A very tricky Question from Stanford University Entrance Exam | Find the Value of a=? & b=?

[@sc2cooptutorials679]

Since both a and b are positive integers, a can only have values from 1 to 6, otherwise a^2 is too large. Testing each of them takes 10 times less time (and infinite time less paper) than this "solution".

[@davidseed2939]

Since the equation is linear in b and quadratic in a, obviously set up the solution as b=f(a) ie b=(44-a^2)/(2a+1), a=1, b=43/3 no a=2, b=40/5 =8 a=3, b=35/7=5 a=4, b=(44-16)/9 =28/9 no a=5, b=(44-25)/11 =19/11 no a=6, b=(44-36)/(13) = 8/13

[@denisrenaldo3506]

Sorry a can only be from 1 to 5 as a^2+2ab shall be inferior or equal to 43 (44-b). Testing the 5 values of a is done in one minute with b=(44-a^2)(1+2a). Only a=2 with b=8 and a=3 with b=5 are possible.

[@sc2cooptutorials679]

@@denisrenaldo3506, what you wrote right after "a can only be from 1 to 5" is called testing for a=6. And you missed one solution.

[@denisrenaldo3506]

@@sc2cooptutorials679 thanks and rectified.

[@Rohmat-oo3pw]

a²+2ab+b=a²(1+2b/a+b/a,a)=44. 44=1×44=2×22=4×11. a,a=4. A=2. 1+2b/a+b/a,a=11. b=8

[@9허공]

more easy way to derive (2a + 1)(2a + 4b - 1) = 175 a^2 + 2ab + b = 44 => (a + b)^2 - b^2 + b = (a + b )^2 - (b^2 - b + 1/4) + 1/4 = 44 => (a + b)^2 - ( b -1/2)^2 = 44 - 1/4 => (2a + 2b)^2 - (2b - 1)^2 = 175 => (2a + 2b + 2b - 1)(2a + 2b - 2b + 1) = (2a + 4b - 1)(2a + 1) = 175

[@johnwaylandbales]

Exactly. Completing the square and then factorizing is easier and faster.

[@leealex3692]

@@johnwaylandbales a^2 + 2ab + b = 44 a^2 < 44 a < 7 then a= 0, b = 44 a= 2, b = 8 a= 3, b = 5 END easier and faster

[@satrajitghosh8162]

( a + b) ^2 - ( b - 1/2) ^2 = 44 - 1/4 ( 2 a + 2 b) ^2 - (2 b - 1)^2 = 175 = 35 * 5 = 25 * 7 = 31 + 33 + 35 + 37 + 39 = 19 + 21 + 23 + 25 + 27 + 29 + 31 Case I ( 2 a + 2 b) ^2 - (2 b - 1)^2 = 20 ^ 2 - 15 ^2 a + b = 10 2 b = ( 15 + 1) i.e b = 8 and a = 2 Case II 2 a + 2 b) ^2 - (2 b - 1)^2 = 16 ^ 2 - 9 ^2 a + b = 8 2 b = ( 9 + 1) i.e b = 5 and a = 3

[@9허공]

another approach a^2 + 2b*a + b - 44 = 0 => a = (-2b ± √(4b^2 - 4b + 176)/2 = -b ± k/2 Δ = 4b^2 - 4b + 176 = k^2 for some positive integer k => k^2 - (2b - 1)^2 = (k + 2b -1)(k - 2b + 1) = 175 since k + 2b - 1 > k -2b + 1, 175 x 1, 35 x 5, 25 x 7 product mixes are possible case 175 x 1 => b = 44, k = 88, a = 0, -88 since a > 0, rejected case 35 x 5 => b = 8, k =20, a = 2, -18 => (a,b) = (2, 8) case 25 x 7 => b = 5, k = 16, a = 3, -13 => (a,b) = (3, 5)

[@leealex3692]

a^2 + 2ab + b = 44 a^2 < 44 a < 7 then a= 0, b = 44 a= 2, b = 8 a= 3, b = 5 END

[@rainerzufall42]

A little number theory: 44 = 11 * 2 * 2 = 11 * 4. a² + b (2a + 1) = 0 (mod 2, 4, 11, 22, 44) a² + b = 0 (mod 2) a + b = 0 (mod 2), because a² = a (mod 2), not only FLT. For a = 0 (mod 2), we already had b = 0 (mod 2). But even b (2a + 1) = b (4 a/2 + 1) = b = 0 (mod 4) For a = 0 (mod 4), we have a² = 0 (mod 4) => b (2a + 1) = b = 0 (mod 4) For a = +/-1 (mod 4), we have a² = +1 (mod 4), thus b (2a + 1) = -1 (mod 4) Case 1: a = +1 (mod 4) => b * (-1) = -1 (mod 4) => b = 1 (mod 4) Case 2: a = -1 (mod 4) => b * (-1) = -1 (mod 4) => b = 1 (mod 4) So basicly b = 4 * k, if a even. Also b = 4 * k + 1, if a odd. All nice and true, but this leads only to checking all the cases... pretty much useless... For a = 5, a² + b * 11 = 0 = 5² (mod 11) WRONG For a = 3, a² + b * 7 = 0 = 3² - 4 b (mod 11) => b = 9/4 = 10/2 = 5 (mod 11) For a = 1, a² + b * 3 = 0 = 1² - 8 b (mod 11) => b = 1/8 = 4/-1 = 7 (mod 11), b = 7, 18, 29, 40, but only 29 is valid, and way too big!

