BELIEVE IN ALGEBRA, NOT CALCULATOR
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- Опубліковано 4 кві 2019
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blackpenredpen | 曹老師
i did some mental math, but hit a wall at trying to find the square root of 63,252,753,001
that’s some impressive mental math assuming you’re telling the truth . Is there a trick or something
@@iamgroot3615 theres no trick hes probably lying
r/iamverysmart
AngryAxew there's no reason not to be able to mental math those numbers
Like 500(500+1) which is easier which is 250000+500 and it similar to the end
I mean it's OBVIOUSLY 251501.
1990: we'll have flying cars by 2019
2019: 2=1+1, wow I'm a genius
LOL
2+2 is 4, minus one that's three quick maths
Flying cars... you can't even have a sharpie that could change color. Say, red and black.
@@ghotifish1838 topical meme reference
Well it can also be 2=500-498
I just started learning English, but the explanations are clear and interesting even at my levels of English. Thanks a lot 😁👍
Юлия Охременко I am
Glad to hear!
x2
U r indian Chinese korean or ....???
@@harelavv8806 the name may seem obviously Russian to some but not all
@@blackpenredpen hii
My last words whispered in a final breath : "Don't forget the +1"
😂
LMFAOOO
He never resumed the video
He added the one wdym 🧐
Wow... this essentially proved that if you take the product of four consecutive -numbers- integers and add one to it, than it's gone be a square number.
Awesome! Good spot!
Gábor Tóth holy shit you’re right! That’s crazy man
Yup!!
The most pathological case I can think of is -1 thru 2, and yes indeed I get 1, which is a perfect square.
My professor had us prove a more general result: take the product of four numbers in arithmetic sequence, then add the fourth power of their common difference. Show that the result is a perfect square.
Olympic math taught me that insanely hard problems often had elegant solutions, this is no exception.
: ))))
@@leif1075???
@@leif1075 people like these are called problem solvers...
@@leif1075 well it took me about 5min to solve it so I think is not impossible to solve. All of this types of equations where you have 4 consecutive numbers multipled are done like this
Lol,This problem is actually super easy,(Every single Olympiad contestant would have solved this question,at some point of their life) .Insanely hard problems need not have simple solutions . That's a downside of the math Olympiad .They make you expect difficult problems have simple solutions.(Although,most imo contestants don t fall for this fallacy).Real insanely hard problems have not been solved by anyone,yet.
Instead distributing at 4:18
u = x^2 + 3x + 1
(u - 1)(u + 1) + 1 = u^2
So the root is x^2 + 3x + 1 = 251501
Bryan Lu omg that cat!!!!
Or even u=x^2+3x, then u^2+2u+1
digno de nyan cat, jajajajaj
My life is a lie. I thought u subbing was only for integrals
@@mattat3847 nah man, sub whenever it makes the problem simpler
"If you're using a calculator, why are you watching this video?"
Sanity check.
hahahahaa
I am a calculator, not a person
*laughs in Shakuntala Devi
i put it in my calculator and got 251501, that was easy
Boo!
@The Balton American calculators are beasts
@@Netherexio Agreed but Americans aren’t
@@paradox9265 What do you mean?
It was easy, but not so beautiful like this)
no 2=1+1/2+1/4+1/8+1/16...
you have got many misconceptions blackpenredpen!!!
When he writes 1, he's obviously just abbreviating 1/2+1/4+1/8+1/16+...
@@iabervon and when he writes 1/2 he's abbreviating for 1/4+1/8+1/16+...
@@InDstructR And when he writes 1/4 he's abbreviating 1/8 + 1/16 + 1/32+...
@ki kus won't stop me,
And when he writes 1/8 he's abbreviating 1/16+1/32+1/64+1/128+...
Whoa that converged quickly
Dafuq did i jusf watch.i lost it when the 2=1+1
You are not nerdy enough
Jeez thats smart
*proceeds to use the calculator to prove that 251501 is the right answer*
Did you know that 2 = 1 + 1?? I bet not!
jk : )
blackpenredpen No i don’t, i need a calculator to this
That's pretty funnt, but sometimes such things are the most difficult to see, for example: we have f(x)=x^4+8x^3+18x^2+8x+17, and a question, for which x, the function f(x) is a prime. You can check infinitely many cases and never know the answer, but what makes this question easy (but on the other hand is not so obvious), is that 18=17+1. Because then we have (x^2+1)(x^2+8x+17), which has to be a prime. One of them has to be = 1, the other one has to be some prime then... We are left with only 2 cases, because we know, that 18=17+1. : )
No but I knew 2= 0.9+1.1.
I Known 2+2 = 5
What i know is 5/2 = 2 with int data type
1 + 1 = 3
And sinx/n=six=6.
