can do it themselves, and people who don't will just enjoy the bit with the trick.
The kids loved it. I did it a few times with the class & they copied down the examples. After two days, most of the kids, either through trial & error or algebra, found the pattern. A couple of them are now seeing how far they can take it out to find an easy multiplication trick for a longer series of numbers.
I played around with this a little bit and observed the following: To every second number (all on odd positions) there is a point, where all numbers summed together are a multiple of this particulary number: The sum of the first three numbers is always two times the third. The sum of the first six numers is always four times the fifth. The sum of the first ten numbers is always eleven times the seventh. (The subject of the video) The sum of the first fourteen numbers is always 29 times the ninth. The sum of the first eightteen numbers is always 76 times the eleventh. The sum of the first 22 numbers is always 199 times the thirteenth. It feels like the sum of the first 4n-2 numbers is always k times the (2n+1)th number, but I can't find a pattern in the k values. I would be glad if James could make another video about this where he explains the patterns
That's basically the same thing. Times 10 you just add a zero, but then the addition is the slightly slower part, but if you think about it, that's basically exactly what his trick is doing. You can ignore the first digit since it's being added to 0 and won't carry over, so you only care about the middle digit carrying.
You will never get as quick doing it that way as his way. T'is the difference between adding a 3 digit number and a 2 digit number, and adding two 1 digit numbers. It's obvious which is actually easier.
Paused at 2:05 to wok out the answer. Each term is as follows (x,y,x+y,x+2y,2x+3y,3x+5y,5x+8y,8x+13y,13x+21y,21x+34y) which sums to 55x+88y, the fourth term from the bottom is 5x+8y which is 1/11 of the sum. I'm sure there are other things you can do with this because the coefficients are all fibonacci numbers. e.g. Sum to the 6th term is 4 times the 5th term. and sum to 14th is 29 times the 9th term
In your e.g. you note that sum to the 6th term is 4 times the 5th term, and sum to the 14th is 29 times the 9th term. We can also see that sum to the 18th is 76 times the 11th term, sum to the 22nd is 199 times the 13th term and so on. What I find fascinating is that while the coefficients in your nx + my expressions are Fibonacci numbers, 4, 11 (James' example), 29, 76, and 199 are alternate Lucas numbers. And the next alternate Lucas term down from 4, namely 1, can be derived from "sum to the 2nd term is 1 times the 3rd term".
Good trick! The proof is fairly easy once you prove (with induction, as you said) that thing about the sum of Fibonacci numbers. Thank you very much for the videos you make here and with numberphile!
Generalization for any: f a Fibonacci series n a natural number k an odd natural number L the Lucas series (2, 1, 3, 4, 7, 11, 18...) f(n+2k) = L(k+1) × f(n+k) + f(n) For n = 2, you get f(2+2k) = L(k+1) × f(n+k) + f(2) f(2k+2) − f(2) = L(k+1) × f(2+k) Sum of all the elements of f from 1 to 2k = L(k+1) × f(2+k) Example for n = 2, k = 5 : f(2+2×10) = L(5+1) × f(2+5) + f(2) f(12) = L(6) × f(7) + f(2) f(12) − f(2) = 11 × f(7)
The connection between this trick and the Lucas sequence can also be expressed as follows: When your volunteer V gets to the seventh number (S7) in her/his sequence, then you multiply that number by 11 to get the partial sum for S10, as Dr Grime showed. When V gets to S9, you multiply by 29 to get the partial sum for S14. When V gets to S11, you multiply by 76 and get the ps for even further ahead, namely S18. And so on. These multipliers are alternate terms in the Lucas sequence, and if you can memorise them (or just use it openly as a reference) you can still come up with results to impress your friends even though you probably have to use a calculator. You can go in the opposite direction down from S7 too of course: at S5 the multiplier is 4 to get the ps for S6. And at S3 you multiply by 1 to retrodict rather than predict the ps for S2.
Proving the sum formula by linear algebra is a good exercise too. It isn't as straightforward as induction, but it lets you find out the thesis by yourself.
In 1995 when I was in grade 2 I looked like a genius when I could multiply by 11. I tried showing this off to a few friends just then and they just said something along the lines of "can't math. Art degree". I miss primary school. I had morale.
When I was in highschool I developed a system for quickly squaring 2 digit numbers in my head. It was great for my physics and algebra homework, but nobody else really cared "cuz calculators".
Ah the tech generation problem. I mean, I don't think it's a problem like our older generation says, but it's always interesting and admirable to hear about self discoveries. In school I also self-discovered how to add two fractions with unlike denominators. We were learning about LCM at the time using the tedious 'list all factors' method. I discovered that I could just multiply each fraction with the other denominator to produce like fractions. It didn't produce the lowest multiple denominator but I solved my exercises in a less tedious and more efficient way. This got me some grief for not doing the task at hand actually. I didn't know what I was doing at the time but I discovered it worked for all fractions. My mind was blown. Only when we learnt algebra (in the following year?) I proved to myself a/b + c/d = ad/bd + cb/db. Mind blown again!