[@rainerzufall42]

a = 0 (b = 44) should NOT be a solution, if a and b are € IZ^+ = { 1, 2, 3, ... } = |N (or in different notation |N^*, with IN = |N^0). |N^0 = { 0, 1, 2, ... } can be written as IZ^{>=0}, but |Z^+ = |Z^{>0}. Would be okay, if you write |N (including 0) or explicitly |N^0 instead of Z^+...

[@rainerzufall42]

Oh, with a little thinking, another approach. All numbers are integer and positive, let's say, they are even from IN^0 = IZ^{>=0}. a² + 2ab + b = a² + b (2a + 1) = 44 ! => a² = 44 - b (2a + 1) => Question: Which squares a² do satisfy 44 - b (2a + 1)? First, b is odd if a is odd. If a is even, 2a + 1 is odd, so b must be even. (This implies, that a+b is even!) Possible squares less than 44 are 36, 25, 16, 9, 4, 1 and 0. The complements are 8, 19, 28, 35, 40, 43, and 44 for b (2a + 1). Values for b are (division by 2a + 1): 8/13, 19/11, 28/9, 35/7, 40/5, 43/3, 44/1. Only b=35/7=5, 40/5=8, 44/1=44 are integer! The corresponding a values are 3, 2, and 0. So we get once more (3, 5), (2, 8), (0, 44).

[@rainerzufall42]

Solved it immediately, mainly because I misread the problem: I missed, that it was b, not b², so I thought, it couldn't be solved. But if it's b, then it just differs from n² = (a+b)² = 44 by (b² - b). We are working with natural numbers € IN, so I easily guessed the square n² for n=(a+b) as 8² = 64 (7² = 49 obviously doesn't work!). As we need 44 not 64, I've asked myself, which number is 20 away from it's square? Or what is b, if b*(b-1) = 20 = 5*4? Sure, b = 5. (for b*(b-1) = 5 = 49 - 44, there's no solution for n=a+b=7, less so for 37 = 81 - 44, if n=9). On the other hand, if n=10, you get 100 - 44 = 56 = 8*7. But with n=8 and b=5, we have a=3. Case 1! Check: 3² + 2*3*5 + 5 = 9 + 30 + 5 = 44. qed And with n=10 and b=8, we have a=2. Case 2! Check: 2² + 2*2*8 + 8 = 4 + 32 + 8 = 44. qed

[@SivaB-g7h]

I assumed correct answer in less than 5minutes 🥵🥵🥵😁😁😁😎😎😎

[@teraincognita-zq6wu]

Stanford doesn't have entrance exam,

[@andresrebolledobanquet1924]

ab = 44/2 ab= 22 b = 44/1 b = 44 F(x)a^2 = 2a 2a = 44 - 22 - 44 2a = - 22 a = - 22 / 2 a = - 11 Demostrate. - 22 + 22 + 44 = 44

[@AnilSonkar-ly5ex]

Nice video

[@superacademy247]

Thanks. So nice 🥰💕✅🙏👏

[@SGuerra]

A questão é bem bonita. Eu a fiz sabendo que o maior valor de a é igual a 6. Basta testar os valores de a já que ele é um número inteiro. Parabéns pela escolha! Brasil Setembro de 2024. The question is quite beautiful. I did it knowing that the largest value of a is equal to 6. Just test the values ​​of a since it is an integer. Congratulations on your choice! Brazil September 2024.

[@superacademy247]

You're welcome. Thanks for your support and input 💕🙏💡✅🥰

[@avalagum7957]

My experience for doing these things is" converting the equation to x *y = number which a small number of combinations (A, B) such that A*B = that number.

[@paulortega5317]

Or u=2(a+b), v=(2b-1) and u²-v²=175 etc

[@jwang734]

a = -b±√(b(b-1)+44), can take a = -b+√(b(b-1)+44) only and it is decreasing, => b=5, a=3; b=8, a=2. (a=1, b is not good.)

[@klausg1843]

Obvious that a

[@generaclesdey4622]

If it takes you 14 minutes or more to solve, it shoud NOT be on the test!!

