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
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!
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.
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.
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".
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.
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.
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.
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!
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.
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!
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 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?
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).
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.
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!)
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.
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
+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.
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.
What if, at the beginnin I said 5 and 8...not 8 and 5. The 3rd, 4th, 5th etc totals would still be the same, but the second number in the column would now be 8, not 5.
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.
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).
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.
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.
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.
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 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.
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.
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 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?
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.
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.
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"
Just memorize the repeating decimal places of fractions with 17 in the denominator and hope that someday someone needs to know one in decimal form. Then just start spouting accurate digits while somebody gets a calculator.
Also F1 + F2 + ... + F6 = F8 - F2 = 8a + 12b = 4*(2a + 3b) = 4*F5, and (trivially) F1 + F2 = F4 - F2 = a + b - b = 1*a = 1*F1. These three multipliers, 1, 4, 11, are alternate terms in the Lucas sequence, as is 29 in the case of F14 and 76 for F18 and so on.
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?
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 think i found a little mistake in the last explanation. I am well aware of how the fibonacci system works. But in the end he says that F1+F2+....+Fn = F(n+2)-F(2). In his Example it is 8 and 5, which is 13 and so on. And he is correct by the sum of 885. But 8+5 is exactly the same as 5+8. In this case it would be F(n+2)-F(1). So it should be minus the lower number. In all his examples he wrote the lower number as second number, that's why it worked out. Or i am just too bad in understanding him, in case he even said that :P
I've been looking into what happens when you swap round Fibonacci terms (as you happened to do in your first example) but keep adding them the same way. When you begin 8,5 instead of 5,8 and continue 13,18,31,49,80,129,209, 338 . . . , the 7th number after 13 equals twice 13 squared. Start with 13,8 for example and the 8th term after their sum, 21, is 882. That's twice 21 squared. And so on. Moreover 7 and 8 are the original Fibonacci indexes for 13 and 21 respectively. Maybe a trick can be made out of this too. Finally if you do such a swap on the Fibonacci 1,2 you get the Lucas sequence 2,1,3,4,7,11,18 . . . in which the 4th term after 3 is twice 3 squared.
F1 + F2 + F3 +…+ Fn= F(n+2)- F2 F1 + F2 + F3+….+ Fn = F(n+2) - 1 Which one is correct????😅😅😅 Many networks show second formula and prove it! So, can tell me which one is correct?😊😊😊❤
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.
No. If you reverse the order of the first two numbers then the 4th number, in this example, will be 21 instead of 18, and the following numbers will consequently also be different. All of the numbers after the 3rd will be different, and the end total will be different, but it will still be correct.