Or have a race to the 12th number, let them do all the hard work and when they get to the 6th number, multiply by 11, add the second one and write that down. Probably more impressive with big numbers, but it will seem like you were doing in your head faster than they could do it with paper.
I have a question. What do you do for a living? Are you a teacher? How do you make your money? This would be like the best life ever for me. Your videos are great!
He used to work with the Millennium Maths Project at Cambridge, but he is a freelance maths presenter now. He has several different presentations that he gives in the UK and around the world, but the topic he's probably best known for is the cryptography of the Enigma machine. He has addressed it in various videos and on his website.
I love you man, you are amazing! Your vids on numberphile are my favourite as well! The excitement on your eyes everytime you share new facts and curtiosities really brings out my curiosity for math! (Even though i hate algebra!) Shout outs from Brazil!
I love this trick! I am also interested in mathematical tricks such as 1089, which takes a random number, does a few calculations that don't cancel themselves out (like Add 3, multiply by 2, subtract 6, now divide it in half), but I am having a hard time to search out more of these equations. Do you know if these types of equations have a group name or something to help my search?
Let's choose the numbers x and y 1. x 2. y 3. x+y 4. x+2y 5. 2x+3y 6. 3x+5y 7. 5x+8y 8. 8x+13y 9. 13x+21y 10. 21x+34 The sum of all these is 55x+88y=11*(5x+8y) which is the 7th column.
Excellent lesson. Shorter, clearer, funnier and more complete than the average competitors. It would be interesting to show the "minus second term" pattern during the sums using a table in order to show the inductive pattern explicitly.
+Cadde So you have never made a typo? That is impressive! You are definitely the intellectual superior of us. Sleep well tonight knowing that you have corrected a grammer error of a lesser human.
Lemme check, does 2f(n)+f(n-1)-f(2) equal f(n)+f(n+1)-f(2)? 2f(n)+f(n-1)=f(n)+f(n+1) f(n)+f(n-1)=f(n+1) correct(because of the definition of fibonacci numbers, woo!)
This is really easy James; you write the first number as a and the second number as d and the write all the ten numbers in terms of a and d and then calculate their sum in terms of the same, and one would find that it is 11 times the 4th number from the bottom.
Your easy multiply by 11 method is a bit mental! Very creative! I just multiply by ten and add the quantity multiplied to that once, to make 11 x whatever. But seeing yours was intriguing. Alround good vid
Wanna hear a good fib? Dancer says blitzen speaks elven; blitzen says yes; dancer says blitzen speaks 'other' languages as well; blitzen says yes; Dancer wants to know if blitzen is lying.
do you know the 111 trick? it just a cool little design under the number. For Example: 231×111= 25641 - (it's easier to show on paper) --- you write down the last ------ number, 1. Then the last two --- added, 4. Then all three, 6. - Then the first two, 5. And finally, just the 2.
Instead of doing a couple more steps, you can stick with what you have. F(n+2) = F(n+1) + F(n) = (F(n) + F(n-1)) + F(n) = 2F(n) + F(n-1). So you're calculating 2F(n) + F(n-1) - F(2).
There's one small problem with this trick. If they make any mistake while creating that list, your answer will be incorrect, and when they check it eg. with a calculator, and your answer does not match, it becomes clear that you tried to use trickery. In order to make sure that your trick works, you would need to check that they calculated all the numbers correctly.
Can't tell if that question is a joke, but I'll answer anyways. He isn't referring to electromagnetic induction, but mathematical induction. It's a proof technique, if you want to look it up.
If you do 2 numbers between 1 and 10 you could just memorise the answers since there's only 18 or 19 total possibilities depending on if you allow duplicates.
Another way I think of x11s is you add the number to itself plus an extra digit 0 at the end.... for example 11 x 11 = 121 = 11 + 110, or 126163984 x 11 = 126163984 + 1261639840 = 1387903824 It seems to take me roughly about the same amount of time to do it this way as to do it your way, with probably a couple more seconds dedicated to writing the number over before addition.
I have a question. For the last trick, F(n+2) - F(2) is the sum F(1) + F(2) + ... + F(n). But the order of the first 2 numbers doesn't matter, so doesn't that mean the answer would not be always correct? Like what if the example you gave was 5 + 8 + ..., then F(2) is 8 and the sum you get is 885 - 8 = 878? What am I missing?
A more general version of this trick is to say S(4n+2)=L(2n+1)*F(2n+3), where F(n) is the nth number in your generalised Fibonacci sequence, S(n)=F(1)+F(2)+...+F(n) and L(n) is the nth Lucas number, indexed with L(0)=2, L(1)=1. This video covers the case where n=2, so S(10)=L(5)*F(7), and indeed the 5th Lucas number is 11. The remaining congruences classes mod 4 are not quite as neat, with S(4n)=L(2n+1)*F(2n+1)-F(1), S(4n+1)=L(2n+1)*F(2n+2)+F(1)-F(2) and S(4n+3)=L(2n+1)*F(2n+4)+F(1).
