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.

Except I don't know anyone who wouldn't screw up the adding part.

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 don't have any friends that can add 2 numbers together, sorry.

My friend told me a Fibonacci joke, it was as bad as the last two combined.

I did this in school, I now have 3 girlfriends and 7 side-chicks. All of them products of my vast intellect of course.

I tried this at my friends house, they called me a nerd and beat the piss out of me, thanks a bunch

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

this man really just hit us with "the proof is left as an exercise to the reader"

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

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!

Thank you, Dr. Grime! I'll show this to the middle school students in my Intro to Number Theory class today.

multiplying by 11 much easier: n * 10 + n

I'm going to this year's Fibonacci Convention - apparently, it's going to be as big as the last 2 put together!

I realize it's a little off-topic, but it needs to be said that you have the most contagious smile on the internet.

That's smart :D Beats the crap out of those "subtract the number you first thought of, your answer's 9" type 'tricks' :P

Thank you ! I will try this tomorrow with my teacher!

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

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.

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.

I'm so showing this to my discrete math professor. Fibonacci and induction. I don't know what could get him more excited. :D

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)

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)

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)

its crazy how every fibbonacci sequence is unique

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?

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.

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!

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!

Can't wait to see more in this channel and numberphile.. Thank you

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!

This trick just got me layed!

Was really awesome seeing you today at the Institution of Education man! Thoroughly enjoyed the talk :)

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.

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.

Liked the math, loved the little mouse in the corner of the chalkboard.

I love watching your videos. They always leave a smile on my face. Your passion and enthusiasm are infectious! Thank you!

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.

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 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).

You're always excited when you're talking about maths. It's a shame that more people aren't like that.

Do you ever get tired of being such a maths legend?

Thanks brother, starting over again is absolutely awesome. Good thing I'm retired....lol p.s., teaching this to my 6yr old Niece.

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

My friends are going to think I'm so cool!

you are probably the worlds most demonstrative example of "how it is like to be in a passion" (?)

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 )

And then the revived John Von Neumann solves the same sequence before you can times 11 in your head.

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?

My friends can't add. I show them a mirror and they think I am an alien.

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

Hey Amazing vid !! I am just curious about your opinion on Vedic Mathematics

James, try starting out with 133/11 or 12+1/11 for both the numbers, for a nice little surprise

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.

I really like how enthusiastic you are when making this video

I've been using this trick to win beer in proposition bets for years ;^>

Thanks for the video! As usual, your videos are awesome!

absolutely LOVE THIS!! :D I wish I had you as my math teacher :D

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.

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.

Fibonacci numbers were discovered in India centuries before by the art of music

Fn marvelous.

I FOUND HIM. YES.

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.

Hi James nice to have you back on the Tube!

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"

My friend literally showed me this trick a few days before this came out.

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.

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.

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.

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.

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).

x*11 = x*10+x which in many cases is waaaaaaay quicker ie 53*10=530+53=583

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?

Hey I have seen the same trick on the channel "scam school" by Brian brushwood. Any chance you know him?

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

Love all your videos! Thanks so much.

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.

Sittin there thinkin "What the f*ck, this actually works"

Yey, I prove it without looking at the comments before. Thanks, singingbanana, you are making me do math!

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?)

I tried math tricks at a party once. Since then I visited no more parties :/

If I haven't been watching, can I still get thanks?

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.

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.

Curiouser and curiouser!

This is so simple and clever.

I absolutely hate math and I can’t stop watching these videos... help!

So cool! I showed it to my friends, I got beaten up.

I got distracted and missed the first minute of the video. I thought this was a stack of random numbers and it was all a joke.

I kinda feel like calculating 11n by doing 10n + n is easier than doing this digit addition method, personal preference though.

You should go over Pascal's triangle!

I've never known how to quickly multiply by 11, thank you so much!

Fibonacci 0,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765...

if only I had this guy as my math teacher

AND YOURE GONNA LOOK SO COOOOL

I felt like the last part was a trick because you said that they could write their two numbers in any order. So reversing the 8 and the 5 means the function no longer works.

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)