Fifth Root Trick - Numberphile
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- Опубликовано: 23 авг 2024
- A neat trick to quickly calculate fifth roots.
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Featuring Simon Pampena, Australian Numeracy Ambassador.
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Наука
Euler was such a badass. His work is everywhere.
There's a joke that says all things in math are named after the second person that discovered them, because the first one was always Euler
Yeah… .
@Alexis Hazel DeSilva They truly were astonishing.
Hes no joke hes a real genius
He definitely was, but I wonder how much of that was because simply not that many people were able to study mathematics at and before his time.
Ah the classic 69^5
5 parties of 69 bro
Tom Stack genius
of course they would do that
"That was a bit hard, but I got used to it."
... Yeah.... I bet. >__
1.0000000000000000000000000000000000000000000000000000000000000000000000000000000
Dude looks like an evil genius waiting for you to say that big number.
😂!!!
My wife's name is Anushka Sharma
@@ViratKohli-jj3wj I dont get it
Dude interviewing the other guy, if you turn your phone sideways you can use the scientific calculator to get 38^5 directly.
+Numberphile nice attempted save
+Srcsqwrn I'm fine with you being fine with this.
Android says square root 2 x2 is 2.8 so it's not number phile approved😃
sqrt(2)^2 = 2, not sqrt(2)x2 :)
I'm disapointed, rip maths
This channel is simply one of the best to ever come out on RUclips. The amount of time I spend watching these videos is profane.
I knew this trick with the third root. There for the last digit you have to swap 2 with 8 and 3 with 7 (and the other way around) and all the other are the same in the last digit.
Same. Square roots too, but you need to do more work since all last digits have 2 possibilities
Natürlich guckt DorFuchs Numberphile...
Hahaha, DorFuchs du pussyslayer :D
DorFuchs p
ja DorFuchs!
This dude's cool, get him here more often!
Check out his "Epic Circles" contribution. Just fascinating...
Why is there a deer walking around in the background?
Hero of the Beach Because it was open season on Mathematicians.
That's Lulu. Brady's dog.
no, its a cheetah.
how did i not notice that xd
That's dinosaur..-(1000x) smaller😂😁😂😁
5:47 "a lot of kids loved 69 to the power 5" haha, those sassy kids :D
it was a bit hard but i got used to it
XD
There is a trick similar to this with cube roots, and at one point i just remembered the number for 69^3...
at 3:25 I thought that was a deer behind him. LOL
Same lol?
Doggi
Lulu passed away sometime this year iirc. R.I.P. );
This guy is my favorite of all the people you interview. He won me over when he split my brain in half about the number line in one of your previous videos.
When he said, “How’d I do it so quickly?” I said to myself, “Cause your a genius.” Only to immediately hear him say, “Cause I’m a genius.”
He truly is a genius!
I started laughing so hard about the 69 thing.
electrocat1 Liking this comment would be wrong...
I waited for 69 likes for so long!
Euler never seems to stop impressing me... from 300 years ago
Every time I hit "like" when the number was originally Prime, I feel slightly bad.
I've now watched every single numberphile video. Binge watched them over the past week. Nice work Brady! Doing the same now for your other channels :P
Locut0s do you still watch them?
Time to look incredibly clever in front of my friends!
Do it!
And fail miserably and get embarrased in front of ur friends
You mean your top 10 friends?
not to be that guy 6 years later but I feel like intelligent would he a better word
Wow! Time to go win some bets!
This video is fantastic.
Video should end at 1:43 lol
Your likes ended at 143 lol
Orbrun, right now 205.
And growing: 214( not mine because in this prospective It may have ended at 1:39).
304 likes
314
I had a math pattern I found when in freshman year of collage that was like this. It was more thorough compared to Euler's Theorem. My Teachers refused to even look at it because and I quote, "You are just a student, you could never come up with anything of value." or "There is nothing you could have come up with that someone else hasn't already found."