Since many people her like Lucas sequences, I propose a pretty fun problem to examine: Suppose we create a sequence by taking any two real numbers (x and y) and extend the sequence with the Lucas sequence rule (every next term is equal to the sum of the previous two) but also extend it to the other end (with the rule that every previous term is equal to the difference between the next ones. So the general sequence with x and y would have the form: .... -3x+2y, 2x-y, -x+y, x, y, x+y, x+2y, 2x+3y, .... Let's name the terms: .... a(-1)=-x+y, a(0)=x, a(1)=y, .... Show that as n goes to infinity: lim(a(n)/a(n-1))=φ for almost every sequence and that as n goes to minus infinity this limit goes to -1/φ for almost every sequence. The key to my solution to this was finding the small special group of sequences for which this is not true and making a clever observation.
We are currently doing induction in math, and because our teacher is a non-native speaker he always pronounces "assume" as "asshume" and it always cracks me up :P Great video btw.
Shorter if harder version of the trick: ask your volunteer to write a column of just six numbers using the same method. When s/he gets to the fifth you can multiply it by 4 to get the sum of all six numbers. So in James' first example: 8+5+13+18+31+49 = 124 = 4x31. Here's an even shorter one if you're in a hurry: 8+5+13 = 26 = 2x13. (What about longer versions?)
It should be noted that this is not unique to 10 values. There is an equivalent formula thathappens every 4n+2, just with a different constant. For instance, the sum of 30 numbers is 1364 times the 17th number in the sequence.
Someone help me out here - been too long since I've engaged in any kind of mathematics here... Seems to me that each step can be written as pairs of consecutive Fibbonaci numbers, multiplied by your starting numbers a and b , i.e. Term (N) = F(k)*a+F(k+1)*b, and that the sum described can then be termed as Sum(T1-TN) = a(Sum(F1 to F(N-2))+1) + b(Sum(F1 to F(N-1))), where F1 = 1, F2=1, F3=2, F4=3, F5=5 ... How would you frame such a formula in standard notational conventions? Missed out on statistical mathematics during my school days. Also any holes in the thought process, feel free to bring them to my attention, with the appropriate amount of ridicule of course. EDIT: Note that this applies from Step 4 or 5 onwards, forgot to account for the initial conditions. Yes, yes, I know.
Hi There-on your chalkboard did you know that you actually inverted the order of the first two digits in your sequence? I’m no math genius, but I do believe that in a Fibonacci sequence it is the number 5 followed by the number 8. (0, 1, 1, 2, 3, 5, 8, 13...etc.) However, You wrote: 8, 5, 13, 18. This throws off all the subsequent sums. Shouldn’t it be instead: 5, 8, 13, 21, 34, 55, .... etc? (On the other hand, you have a great smile and telegenic personality)
Multiplying by 11 is as simple as this: First multiply by 10. (Just add a freaking 0 to the end.) Then add the original number to that. Multiplying by 9 is almost as easy, you just _subtract_ the original number as the last step. It works for other 10n±1 cases as well. Multiply by 21? First multiply by 2, and tack on the 0. Then add the original number. Multiply by 39? Multiply by 4, tack on the 0, subtract the original number. No matter the number of digits involved, the rules don't change.
The trick is that, with the Fibonacci numbers, is that the n-th number is just the first number times the (n-2)th Fibonacci number plus the second number times the (n-1)th Fibonacci number. So we're just gonna count: 1-1-2-3-5-8-13-21-34-55 So the 7th number is 8 times the second number plus 5 times the first number, while the sum is (34+21+13+8+5+3+2+1)=88 times the second number and (21+13+8+5+3+2+1+1)=55 times the first number, which is just 11 times the 7th number. Ah, and we also see that the sum of those numbers is (the (n+1)th Fibonacci number -1) times the second number, plus the n-th Fibonacci number times the first number.
people like you, matt parker, steve mould etc. live the life i really wanna live, but first i must do a maths degree, i hope i get in to the uni ive chosen
I heard about this trick about 8 years ago when I was 9 from a book called Secrets of Mental Math by Arthur T. Benjamin and Michael Shermer. Have you heard of either of them?
Jack is married looking at Anna that if she is not married the answer is "A" yes. In the event that she is married she is looking at George that is not married and therefore Anna is the married person looking at unmarried George, the answer is a definite "A"
Yes. And sometimes I like to do videos for kids, they're the one's I want to engage with maths. It is also for other people who want to show it to kids. I threw in a 'prove this' for people who wanted a bit more.
Did I miss something? You wrote out a formula at the end as S=F_(n+2)-F_2, but I don't understand how that relates to your original trick of S=11*F_(n-4)
Why it works:
a
b
a + b
a + 2b
2a + 3b
3a + 5b
5a + 8b
8a + 13b
13a + 21b
21a + 34b
----------------
55a + 88b = 11(5a + 8b)
...which is 11 times the 4th number from the bottom.
You're the real MVP. I was left empty after the video.
@Spencer 3035 No, he isn't. He just copied my comment from the day before...
LDHFIntegrale Actually, I worked it out myself. But nice conceit there.
