Can You Pass Harvard College Entrance Exam?

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

[SNOW.]

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).

[anands]

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

[eowmob]

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.

[richardslater]

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.

[johnbrennan]

You can also divide t^3+t^2+t-155 by t-5 to get the other factor

[thejelambar]

We can use long division to factorise it, much easier and intuitive

[kyintegralson]

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,
for θ=cos⁻¹(-3/√31)≅.681π
⇒ log₂(t)=log₂(31)/2±iθ/ln(2) where "ln" is the natural log.

[tombufford]

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.

[jpl]

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 !

[StephenHeinz-pl]

Since t=5, the binomial would be (t-5). At this point why not use Synthetic Division to determine the t^2+6t+31?

[cyruschang]

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)

[matushorvath]

Step 1: I know 5 is a solution
...thanks, that really helps /s

[Sam_on_YouTube]

This was a very unintuitive way to solve it. I managed it in my head. Here's how.

Pick it up from t^3+t^2+t=155
Add 1 to set up something with better symmetry for factoring:
t^3+t^2+t+1=156
(t^3+t)+(t^2+1)=156
t*(t^2+1)+1*(t^2+1)=156
(t+1)(t^2+1)=156
Now we guess that t is a integer. That means we get 2 integer factors of 156. The factors are:
2*78
3*52
4*39
6*26
12*13
Of those, a quick check shows 6*26 works with t=5.

Then you can continue as you did. But just randomly coming up with t=5 from thin air made no sense.

I don't know the step you used to factor out t-5. I would have done that with long division. In reality, I skipped it and stopped when I found one solution. Turns out, the other solutions aren't real, but I stopped without proving that.

[michaeljarmula]

Here's a numerical answer (if you want to check your work): X = approximately 2.32193, according to a Microsoft Excel worksheet.

[richardslater]

So your solution is to guess t=5 and then randomly guess substitutions such as -30t+31t=t.

[monroeclewis]

😮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.

[TimothyLoftin-li]

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

[pbassassinz]

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.

[quantumcat]

I do not think this guy statement (in the title) is true.

[chipthequinn]

I went to Harvard. There is no entrance exam.