I sent you what I was able to work out through twitter. It has been over a decade since I came up with it and I just was remembering it off the top of my head so it may be not formatted correctly. I don't have any connections, so this is literally the best method I can think of to get any feedback on it. Yes this is just a shot in the dark... And now I am realizing that this is a old video and... nevermind... :(
I love EULER
He's made of you anyway
StubbornAtom oiler*
Here is the proof behind Euler's theorem for a^5 = 10m + a. I will use a proof by induction.
1.) Let's assume that a = 1, therefore, 1^5 = 10m + 1. 1 = 10m + 1. 0 = 10m, and therefore, m = 0. For m = 0, 1^5 = 10m + 1.
2.) Let's assume that k^5 = 10m + k. Where m is an element of the set of integers.
(k+1)^5 = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1.
(k+1)^5 = 10p + k + 1, where p is an element of the set of integers.
(k+1)^5 - k - 1 = 10p.
k^5 + 5k^4 + 10k^3 + 10k^2 + 4k = 10p
10m = k^5 - k
10p - 10m = k^5 + 5k^4 + 10k^3 + 10k^2 + 4k - k^5 + k = 5k^4 + 10k^3 + 10k^2 + 5k.
Therefore, 10p - 10m = 10(p-m). (p-m) is an element of the set of integers since the set of integers is closed by subtraction.
From Subproof Awesome, below, we know that 5k^4 +10k^3 +10k^2 + 5k is always divisible by 10 for no matter what integer k.
Therefore, a^5 = 10m + a
End of Proof
Subproof Awesome
We need to prove that 5k^4 + 10k^3 + 10k^2 + 5k is always divisible by 10 for all k in the set of integers.
Let's suppose that k = 1
5(1^4) + 10(1^3) + 10(1^2) + 5(1) = 5 + 10 + 10 + 5 = 30. 30/10 = 3, and 3 is an integer. Therefore, it is true for k = 1.
Let's suppose that for k = q is true, can we assume k = q+1 is true.
5q^4 + 10q^3 + 10q^2 + 5q = 10h where h is an integer.
5(q+1)^4 + 10(q+1)^3 +10(q+1)^2 + 5(q+1) = 5(q^4 + 4q^3+6q^2 + 4q + 1) + 10(q^3 + 3q^2 + 3q + 1) + 10(q^2 + 2q + 1) + 5(q+1) =
5q^4 + 20q^3 +30q^2 + 20q + 5
+ 10q^3 +30q^2 + 30q + 10
+10q^2 + 20q + 10
+ 5q + 5
_____________________________________
5q^4 + 30q^3 + 70q^2 + 75q + 30 = 10r, where r is an integer.
Subtract 5q^4 + 10q^3 + 10q^2 + 5q from 5q^4 + 30q^3 + 60q^2 + 75q + 30 and you get 10r-10h.
20q^3 + 50q^2 + 70q + 30 = 10(r-h)
10(2q^3+5q^2+7q+3) = 10(r-h).
As you can see, they are always divisible by 10.
Therefore, for all k integers, 5k^4 + 10k^3 + 10k^2 + 5k is divisible by 10.
END of Subproof Awesome
You used so many ks the ku klux klan is inviting you to their next lynching
+CopiedOriginality he didnt use a single K !
+Joseph Willes But how is that a proof by induction? If you prove that the base case works (the smallest example of it working or the first step that you take) then you prove that it will work for any variable that comes after. Usually for that step we let n equal a new variable k.
He is inducting on a in the main proof and then k in the subproof. There is no n.
This isnt Eulers theorem.
Eulers theorem is: a^phi (n) = 1 mod n
That's a pretty neat trick. My usual math trick is to get someone to think of a number (while I think of the variable X), have them perform simple math operations (while I do the same to X), tell them to subtract their original number when I have some number plus X in my head, and then tell them what that number is. What's best is to ask what their favorite number is first and make it come out to that.
Example:
Pick a number, add 2, multiply by 3, subtract 3, divide by 3, subtract your original number: You're thinking of 1.