And a liar at that. Sad.
LDHFIntegrale Cool story bro. Too bad it's fiction.
a = first number (F1)
b = second number (F2)
F3 = a + b
F4 = a + 2b
...
F7 = 5a + 8b
...
Sum F1 through F10 = 55a + 88b
F7*11 = F10
Symmetric if you write them in the other order to start.
Except I don't know anyone who wouldn't screw up the adding part.
same
Give them a calculator
My friend told me a Fibonacci joke, it was as bad as the last two combined.
Haha nice. I also have a joke, this one's about Fermat; but the problem is I don't think it will fit in the comment section.
+MegaMrFroggy l have found a truely marvellous joke about Fremat, this comment section is to narrow to contain.
Don't Steal Arthur Benjamin's joke!
I saw this on another video, but I just now understand lol
I don't have any friends that can add 2 numbers together, sorry.
i dont have friends at all
Non Rompere Ddddffd :'( me neither
Lolzords. I wasn't saying I don't have friends. I do. They are just morons.
@@thegrandmuftiofwakanda They probably think a moron is a particle.
I did this in school, I now have 3 girlfriends and 7 side-chicks. All of them products of my vast intellect of course.
so they're imaginary? do you even go to school? what is real anymore!??
+mrBorkD
Is this the real life?
Is this just fantasy?
Causing a landslide
No escape from reality
Face the truth
open your eyes
+Mendel Chow caught in a landslide*
I used to add the fibonacci numbers to get to sleep. numbers and patterns have always brought me peace. Thank you for this trick! It's fun and great knowledge
Well this induction proof is astoundingly simple. You don't even need a piece of paper. Just pause a video and do it in your head. Beautiful piece of mathematics James!
It is - I was considering doing it for the video, but decided people who know induction can do it themselves, and people who don't will just enjoy the bit with the trick.
I tried this at my friends house, they called me a nerd and beat the piss out of me, thanks a bunch
+mattyfox666 You need better friends
@@singingbanana I don't think he was being serious... :)
@@jimmythewig3354 I don't think he was being serious either... :P
@@singingbanana Can we be friends?
Thank you, Dr. Grime! I'll show this to the middle school students in my Intro to Number Theory class today.
Perfect. I thought this would be one for teachers :)
The kids loved it. I did it a few times with the class & they copied down the examples. After two days, most of the kids, either through trial & error or algebra, found the pattern. A couple of them are now seeing how far they can take it out to find an easy multiplication trick for a longer series of numbers.
I'm going to this year's Fibonacci Convention - apparently, it's going to be as big as the last 2 put together!
I played around with this a little bit and observed the following:
To every second number (all on odd positions) there is a point, where all numbers summed together are a multiple of this particulary number:
The sum of the first three numbers is always two times the third.
The sum of the first six numers is always four times the fifth.
The sum of the first ten numbers is always eleven times the seventh. (The subject of the video)
The sum of the first fourteen numbers is always 29 times the ninth.
The sum of the first eightteen numbers is always 76 times the eleventh.
The sum of the first 22 numbers is always 199 times the thirteenth.
It feels like the sum of the first 4n-2 numbers is always k times the (2n+1)th number, but I can't find a pattern in the k values. I would be glad if James could make another video about this where he explains the patterns
this man really just hit us with "the proof is left as an exercise to the reader"
That's smart :D
Beats the crap out of those "subtract the number you first thought of, your answer's 9" type 'tricks' :P
That is Sooooo true
1to 20 countine chosse any 6 number sum is 20 wo konsa number hoge jin 6 number ka sum 20 hoga
multiplying by 11 much easier:
n * 10 + n
i agree lol
That's basically the same thing. Times 10 you just add a zero, but then the addition is the slightly slower part, but if you think about it, that's basically exactly what his trick is doing. You can ignore the first digit since it's being added to 0 and won't carry over, so you only care about the middle digit carrying.
That's the normal way. His way is much faster
This way is the exact same as what is shown in the video. There are no fewer and no extra steps.
You will never get as quick doing it that way as his way. T'is the difference between adding a 3 digit number and a 2 digit number, and adding two 1 digit numbers. It's obvious which is actually easier.
Thank you ! I will try this tomorrow with my teacher!
Yes! :D
Great stuff, James! The multiply-by-11-trick reminded me of old Trachtenberg! His ghost lives on...
Keep the videos coming when you've time.
I realize it's a little off-topic, but it needs to be said that you have the most contagious smile on the internet.
Paused at 2:05 to wok out the answer. Each term is as follows (x,y,x+y,x+2y,2x+3y,3x+5y,5x+8y,8x+13y,13x+21y,21x+34y) which sums to 55x+88y, the fourth term from the bottom is 5x+8y which is 1/11 of the sum. I'm sure there are other things you can do with this because the coefficients are all fibonacci numbers.
e.g. Sum to the 6th term is 4 times the 5th term. and sum to 14th is 29 times the 9th term
well done, but what do you mean by other things that can be done with this?
As shown in the e.g. I'm sure there are other sets of numbers you could use to get something with this.
