Can You Pass Harvard College Entrance Exam?
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- Опубликовано: 1 окт 2024
- Entrance examination and Olympiad Question in 2022. If you're reading this ❤️.
What do you think about this problem?
Hello My Friend ! Welcome to my channel. I really appreciate it!
@higher_mathematics
#maths #math
Maybe this is an easier way:
T = 2^x
T + T ^2 + T^3 = 5 +25 +125
T + T ^2 + T^3 = 5 +5 ^2 +5 ^ 3
So T = 5
So 2^x = 5
x Log 2 = Log 5
x = Log 5/Log 2
Start with the equation 2^X = Y.
Given:
Y + Y^2 + Y^3 = 155
This can be rewritten as:
Y * (Y^2 + Y + 1) = 155
Find the factors of 155:
155 = 155 * 1
155 = 31 * 5
Test the possible values of Y:
Y = 1 does not work.
Try Y = 5:
LHS = 5 * (25 + 5 + 1) = 5 * 31 = 155
Since the left-hand side matches the right-hand side, Y = 5 is correct.
Thus, we have:
2^X = 5
Solve for X:
X = log(5) / log(2)
So, the value of X is log(5) / log(2).
excellent proof.
That's the way I did it. The key is realizing that 155 is the product of two primes (5 and 31), which simplifies matters considerably. I hate it when presenters introduce terms (add this, subtract this, etc) without explaining why they chose the particular values that they did.
Method is correct if you suppose that x is integer.It' s not the case if x is given as real.
@@jacquesbasuiau215It assumes 2^x is an integer l
@@silverhammer7779 Dude the thing he did is simple grade 9 mathematics. He simply did factorization of a cubic equation, any person who has graduated high school should be able to do that. I don't think it needs explaining
Take the step from t^3+t^2+t=155 to t^3-5t^2+6t^2 etc. This seems like a totally random step. Why choose -5 and +6? I love to know the reasoning behind this step.
By the method of staring at the equation he noticed that 5 is one of the roots. He then chose coefficients so that he could factor out (t - 5) to check if there are other real roots.
You can as well group like this:
(t - 5) + (t^2 - 25) + (t^3 - 125) = 0
(t - 5)(1 + (t+5) + (t^2+5t+25) ) = 0
@@georgerodionov5941 Oh OK, thank you.
@@georgerodionov5941 I have another proposal: after guessing that 5 is the root, we can notice that t, t^2 and t^3 are all monotone increasing functions (at least in positive t, which is what we are interested in), so their sum is increasing too. And then there cannot be more than one intersection with the horizontal line at height 155.
@@СергейМакеев-ж2н yes, this is probably the nicest way to do it
Only solution is x = log₂5 😊😊😊😊
As apparently some others, I (with a phd in maths, but out of science for years) am a bit confused about the solution. If you guessed t=5, to prove it you can just enter it into the formula, no need to factorise. To factorise, just do polynomial division, no need to guess something wildly like the separation of t^2 into -5t^2+6t^2 etc. Polynomial division tells you algorithmically how to do it.
Yes, there do exist theorems (and this is well beyond ordinary school maths) that if such a polynomial (with integer coefficients) has a rational (a fraction) solution it must already be an integer and then it will be a divisor of the constant term (155). Hence, it is a good strategy to try divisors of 155 (aka: +/- 1,5,31,155). If your teacher likes you, one of them will work. Once you factorise to degree 2, the other solutions can be found with pq formula or quadratic extension, however you name it. BTW, there do exist (very complicated) methods for degree 3 and 4 too.
BUT: There is no indication that 2^x is actually rational. To state it clearly: this was just guessing and hoping for the best. Of course, if you have no better idea, it is worth a try.
I kinda disgaree with "beyond ordinary school"
Here in Indonesia we learn about rational root theorem in high school. I believe Cambridge Further Math A Level or IB DP also teach about rational root theorem
@@thejelambar82 I sadly believe you. Of course I was talking about Europe and Germany, maybe US, where I strongly doubt you learn this before joining university. Comparing to Cambridge, well... Just lets say this is not what 99.999% of pupils experience.
It is a well known fact here, that the maths education in Asia is much, much better than it is here.
Personally, I'm a bit curious if there are other topics which are teached better in the western world than in Asia in return, or.. if our education is generally worse.
Again, I fear I know the answer.
2^x + 4^x + 8^x = 155
y = 2^x
y + y^2 + y^3 = 155
y^3 + y^2 + y - 155 = 0
(y - 5)(y^2 + 6y + 31) = 0
y = 5, -3 +/- i✓22 (not solutions)
2^x = 5 => x = (log5)/(log2)
Me who just finished 5th grade and have no idea what I just watched😅
Am I supposed to already know it??
bro no this is like grade 11+ stuff 💀
My brain: There’s a 5 in here, right?
