Everyone Failed To Solved This SAT Problem | Can You Solve This?

[@garyesser1007]

just wow! impressively convoltued

[@DJ-nw2ef]

You should call this channel Clickbait Academics. Obviously most of the people commenting here could solve the problem (as did I myself), so the claim that "everyone" failed to solve the problem is complete BS. And by the way, I just tried a simple guess, and immediately got x = 3 and y = 0, because 2³ - 7 = 3⁰. That is 8 - 7 = 1. Nor is guessing an invalid problem-solving technique. Lots of computer algorithms - Newton's Method, for example - make use of initial guesses that are then refined to get a valid answer. The real question here is not just how do you get an answer, but how do you do it as quickly as possible? Both the designers of the SAT and real-world engineers value efficiency, in addition to validity, and your method is ridiculously cumbersome (assuming that it is even correct).

[@FirstLast-n5b]

Let say Even=E and Odd=O. Assumption that if E*O=O*E then E=E and O=O is PLAINLY WRONG and it is no more than JUST ONE possibility or no more than guess. You just get lucky. Assuming (BTW this was not said) that we are seeking positive integer solution only it is not difficult to find 4 and 2. However proving that this is ONLY solution would be as difficult as to solve Last Fermat Theorem. So as far as math going you have solved nothing.

[@vorpal22]

He missed the solution x = 3 and y = 0 anyway, so he didn't solve the problem. A counterexample is sufficient enough to show that E_1 * O_1 = E_2 * O_2 does not imply E_1 = E_2 and O_1 = O_2: 4 * 9 = 12 * 3, but 4 ≠ 12 and 9 ≠ 3.

[@JohnH0130]

@@vorpal22 Even if he had listed that set of coordinates, it wouldn't amount to 'solving' the question, since (4,2) and (3,0) are only two of the infinite set of coordinates that satisfy the relation. I believe they are the only solutions in which both numbers are integers, however.

[@vorpal22]

@@JohnH0130 You are correct. There's no indication we should be forced to only find all the solutions in Z, and we would have to demonstrate that those are all the solutions.

[@GustavoPereyraGranda]

X=4 y=2 just out of my top of my head

[@jtfike]

I admire how convoluted you got to solve this problem. Way more than I would ever have mustered. The 16-9 = 7 relationship is what the SAT/ACT is asking you to identify because 16 and 9 are powers of 2 and 3 respectively. While I admit it can be a difficult problem, it is meant to be solved in 10-20 seconds...plus the SAT would have multiple choice answers so it would be easy enough to plug and play.

[@JohnH0130]

Actually an important point: in the SAT setting all you need to be able to do is quickly plug in answer sets until you find one that satisfies the equation.

[@dougwarner59]

I know one answer I did in my head: x=3 and y=0.

[@vorpal22]

Yes, exactly. That jumped right out at me.

[@artennsa6899]

3 sec to solve

[@vorpal22]

@@artennsa6899 Right? 8 - 1 = 7. Easy as fuck. There's no way "everyone failed to solve this SAT problem." The guy missed that solution and only got the second solution, x = 4 and y = 2. The math was also atrociously written. You can't just write that you're dividing the LHS by 2^2, nor can you assume an even term is equal to an even term and an odd term is equal to an odd term: 4 * 9 = 12 * 3, but 4 ≠ 12 and 9 ≠ 3.

[@edwardcullen1739]

x=4, y=2 also works; 16-9=7

[@JohnH0130]

This is a relation, with an INFINITE number of solutions. E G, x=3 and y=0 is an equally valid solution. Setting the even and odd expressions equal locks you into the one illustrated. This and the solution shown are probably the only two solutions in which both numbers are integers. As y becomes a large negative number, this devolves to 2 to the x power equals 7, which has the solution, x=(ln7)/ln2, or 2.807......... Saying the 'solution' is x=4 and y=2 is mathematically disingenuous, it is 'a' solution, one of an infinite set of mathematically valid solutions.

[@ckahrl]

I agree. It would seem that the solution to this problem would take the form of y=f(x).

[@FirstLast-n5b]

Such stupid problems usually assume that we are looking for positive integer solutions greater than zero. "Stupid" because it must be stipulated and almost always those KZbin "mathematician" never stipulate it. Or as very least it should be written as 2^n-3^m=7. Any such equation has indefinite solution in real number realms - it is just function in form y=ln(2^x-7)/ln(3), but I suspect that this particular one has only one solution in positive integers above zero. Pretty much as Last Fermat Theorem. Prove me that I am wrong.