In your e.g. you note that sum to the 6th term is 4 times the 5th term, and sum to the 14th is 29 times the 9th term. We can also see that sum to the 18th is 76 times the 11th term, sum to the 22nd is 199 times the 13th term and so on. What I find fascinating is that while the coefficients in your nx + my expressions are Fibonacci numbers, 4, 11 (James' example), 29, 76, and 199 are alternate Lucas numbers. And the next alternate Lucas term down from 4, namely 1, can be derived from "sum to the 2nd term is 1 times the 3rd term".
Good trick! The proof is fairly easy once you prove (with induction, as you said) that thing about the sum of Fibonacci numbers. Thank you very much for the videos you make here and with numberphile!
I'm so showing this to my discrete math professor. Fibonacci and induction. I don't know what could get him more excited. :D
Generalization for any:
f a Fibonacci series
n a natural number
k an odd natural number
L the Lucas series (2, 1, 3, 4, 7, 11, 18...)
f(n+2k) = L(k+1) × f(n+k) + f(n)
For n = 2, you get
f(2+2k) = L(k+1) × f(n+k) + f(2)
f(2k+2) − f(2) = L(k+1) × f(2+k)
Sum of all the elements of f from 1 to 2k = L(k+1) × f(2+k)
Example for n = 2, k = 5 :
f(2+2×10) = L(5+1) × f(2+5) + f(2)
f(12) = L(6) × f(7) + f(2)
f(12) − f(2) = 11 × f(7)
The connection between this trick and the Lucas sequence can also be expressed as follows: When your volunteer V gets to the seventh number (S7) in her/his sequence, then you multiply that number by 11 to get the partial sum for S10, as Dr Grime showed. When V gets to S9, you multiply by 29 to get the partial sum for S14. When V gets to S11, you multiply by 76 and get the ps for even further ahead, namely S18. And so on. These multipliers are alternate terms in the Lucas sequence, and if you can memorise them (or just use it openly as a reference) you can still come up with results to impress your friends even though you probably have to use a calculator. You can go in the opposite direction down from S7 too of course: at S5 the multiplier is 4 to get the ps for S6. And at S3 you multiply by 1 to retrodict rather than predict the ps for S2.
Proving the sum formula by linear algebra is a good exercise too. It isn't as straightforward as induction, but it lets you find out the thesis by yourself.
In 1995 when I was in grade 2 I looked like a genius when I could multiply by 11.
I tried showing this off to a few friends just then and they just said something along the lines of "can't math. Art degree".
I miss primary school. I had morale.
When I was in highschool I developed a system for quickly squaring 2 digit numbers in my head. It was great for my physics and algebra homework, but nobody else really cared "cuz calculators".
Ah the tech generation problem. I mean, I don't think it's a problem like our older generation says, but it's always interesting and admirable to hear about self discoveries.
In school I also self-discovered how to add two fractions with unlike denominators. We were learning about LCM at the time using the tedious 'list all factors' method. I discovered that I could just multiply each fraction with the other denominator to produce like fractions. It didn't produce the lowest multiple denominator but I solved my exercises in a less tedious and more efficient way. This got me some grief for not doing the task at hand actually.
I didn't know what I was doing at the time but I discovered it worked for all fractions. My mind was blown. Only when we learnt algebra (in the following year?) I proved to myself a/b + c/d = ad/bd + cb/db. Mind blown again!
Or have a race to the 12th number, let them do all the hard work and when they get to the 6th number, multiply by 11, add the second one and write that down. Probably more impressive with big numbers, but it will seem like you were doing in your head faster than they could do it with paper.
I have a question. What do you do for a living? Are you a teacher? How do you make your money? This would be like the best life ever for me. Your videos are great!
I think he might be a professor at Cambridge. Not sure though
I thought it was Nottingham.
proffessor at nottingham
+tman301j He taught courses on Cryptography at Cambridge
Link: www.ice.cam.ac.uk/components/tutors/?view=tutor&id=1892&cid=6842
He used to work with the Millennium Maths Project at Cambridge, but he is a freelance maths presenter now. He has several different presentations that he gives in the UK and around the world, but the topic he's probably best known for is the cryptography of the Enigma machine. He has addressed it in various videos and on his website.
I love you man, you are amazing! Your vids on numberphile are my favourite as well! The excitement on your eyes everytime you share new facts and curtiosities really brings out my curiosity for math! (Even though i hate algebra!) Shout outs from Brazil!
I love this trick! I am also interested in mathematical tricks such as 1089, which takes a random number, does a few calculations that don't cancel themselves out (like Add 3, multiply by 2, subtract 6, now divide it in half), but I am having a hard time to search out more of these equations. Do you know if these types of equations have a group name or something to help my search?
What you described cancels itself out... x (add 3) x+3 (multiply by 2) 2x+6 (subtract 6) 2x (divide in half) x. 😁
Let's choose the numbers x and y
1. x
2. y
3. x+y
4. x+2y
5. 2x+3y
6. 3x+5y
7. 5x+8y
8. 8x+13y
9. 13x+21y
10. 21x+34
The sum of all these is 55x+88y=11*(5x+8y) which is the 7th column.