You stupid, 1 + 1 ≠ 3
3 = 2 + 1
π = 2 + 1
π - 1 = 2
@@rio_agustian_ so π = 3 lol nice discovery
@@CookieJar2025 e = 3 = π
@@rio_agustian_ noob
@@Kevin-14 lool nooob
5:48 “Back in my day kids would use *ALGEBRA* but now their brains are rotting from these darn *CALCULATORS* ”
I remember solving this exact question in my JEE ( Mains ) exam.
@Sanat R mains usually has easy questions
@Sanat R - Study Vlogs Sure, yeah, "easy question" 😬
@Sanat R - Study Vlogs Woah, really? What kinda questions do they ask? Could you send me a link?
Take x ^2 + 3x = a
Then in step 2
a(a+2) + 1
a^2 + 2a + 1
= (a+1)^2
Yes much easier, and you see it imediatly too.
was about to say this, i think it's a lot more intuitive
Isn't it beautiful how one problem can be solved in diffrent ways, even if the idea and the method are nearly the same. That's why we love maths.
@@milanmitreski7657 Yes
You extra smart boy the time required here will be same
0:10 is this a pewdiepie reference? 😂😂
He liked it!!!
@@albel2094 yupp
Yupp
Nice. I did it this way:
Assume that the expression is a square number so:
x(x+1)(x+2)(x+3)+1 = n^2
x(x+1)(x+2)(x+3) = n^2 - 1
x(x+1)(x+2)(x+3) = (n+1)(n-1)
What I did then is realise that the factors of the product on the right differ by 2. Playing around you can find that:
x(x+3) = x^2+3x = n-1
(x+1)(x+2)=x^2+3x+2 = n+1
So n = x^2 + 3x + 1
Not as neat as your method though!
Thanks for the video
OMG ive almost done it completely. i just stopped at (n+1)(n-1) lol WD! i mean 'not that almost' lmao
How do I play around with it
@@joshuamason2227 Well you have the product of 3 binomials and a monomial for which we can multiply in any order. If you try a few cases, or think about it you spot that x(x+3) and (x+1)(x+2) have a difference of 2.
@@dr3w199 okie
Damn... It's a great method. Neat work. 💯
Not only 2 = 1+1, but also 0 = 1-1.
From the second row:
(x^2+3x+1-1)(x^2+3x+1+1)+1 and per the third binomial equation
= (x^2+3x+1)^2 -1^2 +1
= (x^2+3x+1)^2
I didn't understand. Where did you use binomial expansion
0 = 1-1
1 = 1*1
2= 1+1
I didn't see any binomial here... But what i see is that you used the form (a+1)(a-1) + 1 = a^2 - 1 +1= a^2
@@yogeshpathak73 I think Tibor Grün is from Germany. In German school curriculum the formula (a+b)*(a-b) = a^2 - b^2 is known as 3. binomial formula. b=1 is a special case.
Oh ok... Didn't know that. Thanks.
0:09 That was PowerFul
now do it with CALCULUS
Why do you think this is funny?
@@davidappell3105 because suffering is funny
@@davidappell3105 Its FUNI
This is beautiful. I've been looking at it for five hours now
I love this. You did a great job of laying out a good challenge.
Sorry...Time over! give me your exam!
I'm only in 8th grade Algebra 1 but I was using variables to find how some of your factorizations works.
You went from (x^2+3x)(x^2+3x+2)+1 to (x^2+3x)(x^2+3x+1)+(x^2+3x+1).
What I did was set (x^2+3x) to a variable (a).
(a)(a+2)+1
a^2+2a+1
(a+1)(a+1)
Now substitute back in.
(x^2+3x+1)(x^2+3x+1)
When in doubt use variables..
Yes, that's using even more algebra than BPRP did.
Well yes because using the variables is actually the logic behind the solution, it's just that it was invisible throughout the process :D
Where r u from?
Agreed! Variables always help to proceed the solution.
I put x=500 but multiplied everything. In the end i got to sqrt((x+y)^2) with x=500 and y=501^2
a very good perspective and a very good solution. thank you!!!
Blew my mind! Earned yourself a new subscriber! Keep up the good work!👍
0:01 that's my life philosophy now
Continueing from: sqrt( (x^2 + 3x) (x^2 + 3x + 2) + 1 )
let y = x^2 + 3x
sqrt( y * (y + 1) + 1 )
= sqrt( y^2 + y + 1 )
= sqrt( (y+1)^2 )
= y + 1
= x^2 + 3x + 1
= (x + 1) (x + 2) - 1
= 501 * 502 - 1
= 251501
much easier to multiply :p
that is what i wanted ti type nice 👍
Why are your videos so entertaining? I'm so glad I came across this channel.
i remember my math teacher asking me to prove that n(n+1)(n+2)(n+3) + 1 is always a perfect square given that n is an integer
The world needs more teachers like you. I'm more impressed by your teaching skills than any math. Much respect.