[@sc2cooptutorials679]

It takes 1,5 minutes to solve if you're not a masochist. Just not the case with this guy.

[@joyhuang7845]

式子能因式分解，這過程漂亮。

[@superacademy247]

Thanks 👍💯

[@ASAPSquatterRemoval]

Why is 2a+1 < 2a+4b-1? Couldn't b = a negative number? I dont remeber you saying the variables needed to be positive. You just mention it at the end when I was already in the middle of solving the problem.

[@sc2cooptutorials679]

@@ASAPSquatterRemoval, he says it at the start of the video, and Z+ which is written on the paper literally means positive integer.

[@ASAPSquatterRemoval]

@@sc2cooptutorials679 I completely missed all of that, thanks. Was trying to solve a much more difficult problem.

[@RajeshKumar-rd7xn]

a = √44 -1 and b = 1

[@davidpatterson5426]

(2,8) & (3,5)

[@superacademy247]

That's it

[@anestismoutafidis4575]

(44)^1/2=6,633' (3+3,633') (3+3,633')^2= 9+10,9+10,9+13,20 (3+3,633')^2=9+21,8+13,20=44 a=3; b=3633'

[@Emmashar]

(A, B)=(-1; -43).

[@dsl77354]

a=2 b=8

[@tontonbeber4555]

I propose you multiply first by 1543433456777632454363364745754377436 to find the solution much faster

[@KenPaulsenArchitect]

a=2, b=8

[@walterwen2975]

Stanford University Entrance Exam: a² + 2ab + b = 44, a, b ϵ ℤ; a, b =? a² + 2ab + b = (a² + 2ab + b²) - b² + b = (a + b)² - b(b - 1) = 44, b > a ≥ 0; a, b ϵ ℤ (a + b)² = 44 > b(b - 1) > 0 (a + b)² - b(b - 1) = 44² = (0 + 44)² - 44(44 - 1) = 44(44 - 43) = 44; a = 0, b = 44 (a + b)² > 44 > b(b - 1) > 0 (a + b)² - b(b - 1) = 8² - 20 = (3 + 5)² - 5(5 - 1) = 44; a = 3, b = 5 (a + b)² > b(b - 1) > 44 (a + b)² - b(b - 1) = 10² - 56 = (2 + 8)² - 8(8 - 1) = 44; a = 2, b = 8 Answer check: a = 0, b = 44, a = 2, b = 8 or a = 3, b = 5: a² + 2ab + b = (a + b)² - b(b - 1) (a + b)² - b(b - 1) = 44; Confirmed as shown Final answer: a = 0, b = 44, a = 2, b = 8 or a = 3, b = 5

[@leealex3692]

a^2 + 2ab + b = 44 a^2 < 44 a < 7 then a= 0, b = 44 a= 2, b = 8 a= 3, b = 5 END

[@shelonnikgrumantov5061]

You must be kidding. My 10 yo son found a solution within 1 minute by simply using logic. Does he qualify for Stanford?

[@superacademy247]

Yes

[@shelonnikgrumantov5061]

@@superacademy247 )

[@somnathchattopadhyay1596]

Wrong method

[@superacademy247]

In what way?

[@leealex3692]

right but stupid a^2 + 2ab + b = 44 a^2 < 44 a < 7 then a= 0, b = 44 a= 2, b = 8 a= 3, b = 5 END

[@mihaihaisan5564]

😏 complicated variant presented. Easiest general way is this: a^2+2ab+b=44; (a+b)^2-b^2+b=44 We give values to a: a=0 =>b=44 a>=1; if a=n => b=(44-n^2):2n+1 (If we need only Z solutions we go to check max a=6)

[@lewisng4606]

Someone that can take Stanford exam would not need such slow and clumsy explanation. Clicking bait.

[@46swa]

What a nonsense! I have the solution after one minute. b=(44-a^2)/(2a+1) and a < 7

[@cosimo7770]

Universities look for insight and imagination, not blind application of absurd rules.

[@superacademy247]

Are you aware Ramanujan didn't have formal training in Math ?

[@ΓΕΩΡΓΙΟΣΑΡΣΕΝΙΑΔΗΣ]

It's not a proof it's bullshitt

[@yangchong4680]

Stupid method

[@petechen794]

Just is a boring question

[@julianbrown7976]

Yet another embarassingly bad YT maths video. Slow maths is bad maths. Here is a fast solution. See that there are only six possible vales fir a: 1,2,3,4,5,6. Try each one in turn a=1 doesnt work because 43 is prime. Try 2 and you get b=8 and bingo that is a solution.

[@superacademy247]

Math doesn't depend on guesswork but established principles based on validity.

[@AgricolaMACdz]

Du bricolage !