Or the last 4th...
@@Luffy_wastaken and extend it for the other fact:
11. 34x+55y
12. 55x+89y
55x+88y=55x+89y-y
Excellent lesson. Shorter, clearer, funnier and more complete than the average competitors. It would be interesting to show the "minus second term" pattern during the sums using a table in order to show the inductive pattern explicitly.
Instead of using induction I rewrote it as a telescoping sum.
F(1) + F(2) + ... + F(n-1) + F(n) = (F(3) - F(2)) + (F(4) - F(3)) + ... + (F(n+1) - F(n)) + (F(n+2) - F(n+1)) = -F(2) + F(n+2)
Was really awesome seeing you today at the Institution of Education man! Thoroughly enjoyed the talk :)
This trick just got me layed!
Your spelling suggests you are in kindergarten. That's dirty.
Cadde Cant you go torture someone on a grammar channel?
bryphi77 Can't you go learn how to spell?
Or go play in your sandpit kid.
+Cadde So you have never made a typo? That is impressive! You are definitely the intellectual superior of us. Sleep well tonight knowing that you have corrected a grammer error of a lesser human.
*grammar
An easier way to do the n case is to double the last number, then add the second to last number, then subtract the second number.
Lemme check,
does 2f(n)+f(n-1)-f(2) equal f(n)+f(n+1)-f(2)?
2f(n)+f(n-1)=f(n)+f(n+1)
f(n)+f(n-1)=f(n+1)
correct(because of the definition of fibonacci numbers, woo!)
its crazy how every fibbonacci sequence is unique
I love watching your videos. They always leave a smile on my face. Your passion and enthusiasm are infectious! Thank you!
thanks! my 10 yo son loved it - and ended up learning how to multiply by 11. He was even able to prove it: 55a+88b = 11(5a+8b)
+DitDede This has made my day!
This is really easy James; you write the first number as a and the second number as d and the write all the ten numbers in terms of a and d and then calculate their sum in terms of the same, and one would find that it is 11 times the 4th number from the bottom.
Do you ever get tired of being such a maths legend?
Who would get tired of being a maths legend?
Can't wait to see more in this channel and numberphile..
Thank you
Your easy multiply by 11 method is a bit mental! Very creative! I just multiply by ten and add the quantity multiplied to that once, to make 11 x whatever. But seeing yours was intriguing. Alround good vid
Wanna hear a good fib?
Dancer says blitzen speaks elven; blitzen says yes; dancer says blitzen speaks 'other' languages as well; blitzen says yes; Dancer wants to know if blitzen is lying.
do you know the 111 trick? it just a cool little design under the number. For Example:
231×111= 25641
- (it's easier to show on paper)
--- you write down the last
------ number, 1. Then the last two
--- added, 4. Then all three, 6.
- Then the first two, 5. And finally, just the 2.
Thanks brother, starting over again is absolutely awesome. Good thing I'm retired....lol
p.s., teaching this to my 6yr old Niece.
Instead of doing a couple more steps, you can stick with what you have.
F(n+2) = F(n+1) + F(n) = (F(n) + F(n-1)) + F(n) = 2F(n) + F(n-1).
So you're calculating 2F(n) + F(n-1) - F(2).
Liked the math, loved the little mouse in the corner of the chalkboard.
You're always excited when you're talking about maths. It's a shame that more people aren't like that.
nth term formula for the fibonacci sequence derivation
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ..............
F_n = F_(n - 1 ) + F_(n - 2 )
I like to call the first term as the 0th term.
Let F_n = r^n
so F_(n - 1 ) = r^( n - 1 )
and F_(n - 2 ) = r^( n - 2 )
That means that r^n = r^( n - 1 ) + r^( n - 2 )
1 = ( 1 / r ) + ( 1 / r^2 )
r^2 = r + 1
r^2 - r - 1 = 0
using the quadratic formula we obtain :
r_1 = ( 1 + sqrt ( 5 ) ) / 2
and r_2 = ( 1 - sqrt ( 5 ) ) / 2
Now let F_n = a*( r_1 )^n + b*(r_2 )^n
F_n = a * [ (1 + sqrt 5 ) / 2 ]^n + b * [ ( 1 - sqrt 5 ) / 2]^n
Let's try it for the first two terms since it is going to be the easiest values.
1 = a + b
1 = a * [ ( 1 + sqrt 5 ) / 2 ] + b * [ ( 1 - sqrt 5 ) / 2 ]
solving the system of two equations yields :
b = ( - 1 / sqrt 5) [ ( 1 - sqrt 5 ) / 2]
a = ( 1 / sqrt 5 ) * [ (1 + sqrt 5 ) / 2 ]
Now to back substitute : F_n = ( 1 / sqrt 5 ) * [ ( 1 + sqrt 5 ) / 2 ]^( n + 1 ) - ( 1 / sqrt 5 )* [ ( 1 - sqrt 5 ) / 2 ]^( n + 1 )
James, try starting out with 133/11 or 12+1/11 for both the numbers, for a nice little surprise
My friends are going to think I'm so cool!