In my head:
X -> X+2 -> 3X+6 -> 3X+3 -> X+1 -> 1
The same works for cubing/taking cube roots you just need to remember the answers to 1-9 cubed. The difference is that if the last digit is an 8 (e.g 74088) the last digit of the number you've cubed is 2 and vice versa (in this case the answer is 42) and if the last digit is 7 then the last digit of the number you have cubed is 3 and vice versa.
Euler's theorem can be easily proved:
The Little Fermat theorem says
a^p - a is divideable by p, IF p is a prime.
That means a^5 - a is divideable by 5.
If it's divideable by 5, then the last number must be 0 or 5.
If it ends with 5 then of course it's an odd number. If it ends with 0 then it's even. We just have to proof that it's even, so it ends in 0 every time.
The way we show this is by doing this (i don't know how you say it in english):
a^5 -a = a*(a^4 -1)
If a is an even number then of course a*(a^4 -1) is even.
If a is odd then a^4 is odd too, and a^4 -1 is even so our number is even again.
We proved that it's even and divideable by 5, so it means it ends with 0.
But if a^5 -a ends with 0, then a^5 ends with a.
Ooops, I mean a^5 ends with a 's last number
+Shri harsha Nayak Yeah, you're right sorry :/
Fun tip... @ 1:04 on the calculator, you don't have to keep typing *38 = .. *38 = ... *38= .... --------- You can just type *38 once then press the "=" sign 4 times in a row, it will automatically preform the last operation (that being *38)
I think he was trying to make sure people knew he was doing 38^5
Albert Renshaw or you could just swipe the screen to the right....you will get a scientific calculator
Or you could just write the result directly.
Never take a pub bet against a Nottingham mathematician. :-)
Love this channel. Really nice work. One note: you guys probably don't realize how a sharpie on brown paper sounds in a recording. To many people, it's worse than nails on a chalkboard.
never thought it was annoying to hear. then again.. i don't care too much about nails on a chalkboard either
Personal preference. I quite like it.
I actually quite enjoy the sharpie sound. It's not grating like a chalkboard.
Very cool. This reminded me of when I was first learning the times tables. I didn't enjoy memorizing them so I just memorized the 12 times table and convinced people that I new all the rest. Because I must if I can do 12x, right?
69 was a bit hard. But i got used to it.
please no
+COLW321 Gaming lennyface.jpg
( ͡° ͜ʖ ͡°).tiff
nope.avi
69 likes
no one in this world gets to business as quickly as Numberphile.. legit
"Believe it or not, kids pick 69^5 a lot"
I believe it....
Mentally calculating 20^5 is easy: just apply the distributive property to exponentiation the same way you would with multiplication: 20^5 = (2 * 10)^5 = 2^5 * 10^5 = 32 * 100000 = 3200000
30^5: 3^5 = 9 * 9 * 3 = 81 * 3 = 3(80 + 1) = 24 * 10 + 3 = 243 = 24300000
40^5 = (4 * 10)^5 = 4^5 = 16 * 16 * 4 = 256 * 4 = 4(200 + 50 + 6) = 800 + 200 + 24 = 1024 * 10^5 = 102400000
That's all mental calculation, except I have a trick for 4^5. I know that sqrt(4) = 2, so I can just double the magnitude and get 2^10, which I happen to know is 1,024. Or you can square 4, subtract 1 from and halve the magnitude to get 4^5 = 4^4 * 4^1 = 16^2 * 4
Brilliant 😍
"69 is a bit hard, but I got used to it." - Simon Pampena
Damn it why couldn't all my math teachers in school be as enthusiastic as you?
Equis Igriegazeta teachers are enthusiastic when they don’t crash into unmotivated students... (as you?)
I can seem like I'm smart now
JustAnotherSunny one thing I'd point out is that for the X0^5 stuff you can just know X ^5 and how many zeroes it would have at the end times 5 are added on. in fact this potentially allows you to if you pay attention to do any number to the power of 5 in theory except they overlap at times.