Really good solution! GOOD Teacher👍
This just blowed my mind!!! Love this
That was beautiful. Thank you.
You can also put a +1-1 inside the x^2+3x bracket and it'll be in the form of (a+b)(a-b).
that's what i thought he was gonna do as well but what he did was cool as well.
why?
Yeah, (x-1)(x+1)+1=x^2-1^2+1 seems easier to find than multiplying out exactly the right portion of the big expression.
What a incredible content. Im a student of math (i'll be a teacher in the future) from Brazil. Thank you so much for sharing knowledge!
Your explanation is awesome . I like your teaching very much. Thanks
Really good video. Thanks for inspiring students. Keep it up
I chose to make x = 502, which ends up yielding a nice difference of squares and a two term quadratic, which is much easier to distribute. The quartic you get has a palindromic pattern reminiscent of pure binomial coefficients, making it tempting to say the golden ratio is a root. It is, in fact, a root, so synthetically divide the quartic by the golden ratio identifying polynomial, x² - x - 1. You end up with the golden ratio identifying polynomial again, meaning that the original quartic in that square root is (x² - x - 1)², so cancel the power and the root. Plug 502 back in for x, some quick multiplying and subtracting by hand and you've got 251501.
I love the explanation, though I did it a bit differently. When I got to the second line, I substituted (x²+3x) as y and found that that worked much simpler than distributing 2 as 1+1.
This was fantastic. I wish to thank you for your videos.
I love you, you make me remember stuff I had forgotten!
3:20 I solved it differently.
Let y= x^2+3x. Then substitute y into the expression making y(y+2)+1, distribute so y^2+2y+1 and that is a perfect square of (y+1)^2.
Here, the square root and exponent cancel each other leaving y+1, sub back in x and then easily find the answer :)
this is also what I did and I think that this is a bit better because you don't have to split 2 into 1 + 1 and do the rest
That's what I did as well
I am so impressed with myself, I actually used the same method you did before watching the video =D
Jan Wrobel nice!!!!
i watched this video this video right before my math competition and the same type of question came up on the task sheet. Thank you very much!
for those wondering the question was
202120212019(202120212021)(202120212023) all over 100010001 x (202120212021 squared +4)
Seemingly elementary problems can have wonderfully elegant solutions! All we need is to substitute a number with x, and the magic begins.
3:20 just put y = x^2 + 3x, then you have y(y+2) + 1 = y^2 + 2y + 1 = (y+1)^2. So the answer is y + 1, or x^2 + 3x + 1.
I also did it in this way...but that way was also fine...its all about which method comes in your head first
Dang that’s genius
Every body knows 1+1=2 but i know 1+1 =/= 3
♫♪Ludwig van Beethoven♪♫ Hahahhaha
I’ve got you all beat with 1+1 > 0
@@jgsh8062 nah mine's better 1+1≠1+1
Actually trying it out was fun, and there was some surprisingly elegant generalization going on behind the scenes.
man this was amazing to watch! so clever!
There are things to learn from each of your videos 😁❤️
Well yes, that's the point.
I did assume there was a nice solution, but expanding under the root to get x^4 + 6x^3 + 11x^2 + 6x + 1 was pretty easy, and then matching coefficients in (x^2 + ax + 1)^2 was straightforward too.
But yeah, the main thing is to replace 500 by x. I don't think I could intuitively see which two of the brackets would make it easier, and I'm not sure that's a better method than expanding the whole thing to only 4 terms (plus the one on the outside).
The entry was epic and the whole video is interesting.
You are an absolute genius bro🤟🤟
I was surprised by your last step result 😱😱😱
please give an example differentiation of complex functions
Doing a PhD in Literary Studies, but stuff like this is why I absolutely love maths ♥
It is so interesting and good to remember. Thanks.
This is very useful, I’m trying to think more outside the the box for hard mathematical equations either for proofs or something like this, I didn’t thing to make 500 = x, but it simplified it so much. This is a useful skill and I will be sure to use it
You can also this as x^2+3x=t and expression would become t(t+2)+1 =(t+1)^2 this que came in practice test for jee last week And guess what i solved that 😎😎😎👍👍
I love your videos about not using calculators (Like the Wolfram-Alpha video)! They're the best! Keep up the good work! It's nice going back to algebra sometimes...
Why It?
Yea me too. I try to mix things up a bit.
@4:15 wow I did not see it coming! Really good thank you!