There's one small problem with this trick. If they make any mistake while creating that list, your answer will be incorrect, and when they check it eg. with a calculator, and your answer does not match, it becomes clear that you tried to use trickery. In order to make sure that your trick works, you would need to check that they calculated all the numbers correctly.
Aww, there's a little mouse in the bottom corner of your chalkboard! Also, what do magnets have to do with proving the F(n+2)-F(2) thing?
Can't tell if that question is a joke, but I'll answer anyways.
He isn't referring to electromagnetic induction, but mathematical induction. It's a proof technique, if you want to look it up.
I was thinking the same, with the induction....
If you do 2 numbers between 1 and 10 you could just memorise the answers since there's only 18 or 19 total possibilities depending on if you allow duplicates.
Another way I think of x11s is you add the number to itself plus an extra digit 0 at the end.... for example 11 x 11 = 121 = 11 + 110, or 126163984 x 11 = 126163984 + 1261639840 = 1387903824
It seems to take me roughly about the same amount of time to do it this way as to do it your way, with probably a couple more seconds dedicated to writing the number over before addition.
you are probably the worlds most demonstrative example of "how it is like to be in a passion" (?)
I really like how enthusiastic you are when making this video
Hey Amazing vid !!
I am just curious about your opinion on Vedic Mathematics
I have a question. For the last trick, F(n+2) - F(2) is the sum F(1) + F(2) + ... + F(n). But the order of the first 2 numbers doesn't matter, so doesn't that mean the answer would not be always correct? Like what if the example you gave was 5 + 8 + ..., then F(2) is 8 and the sum you get is 885 - 8 = 878? What am I missing?
But then F(4) would be 13 + 8 instead of 13 + 5 and so it would lead to a whole different sequence of numbers.
x*11 = x*10+x which in many cases is waaaaaaay quicker ie 53*10=530+53=583
absolutely LOVE THIS!! :D I wish I had you as my math teacher :D
Thanks for the video! As usual, your videos are awesome!
My friend literally showed me this trick a few days before this came out.
A more general version of this trick is to say S(4n+2)=L(2n+1)*F(2n+3), where F(n) is the nth number in your generalised Fibonacci sequence, S(n)=F(1)+F(2)+...+F(n) and L(n) is the nth Lucas number, indexed with L(0)=2, L(1)=1. This video covers the case where n=2, so S(10)=L(5)*F(7), and indeed the 5th Lucas number is 11.
The remaining congruences classes mod 4 are not quite as neat, with S(4n)=L(2n+1)*F(2n+1)-F(1), S(4n+1)=L(2n+1)*F(2n+2)+F(1)-F(2) and S(4n+3)=L(2n+1)*F(2n+4)+F(1).
Since many people her like Lucas sequences, I propose a pretty fun problem to examine:
Suppose we create a sequence by taking any two real numbers (x and y) and extend the sequence with the Lucas sequence rule (every next term is equal to the sum of the previous two) but also extend it to the other end (with the rule that every previous term is equal to the difference between the next ones.
So the general sequence with x and y would have the form: .... -3x+2y, 2x-y, -x+y, x, y, x+y, x+2y, 2x+3y, ....
Let's name the terms: .... a(-1)=-x+y, a(0)=x, a(1)=y, ....
Show that as n goes to infinity: lim(a(n)/a(n-1))=φ for almost every sequence and that as n goes to minus infinity this limit goes to -1/φ for almost every sequence.
The key to my solution to this was finding the small special group of sequences for which this is not true and making a clever observation.
Sittin there thinkin "What the f*ck, this actually works"
My friends can't add. I show them a mirror and they think I am an alien.
Love all your videos! Thanks so much.
I FOUND HIM. YES.
If you had given the multiply-by-11 trick it's own video I would have been perfect fine with that lol. That's a nice trick.
Hi James nice to have you back on the Tube!
We are currently doing induction in math, and because our teacher is a non-native speaker he always pronounces "assume" as "asshume" and it always cracks me up :P Great video btw.
Yey, I prove it without looking at the comments before. Thanks, singingbanana, you are making me do math!
I've been using this trick to win beer in proposition bets for years ;^>
Shorter if harder version of the trick: ask your volunteer to write a column of just six numbers using the same method. When s/he gets to the fifth you can multiply it by 4 to get the sum of all six numbers. So in James' first example: 8+5+13+18+31+49 = 124 = 4x31. Here's an even shorter one if you're in a hurry: 8+5+13 = 26 = 2x13. (What about longer versions?)
If I haven't been watching, can I still get thanks?
I did this a a magic trick. Finally won me a beer, by betting my friend, my math teacher wouldn't know F7 * 11 = E (1-10) Fi
It should be noted that this is not unique to 10 values. There is an equivalent formula thathappens every 4n+2, just with a different constant. For instance, the sum of 30 numbers is 1364 times the 17th number in the sequence.