Awesome trick, thanks for sharing! ;)
I love this guy!
I get the feeling he is really smart, really really smart, and I always feel he could probably work it out in long hand if he wanted. Great video, thanks
ok , i will admit that i am more of a numberphile than i was before watching this video, this channel is not only for someone who is a numberphile to enjoy, it will slowly make you in to one, just give it some time . thanks a lot for everything :)
Question: "It's a big task to remember that"
Response: "Is it?"
I love it! :)
Great minds are always interesting. Thanks! :)
You can extend this trick to any odd power (my dad used to teach me how to do it to find cube roots). The basic idea is to remember 10³, 20³, 30³ and so on. And obviously know the last digit of any digit cubed (as the euler's theorem only works witha power of 5). It can also work for power 7, 9, and 11 (I have not checked other powers)
I didn't even know something called a fifth root existed.. :)
I gues you don t know the solution of X^5 + 5t*X^3 + 5t^2*X = -2q = 0 then.
Dude Im knows as the smart kid in my class and when I go back to school monday I'm gonna blow everyone's minds. You've done it again numberphile!
Great number tricks, thanks for sharing.
Jjijijo
I didn't know Russell Brand was a math genius!!
Prazkat Reviews He looks more like Matthew Santoro to me.
the first sword of bravos does not run
He really does not look all that much like Russell Brand.....Doesn't have enough of an aloof look about him for one.
He doesn't look enough like a slimy, bohemian socialist
I wonder if he knows why kids choose 69^5...
likemynewname lol that's what I've been thinking
Pretty sure he does
He’s a mathematician, therefore he doesn’t know.
I’m a mathematician in the making so don’t take it too personally I’m just joking lol.
its because its used in lots of memes and stuff
They probably dont know the true meaning
The true meaning i-
Dude this guy is legit, MORE OF HIM!!
I wish numberphile was around when I was at school. I was ok at maths, but I think if I saw the beauty in it I would have been far more interested.
Haha, 69^5. Kids. XD
This will forever be the nicest ending to a Numberphile video
“That’s STILL a big ask, to memorize all that” LOL - I too, was here hoping that the trick were a lot easier than this- 😅
Haha! It would've been awesome if the video ended at 1:39 right after he says "I'm a genius" And we were all like whaaaat? xD
Yay, new party trick!
...yes, my parties are frequented by drunk geeks, why do you ask?
No one is asking
That's a neat trick, but how likely am I ever to need to know the fifth root? It isn't like square roots or even cube roots, which come up all the time!
This is one of the loveliest videos on the internet.
I really love when you post this videos with captions. I'm brasilian and i really don't understand much things. The math is ok, but i lost all jokes :/
I've seen this repeats with all (1+multiple of 4) powers like ^(1+4) or ^(1+8). The last digit stays the same.
To use a similar trick, you just have to learn the powers of the numbers from 0 to 9 (I guess is because 10 base numerical system.).
It does because a^5=10m+a, times this by a^4, and get a^9=10ma^4+a^5=10ma^4+10m+a, and same for a^8, etc
A slight variation of the trick allows for all 3+multiple of 4 powers.
well, memoization may work for these small numbers, but what to do, if I am handling sextillions ??
These guys never stop surprising me. Great job.
Now this I could see trying to use in social situations. Very cool.
This is really cool, can't really share because I don't want my friends to know about this :)
Simple way to memorize the first digit.
10^5 is the same as 1^5 with 5 0s after
20^5 is the same as 2^5 with 5 0s after
and so on. So just learn your powers of 5
Kumartheffar that’s quite obvious, isn’t it?
Please give us more calculator unboxings. I need more. I need them!
Here's a proof of the property that makes it work:
Since phi(10) = 4, Euler's theorem states that a^4 ≡ 1 (mod 10), multiplying both sides by a gives that a^5 ≡ a (mod 10), which means that 10 divides a^5 - a. This means that there exists a number q such that 10q = a^5 - a, which gives us that a^5 = 10q + a.