Me: Maybe. But you’re 40 years old, you might have epilepsy and you had an actual seizure 4 weeks ago.
My brain: So that’s why I understand absolutely nothing else 😅
Saw the problem at first and thought how hard can it be 2^x + 4^x + 8^x=155 answer: x= log5/log2 FML SMH I give up on life.
I do not think this guy statement (in the title) is true.
You can also divide t^3+t^2+t-155 by t-5 to get the other factor
Yes, and solve the quadratic equation in 2^x as complex conjugates. Then take the log base 2 of them to yield an infinite number of solutions related to 2 main values.
Step 1: I know 5 is a solution
...thanks, that really helps /s
Just guess that the answer must be a bit bigger than 2 and then iterate. Much easier and faster than his baloney method
I think that you are making these up as you go along. Is this realy from Harvard, was the other one realy from Cambridge? They are good fun but perhaps not as hard as your titles suggest?
@alphalunamare True 💯, I too doubt the same because it is a Question of JEE Mains (Easy to Medium Toughness)
We can use long division to factorise it, much easier and intuitive
Yeah, exactly.
33% the question said nothing about “real solutions” 😜
You use a solution to a cubic equation which is quite difficult after forming the equation t^3 + t^2 + t -155 = 0. On inspection I would set t to 5 and solve. 125 + 25 + 5 - 155 = 0.
Ah, yea....let's just guess 4, no 6...no let's go with 5.
This could be a Dilbert solution....where the next to last step is:
"Then a miracle happens "
As stated ....there is NO Harvard entrance math problem
Clearly not good
Need to eliminate fake intelligence.
Teach life skills to high schools.
Complete teachings
Buying insurance.
Car home life insurance
And different types
Something actually helpful?!!!!??
I went to Harvard. There is no entrance exam.
Once yo make the change of variables you can get the real 5 solution by examination. You do the remaining part to show that the other solutions are complex and that you have not overlooked some not-so-obvious real solution
Write 155 in binary, then use the appropriate exponents, lol...
I fully agree with @fibonacci_fn ! Since the equation is degree 3, you have to "guess" an integer solution for Y, and Y = 5 comes rapidly (especially when you notice that 155 = 31 x 5) !
My remark is : we have to show that 5 is the only real solution (and then ln 5 / ln 2 is the only x)...
For that purpose, we see that f (x) = 2^x + 4^x + 8^x is strictly increasing on R, and the TVI says that it takes a given value 155 for at most a single x.
(By the way, this avoids the calculation of (Y + Y^2 + Y^3 - 155) / (Y - 5), and the "∆ < 0" of the quotient...). Thanks for your videos !
How can you justify not finding the complex solutions too?
Extremely simply "hard task".
ln.
5³ + 5² + 5 = 125+25+5 = 155
x = ln(5)/ln(2)
can use polynomial theorem? Horner scheme? btw it's same with log2(5) right?
Start converting 155 to binary
Lets not throw out the complex babies w/ the bath water! So, complex roots are given by
t²+6t+31=0 ⇒ t=√31·[(-3±i√22)/√31].
To evaluate log of t, we choose the branch cut of the log to be the non-positive real half-line & restrict the arguments of the complex numbers to lie between -π & π. Then,
log₂(t)=log₂(√31·[(-3±i√22)/√31])=log₂(√31·e^(±iθ)) for θ=cos⁻¹(-3/√31)≅.681π
⇒ log₂(t)=log₂(31)/2±iθ/ln(2) where "ln" is the natural log.
Why don't you write it to show that there are an infinite number of solutions?
@@oahuhawaii2141 You can generalize the solution I mentioned by adding 2nπ, for n any integer, to the expression for θ, but you have to consider consistency & functions becoming multi-valued. For one, the real solution mentioned in the video, should then be generalized to
t=5 → t=5e^(2ikπ) ⇒ x=log₂5 → x=log₂5+2ikπ/ln(2) for k∈ℤ
& you should choose k=n, so that all 3 solutions are consistently in the same interval. And, why not generalize further
t=2^x → t=[2e^(2iπm)]^x for m∈ℤ ⇒ x=ln(t)/(ln2+2iπm)
So, to at least be consistent w/ the real solution presented in the video, I chose the fundamental interval for the complex plane.
Here in India we do this kind of problem even harder in 7th grade even in remote villages schools 😂😂 "VANISHING METHOD " ASSUMING AND PUTTING, X=5😂😂😂😂😂😂😂😂😂😂😂
Womp womp my father is a government school teacher.....most village kids suck at calculation
Aaaah I wanted to see the complex solution as well
Oh this famous Harvard college in Zimbabwe! Veri diffycult equation.
Forget the complex bullshit.