[@lesking6541]

Dividing by (5-5) sounds like dividing by zero to me. I can prove a lot of things if I do that

[@timirbiswas3834]

What you are doing is totally wrong. You are saying if in an equation like odd1 × even1= odd 2×even2 then odd1 = odd2 and even 1= even 2. But it is wrong. For example 2×15 = 6×5 then according to your theory 2=6 and 15= 5 . Don't teach maths without proper thinking.

[@JohnH0130]

I also saw that he used this shortcut to find one solution, but you are right it is not a mathematically rigorous treatment.

[@danielleza908]

Some issues with this video: 1. You can't equate the even parts and odd parts, it is simply not true. For example, take the number 36, it can be written as 4*9 or as 12*3, but the even and odd parts are not equal. The even and odd parts being equal is only one possibility, but it is not the only one. As a result of this, you've missed the solution x=3, y=0. 2. You assumed that 2^(x-2) - 1 is odd, but you should have tested the case where x=2 seperately (because the expression is even for that case). 3. You did not state that x and y are natural numbers, but you seem to assume that in your proof. That adds a layer of confusion to the video.

[@mirandahotspring4019]

I did it in my head and got 16 - 9 =7 so 2^4 - 3^2 = 7

[@helenamcginty4920]

Sad. I could have done this when I left school in 1966. I loved algebra. But never used it since. Used tons of basic arithmetic of course.

[@vorpal22]

If you loved algebra, you really should consider looking into abstract algebra if you haven't... groups, rings, and fields (especially finite fields) are really fascinating.

[@florian_z_8735]

This is not! an SAT-problem!

[@victorrobison5069]

It would seem simple that 2 to the 4th would be 16, 3 to the 2nd would be 9, and 16 - 9 =7. I've got no idea where the hell you went with that other stuff. x=4 &y=2 . Or is that too simple to be called math?

[@vorpal22]

Also x = 3 and y = 0.

[@Karlem-Medias]

2⁴ = 2×2×2×2=16 3² = 3×3=9 16-9=7 X=4 and y=2

[@ottonormalverbrauch3794]

I had that in less than ten seconds but it gives no explanation

[@DJ-nw2ef]

@@ottonormalverbrauch3794 SAT problems like this are not looking for an explanation, but rather just the ability to get the right answer as quickly as possible. So, guessing at a few small powers of 3, starting with 3 to the 0, and then adding 7 will get you a valid answer (x = 3, y = 0) in a few seconds. No explanation needed. Point is, the goal here is general problem-solving ability, and not necessarily huge mathematical sophistication. Also, wasting time answering questions like this will mean you will answer fewer questions overall, which will probably lower your total score a lot.

[@ottonormalverbrauch3794]

@DJ-nw2ef Clear!

[@jtbaker2674]

x=3 and y=1 works also 8-1=7

[@bafbl]

(I think you mean y=0)

[@vegontoastog]

X is 5 y is 1

[@JohnH0130]

Sorry, not, plug these in you get 32-3=7.

[@revanwallace]

At 2:00, he divides one side of the equation by 2 squared and the other side by 3. That is not a valid operation.

[@vorpal22]

Right? He is trying to factor our 2^2 on the LHS and 3 on the RHS but it's a complete abuse of notation.

[@JohnH0130]

He called it 'dividing', and that fooled me at first also, but if you watch closely, you will see that what he did was extract common factors from terms, and it's not necessary to do that the same way on each side to be correct. For example, 32=32 can be written 8x4=16x2. That's all he did.

[@vorpal22]

@@JohnH0130 It's still incorrect. What he should have done is separated the equation or multiplied the numerator by the same value.

[@JohnH0130]

@@vorpal22 No. it's fine. He used the expression 'dividing' when describing his method, creating confusion, but he is only rewriting each side of the equation by extracting common factors from each term on that side. In so doing, he can factor each side differently, because in this step he is NOT performing any sort of mathematical operation to both sides. (8+4)=(6+6) 4x(2+1)=3x(2+2) See what I did there? Extracted different factors from each side, but did nothing that invalidates the equality. That's all he did.

[@vorpal22]

@@JohnH0130 Of course he can factor each side differently, but he goes from: 2^x - 2^2 = 3^y + 3 to: (2^x - 2^2)/2^2 = (3^y - 3)/3 He started off right when he said he was going to factor out 2^2, but saying he is dividing by 2^2 is a complete abuse of notation. Watch from 1:41 to 2:37, where he changes from an equation to something which is complete nonsense. He should have written: (2^x - 2^2) * (2^2/2^2) = (3^y - 3) * (3/3) if he wanted the equation to make sense. You cannot just arbitrarily divide sides by different values. I know what he *means*, but his notation is completely wrong.