5:50 I wonder why the children say 69 :3 koff koff
***** I have no idea what you're trying to say.
Samurai Nakruf Because they are immature?
WarpRulez no, to trick him. :D
***** 69, 9 is just 6 upside down so its neat and easy to remember pattern.
GameDogLeader21 Sure, but that's not the reason the children find it amusing to suggest it and get excited about it
I personally really hate the "root" notation. It just obscures the relationship with exponentiation. Is it really that much harder to say to the 1/5 power instead of 5th root?
yes, however my advice is: deal with it. that's the least of problems a mathematician could have.
also, is anyone stopping you from using the power notation and not using the root-notation at all?
That's really only a problem when kids first learn about them (like 2pi vs tau) and with little practice it becomes hard to tell if it was ever a problem to begin with. If you really want you could mess with people by saying "the 1/2th root of X" for X^2.
how much harder is it to multiply by a fraction to divide, or add to a negative number to subtract?
i was told the equation
X^5-X will always be divisable by 30. i tried a lot of numbers and it all worked out to me true. now can you prove it? and is this related to the above.
Example 2^5-2= 30
Yes it can be proved .. Factorising gives (x-1)x(x+1)(x^2+1) .. The first three factors guarantee an even number and a multiple of 3. If none of the first three factors divides by 5 then either x-1 or x+1 must be adjacent to a multiple of 5. So either x-1=5n+1 or x+1=5n-1 and in either case we find that 5 divides x^2+1. QED
LOL! Every kid knows the infamous 69. EVERYONE OF THEM!
I dared my maths teacher to calculate in his head "69 x 69". My ten year old self believed no one can do it because of the 9's at the end. Too bad he worked it out in like 10 seconds.
Turns out you can multiply the number with the 9 unit to the next number. So he calculated 69 x 70 by multiplying then 10 then minus 69. I don't know why he said that way is easy! But thanks to him, I appreciated the 9 multiples and gained more insight in number manipulation. Awesome maths teacher!
Why does the number have to be in English?
Maybe he doesn't speak any other language...
Because in languages like German or Arabic, you don't say "sixty-eight", but "eight and sixty".
In German, for instance, it's "achtundsechzig" (8 and 60).
Hope that answers your question... albeit 4 years later.
Also was that a deer?
looked like it
Presumably the key figures here are 5 and 2 because their product is 10 (the base of our number system), so if it was based on 12 numbers the trick would work for 4th roots which are 3 digits in length and 6th roots of 2 digits in length.
i love watching this man talk about maths.
does this work in other bases other than base 10 such as base 12? I'm too lazy to figure it out myself.
hey Braidy ask your proffesors about spheres. more precisely about
1. turning sphere inside out
2. makeing 2 spheres (or actualy any number of spheres) from one sphere (you can cut a sphere into very small puzzle pieces and then when you put these pieces back together you can do so in such a way that you can make two or more spheres)
Unpronouncable talk shite pal
#2 is a Vsauce video
@@zashtozabogaepisode on Banach Tarski paradox
I tried to figure out a rule before he explained his trick. I came up with something similar, to guess the leftmost number. Since it was between 100,000 and 10,000,000,000, the root had to be between 10 and 100. Dividing by 100,000, I found out the number was between 3^5=243 and 4^5=1024, so the leftmost number had to be 3. For the rightmost, I tried most digits (some were obvious) to see how the 5th power ended, and was a bit surprised to find out they were always identical to the original number, so that gave the 8. Very cool !
I did some experimenting and found this same method works for every other odd root. a^5, a^9, a^13, a^17, a^21, etc. The other odd roots (a^7, a^11, a^15, etc.) also work, but you must memorize the possible outcomes (Ex: ending in 6 predicts the root ends in 4). Evens (a^6, a^8, a^10...) do follow patterns, but cannot be summarized in a predictive function (1-in, 1-out).