I expressed it as sqrt((501.5-1.5)(501.5-0.5)(501.5+0.5)(501.5+1.5)+1)
You get two a^2-b^2 expressions that you can multiply out, add the 1, and then factor into a squared quadratic expression
Very neat and, as someone mentioned elsewhere, it generalizes to “1 plus the product of any four consecutive integers is a perfect square”
Now this video makes me like algebra
ua-cam.com/video/XX2DI9E1zV8/v-deo.html
Everybody knows e^{iτ}=1
.
.
.
.
But I know 1=e^{iτ}
Nice!!!
@@user-bd9mu3ee1i That's e^{iπ}. τ=2π
My mans using tau! Up top!
@@user-bd9mu3ee1i he's using tau not pi
Nice video . I learnt a lot
Nice... Great method of solving
I know this is not related to this video but I wanted to post this on a new video so you might see it :) your trick for integrals of thinking "wouldn't it be nice if..." has helped me so so much, so thank you :) love your videos!
Awww thank you!!!!!
I know this kind of the prob, i use (n+1)(n+2)-1
Man, you deserve more subs!
More impressed with how someone came up with the question
Nice factoring method but it might have taken me a while to spot. Multiplying out and factoring isn't so bad
(x - 1)x(x + 1)(x + 2) + 1 = (x^2 - 1)(x^2 + 2x) + 1 = x^4 + 2x^3 - x^2 - 2x + 1 = (x^2 + bx +- 1)^2
= x^4 + 2bx^3 + (b^2 +- 2)x^2 +- 2bx + 1
We see this works if b = 1 and c = -1 so the answer is 501^2 + 501 - 1 = 500^2 + 2*500 + 1 + 500 = 251501
"And now, here's the deal"… You know that when he pronounces that phrase things are 'bout to get complicated.
WOW! Now that's an amazing way to solve it!
When he drops the "Check this out", you know crazy stuff will happen on the board
Dude, I always had a good grasp on algebra as a kid and in highschool I always aced most algebra, but somehow my teachers (and I) missed this property of algebraic equations. So freaking cool. It has been nigh on 15 years since high school, but I am still learning new and cool algebra. Thanks so much blackpenredpen!
Shout-out to my colorblind fam who can never tell when he switches pens
3:32 perhaps an easier method to spot is if you partially expand the brackets so that you end up with (x²+3x)² + 2(x²+3x) + 1, which is precisely (x²+3x+1)²
Awesome solution to the question
Everybody know e^2.pi.i = 1
.
.
But I know 1 = e^2.pi.i
Wrong it's
- (e ^ pi × i)
@@mundane3809 Nice try but thats -1
ignoring your name
@@nikolas9105 no
e ^ ( pi × i ) = -1
So if you make -1 negative, it become positive.
@@mundane3809 you are correct but the original comment was also correct. e^2πi = 1.
@@RunstarHomer oof yea it's actually correct. sorry for the mistake!
this guy is a genius!
Very interesting calculation!
I multiplied everything out and got stuck... This is a brilliant solution. One day I will achieve this type of mathematical intuition. Lead the way, blackpenredpen!!!
Wow your method and my method are similar .....I had take the whole expression as y and then square it and then assume x to be 500 and multiplied and had taken x^2+3x to be z and at end I got y=z+1 that is y = x^2 + 3x + 1 and it's done
me : multiplies it out, then square root it
#bprp #YIAY
wow, just wow!! Fascinating !!
if only most teachers were like this guy, it actually makes me wanna learn math again and I'm 29 years old! not gonna lie that did look fun for some reason.
What Everybody knows : 1+1=2
What BPRP knows : 2=1+1
.
.
.
.
.
What I know : 1+1=2 and 2=1+1
😇😇😇😇😇😇😇😇😇😇😇
Молодец! Хорошо объясняет, все понятно. Спасибо
Ага, он знаток
3:19 The "WOW". LOL
this is the best channel of the platform. no doubts.
Him: Starts the video with 1+1=2
Me: *ok we are getting somewhere now
I'm doing this on the toilet, so I only hope I'm starting correctly, with (500)(502)=(501²-1) and (501)(503)=(502²-1)
But then again, we could cheat and go with (501)(502)=(501½²-¼) and (500)(503)=(501½²-9/4)?
I need to start a tradition of watching this video every year. It's just that good
Your way to solve this is pretty AWESOME!! First I multiplied all together, i got x^4 + 6x^3 + 11x^2 + 6x + 1 then i calculate this polynomial for x=1 x=2 x=3 ....all the time i got a square!! I was really surprise!! I didn't expect x(x+1)(x+2)(x+3)+1 to be a perfect square for all x at all!!! This is incredible!! Thanks for sharing your knowledge you are very inspiring to me
Just do the multiplication by hand.
I have never been so hyped at 2 = 1+1 before.
Thats reall nicee man!!
This was a great explanation