X=2.32 approximately
That’s what I came up with doing it in my head like my dad taught me at 7 😅
Did he guess the answer like I did?
x=log₂5
y u splitting 155 so tough just do 5 + 5^2 + 5^3 = 5 + 25 + 125 = 155 which makes it clear that its 5
Your math work isn't rigid. You should be writing that 2^x = 5 is one root to the cubic form of (2^x)^3 + (2^x)^2 + (2^x) = 155 . You still have to find x, and see if the other 2 solutions to the cubic equation are real or complex conjugates.
Can we do it like this for faster answer but the ans would be near the original ans not exact
2^x + 2^2x + 2^3x = 155
2^6x = 2^7.3( approx )
On comparing,
6x = 7.3
x = 1.21 ( this is my ans )
😮Independently followed your methods through the cubic equation, but then quickly realized t must equal 5 because only multiples of 5 add up to 155. Synthetic division could have been employed, but was unnecessary. Then solved with logs as you did.
2ⁿ+4ⁿ+8ⁿ=155
(2ⁿ)³+(2ⁿ)²+2ⁿ-155=0
Let h=2ⁿ
h³+h²+h-155=0
h³-125+h²-25+h-5=0
(h-5)(h²+5h+25)+(h-5)(h+5)+1(h-5)=0
(h-5)(h²+6h+31)=0
h²+6h+31=0
h²+6h+9=-22
(h+3)²=-22
|h+3|=i√22
h+3=±i√22
h=-3±i√22
2ⁿ=-3±i√22 reject
h-5=0
h=5
2ⁿ=5
n=log_2(5) ❤
Here's a numerical answer (if you want to check your work): X = approximately 2.32193, according to a Microsoft Excel worksheet.
Why not simply divide the cubic equation by (x - 5) to find the quadratic equation?
About 2.5
x=log2(5)
155/14 = 11 2/3
So your solution is to guess t=5 and then randomly guess substitutions such as -30t+31t=t.
Try speaking a bit slower and we might understand what you are saying!
x=ln5/ln2. оставшееся уравнение m^2+ 6m + 31 = 0 не имеет действительных корней.
Si poteva risolvere l'equazione cubica anche scomponendo il polinomio di terzo grado con il metodo di Ruffini, notando che tra i divisori del termine noto, 155, vi è proprio il valore t=5
...and that is how you calculate the net worth of Jeff Bezos
Let 2^x= y. So y+y^2+y^3=155
So y(y^2+y+1)=5*31
After close observation,Equating y=5, gives (y^2+y+1)=31. So conclusion y=5 =2^x
x = ln 5/ln 2. Very easy to get real solution 😂
If it is this easy to get to Harvard I need to change colleges right now
I've never seen someone write an x that way in my life
Yes. Thanks for asking😂
By taking log it will easily solved😂😂😂😂😂
We can frame a quadratic equation in t^2+ 6t+ 31 =0
Interesting would be if you would allow complex numbers as solutions. What would be the answer then?
See my other comment under this video.
log5/log2..... 5 plus 25 plus 125....
After a step t^3+t^2+t-155=0
You should take a deriative . D by the monotonicity theorem you can get no more than 1 solution
By taking derivatives we can't find the value of t . instead of the values obtained for critical point which is in complex from
@@keerthany7556 one of the root can be found orally , this is why you just need to prove , that this is the only solution
@@keerthany7556 in this problem this is one of the way of solving, but you can also use a Cardano formula , 2 other roots are complex
155 ÷14x =
11.071
That's what I got. What have I missed?
@@damienmorris2903
I hear you Bro 😐👍
I used Ruffini method
I'm the first to comment😊😊😊.
2^x=5
If I could pass the entrance exam, I wouldn't need to attend Harvard.
Given the choice I'd deliberately fail the entrance exam.
Math 55 covers essentially an entire undergraduate curriculum. And people have been known to place out of it. So there would still be plenty of things to keep you entertained for the next few years.
I don't wanna be member of a club that wants me as a member.
I'm black.
Harvard: "You're in!!!"
x = log2(5) is an immediately obvious solution
this is not an olympiad problem, this is way too easy
Thanks. Very nicely explained ! 👏👏
令a=2^x
a+a^2+a^3=155
a=5 則x=log5/log2
how about this?
10 print "higher mathematics-can you pass harvard college entrance exam?"
20 sw=.1:x=sw:goto 50
30 dgu1=(2^x+4^x+8^x)/8^x:dgu2=155/8^x:dg=dgu1-dgu2:return
50 gosub 30
60 dg1=dg:x1=x:x=x+sw:x2=x:gosub 30:if dg1*dg>0 then 60
70 x=(x1+x2)/2:gosub 30:if dg1*dg>0 then x1=x else x2=x
80 if abs(dg)>1E-10 then 70
print x
higher mathematics-can you pass harvard college entrance exam?
2.32192809
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run in bbc basic sdl and hit ctrl tab to copy from the results window