Whys there a deer in the back!!!?
do a video on 1,000,000 factorial
The engineer joke at 4:53 made my day lol thanks 😊
Love that stuff! Enjoyed it a ton.
The only issue is giving him a non-perfect fifth root, such as taking the fifth root of a random number like 766445.
That's the trick, he said think of any two digit number. This implies it's an integer between 10 and 99. Then he told that person to multiply it by itself 5 times and tell him the result.
They didn't start by giving him a large number and asking for it's fifth root, that would likely not be an integer result.
1 min in when he is on the calculator; Simon goes into "Gollum mode" :) hahahahahaha
If anyone cares: Eulers theorem states that a^(φ(n)) ≡ 1 (mod n) where n and a are coprime and φ(n) is number of numbers that are smaller than n and are coprime with n. So for example for n=10: φ(n)=4 (1,3,7 and 9). Therefore:
a^4 ≡ 1 (mod 10)
We multiply equation by a:
a^5 ≡ a (mod 10)
So any number to the power of five ends with the same digit as the number
This dude is just so... freakin... competent!!
Great job! You make me proud! 🤗
phi(10) = phi(2)*phi(5) = (2-1)(5-1) = 4
WHO are the cretins who thumbed this down?
Bro you are greatest mathematician for me on youtube
I shall blow some minds now , Thanks Brady and Mr. Simon Pampena :D
I have to say I enjoyed this video much more because it featured russell brand.
If it was Russell Brand he would have taken four times as long explaining it. :P
I do like Russell actually, but Simon looks nothing like him IMHO.
ShamelessHorse woah, so Russel Brand does mathematics even? ffs missing out on so much
3:25 a dog!
Because the 5th powers of the multiples of 10 are just used for rough comparison, you don't need to memorize all the digits. The code I use (in terms of millions) is 10=.1, 20=3, 30=25, 40=100, 50=300, 60=800, 70=1600, 80=3200, 90=6000
A lot of this stuff goes over my head, but when he showed the formula that included "ten times something plus the original number", I was at least able to reason out that it meant the second digit of a two-digit whole number on the input would be the last digit in the output. Small blessings. Been about twenty years from the last time I was in a math class. Nice to know I haven't lost everything.
And actually, looking at the way it's written one last time, I see now that that's... that's pretty much the entirety of what that formula means, isn't it? On its own, the sole conclusion is that the "ones" digit is persistent. That is simply a translation from the thing I said in English (well, from Euler's perspective probably either German or French) into math. So I didn't really divine anything out of the tea leaves. I just translated a mathematical sentence into English. Which isn't _nothing,_ but I guess all I've accomplished here is a basic literacy test.
0:37 O-oooooooooo AAAAE-A-A-I-A-U-
JO-oooooooooooo AAE-O-A-A-U-U-A-
E-eee-ee-eee AAAAE-A-E-I-E-A-
JO-ooo-oo-oo-oo EEEEO-A-AAA-AAAA
…What?
NoriMori Brain power!
Osu! somebody? Please?
+Nano poison What?
Oooohhhh ahhhhhh Oooohhhh ahhhhhh I love Tzuyu
Is it big?
Yea it's big, are you ready?
°¿°
`
oh how I would have loved maths with an ambitions teacher like him
You can do the same thing with other roots too, like 9th, 17th, 21st, 25th, 33rd and 41st. 41st roots are really cool because the last TWO DIGITS are the same as the given number. For example, 23^41 = 67739389260745218861137988047774370539553852007909099223
You know that you can turn an iPhone calculator into a scientific calculator (roots and exponents and etc) by rotating it 90°?
Scientific calculator :D
Pause at 1:05. Scary.
If a number is divisible by three, the sum of its base 10 digits is also divisible by three. That's just a fun trick I use all the time.
"I wasn't paying attention, don't take it so seriously". I'd like to say that to my boss after the meeting.