I should elaborate that the constant will be ϕ (1 + ϕ)^(4n+2) + (1 - ϕ) (2 - ϕ)^(4n+2), where ϕ is the golden ratio.
This is so simple and clever.
Someone help me out here - been too long since I've engaged in any kind of mathematics here...
Seems to me that each step can be written as pairs of consecutive Fibbonaci numbers, multiplied by your starting numbers a and b , i.e. Term (N) = F(k)*a+F(k+1)*b, and that the sum described can then be termed as Sum(T1-TN) = a(Sum(F1 to F(N-2))+1) + b(Sum(F1 to F(N-1))), where F1 = 1, F2=1, F3=2, F4=3, F5=5 ...
How would you frame such a formula in standard notational conventions? Missed out on statistical mathematics during my school days.
Also any holes in the thought process, feel free to bring them to my attention, with the appropriate amount of ridicule of course.
EDIT: Note that this applies from Step 4 or 5 onwards, forgot to account for the initial conditions. Yes, yes, I know.
And then the revived John Von Neumann solves the same sequence before you can times 11 in your head.
Hi There-on your chalkboard did you know that you actually inverted the order of the first two digits in your sequence? I’m no math genius, but I do believe that in a Fibonacci sequence it is the number 5 followed by the number 8. (0, 1, 1, 2, 3, 5, 8, 13...etc.) However, You wrote: 8, 5, 13, 18. This throws off all the subsequent sums. Shouldn’t it be instead: 5, 8, 13, 21, 34, 55, .... etc? (On the other hand, you have a great smile and telegenic personality)
I kinda feel like calculating 11n by doing 10n + n is easier than doing this digit addition method, personal preference though.
Multiplying by 11 is as simple as this:
First multiply by 10. (Just add a freaking 0 to the end.)
Then add the original number to that.
Multiplying by 9 is almost as easy, you just _subtract_ the original number as the last step. It works for other 10n±1 cases as well. Multiply by 21? First multiply by 2, and tack on the 0. Then add the original number. Multiply by 39? Multiply by 4, tack on the 0, subtract the original number.
No matter the number of digits involved, the rules don't change.
Fibonacci numbers were discovered in India centuries before by the art of music
The trick is that, with the Fibonacci numbers, is that the n-th number is just the first number times the (n-2)th Fibonacci number plus the second number times the (n-1)th Fibonacci number. So we're just gonna count:
1-1-2-3-5-8-13-21-34-55
So the 7th number is 8 times the second number plus 5 times the first number, while the sum is (34+21+13+8+5+3+2+1)=88 times the second number and (21+13+8+5+3+2+1+1)=55 times the first number, which is just 11 times the 7th number.
Ah, and we also see that the sum of those numbers is (the (n+1)th Fibonacci number -1) times the second number, plus the n-th Fibonacci number times the first number.
I tried math tricks at a party once. Since then I visited no more parties :/
people like you, matt parker, steve mould etc. live the life i really wanna live, but first i must do a maths degree, i hope i get in to the uni ive chosen
You should go over Pascal's triangle!
Hey I have seen the same trick on the channel "scam school" by Brian brushwood. Any chance you know him?
Oh! I had no idea! I do know Brian, we're internet pals. I hope someone will reply with the link.
His delivery is a lot more forced. Plus, that hairstyle.
Thanks. Ah, it was five years ago - I think that's acceptable. Good old Brian.
I heard about this trick about 8 years ago when I was 9 from a book called Secrets of Mental Math by Arthur T. Benjamin and Michael Shermer. Have you heard of either of them?
Ishaan Sabnis I have, that's where I first saw this trick too. Arthur Benjamin's amazing.
Jack is married looking at Anna that if she is not married the answer is "A" yes.
In the event that she is married she is looking at George that is not married and therefore Anna is the married person looking at unmarried George, the answer is a definite "A"
Fibonacci
0,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765...
Is this for kids...? (Except the figuring out why part of course)
Yes. And sometimes I like to do videos for kids, they're the one's I want to engage with maths. It is also for other people who want to show it to kids. I threw in a 'prove this' for people who wanted a bit more.
That's awesome some kids will have a better interest at maths and has a little fun puzzle for more mature audience!
Thanks, that's the idea :)
...kids of all ages
Get a kid to start early enough, they won't be scared of math :D
Hi, I really love your videos. Obe question... Can you solve olympic math problems? Could you solve an interesting one in one video?
You weren't lying when you said the proof was easy. That's the only time I think the basis was harder to prove than the induction hypothesis.
+William Rutherford I agree!
Did I miss something? You wrote out a formula at the end as S=F_(n+2)-F_2, but I don't understand how that relates to your original trick of S=11*F_(n-4)
I absolutely hate math and I can’t stop watching these videos... help!
if only I had this guy as my math teacher
I've never known how to quickly multiply by 11, thank you so much!
Fn marvelous.
+Castor Quinn Ha!
That was a really good video, I found it interesting and I learnt from it too 😊
can u pls do a vid on general formular for fibonacci numbers(yes that giant fomular containing calculus and stuff)