The Hardest Exam Question | Solve for integers x,y
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- čas přidán 11. 09. 2024
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X is 9, y is 4. Solved it in 2 seconds
dam wow ur good can u pls tell me how to solve it in a shorter method
It is not clear why 4 can only be factored into integer factors? Why can't there be other solutions?
exactly !
why do not assume:
(√x-√y)= 4/3
(√x+√y-1)=3
?
Because of sqare root,so x and y larger than 0,y smaller 7, x smaller than 11, assume x and y are integers and square rootable, then,x equal to 9 and y equal to 4. Sloved within 20 seconds.
I respect your explanation but just to say if we are looking for integers then for y we already have only 6 possibilities so we can check them all and thats it since you used thus method at the end of your sophisticated solution
Very easy and I don't know why you solved it so complicated. You are looking for an integer solution so you can see the solution immediatly.
no, solving a math it means practice logical mind, not depend on difficulty level, if you have good knowledge and logical mind, solution will come on time
√x+y=7
x+√y=11
Let a=√x and b=√y
a+b²=7
a²+b=11
a²+b-a-b²=11-7
a²-b²+b-a=4
a²-b²-1(a-b)=4
(a-b)(a+b)-1(a-b)=4
(a-b)(a+b-1)=4
Case 1
a-b=1
a+b-1=4
a-b=1
a+b=5
2a=6
a=3
√x=3
x=9
9+b=5
b=-4
√y=-4
y=16
(9, 16) extraneous
Case 2
a-b=2
a+b-1=2
a-b=2
a+b=3
2a=5
a=2.5
√x=2.5
x=6.25
2.5+b=3
b=0.5
√y=0.5
y=0.25
(6.25, 0.25) ❤
Case 3
a-b=4
a+b-1=1
a-b=4
a+b=2
2a=6
a=3
√x=3
x=9
3+b=2
b=-1
√y=-1
y=1
(9, 1) extraneous
The variables must both be perfect squares. I immediately thought of x=9, y=4.
why must they be both perfect squares? pleases explain
@@scpmrbecause for integers x and y their squareroots are integers only if x and y are squares.
@@YAWTon and why x and y must be integers? Why can't they be real numbers? The sum of real numbers can be integer.
@@scpmrBecause the problem says "solve for integers". That's why he has to consider only integer factorisations of 4. For other factorisations he would get non-integer solutions.
It is nice. Here we have ignored another 3 set of factors of 4 (right side value) such as such as (-2, -2), (-4, -1), (-1,-4) which could have been tested too. Of course, they will not yield any integer solutions x and y.
Given that the solutions are integers, by inspection, it's clear that x must be less than 11, and a square of an integer. This means x can only be 9 4 or 1. Very quickly, it's clear x= 9, y =4. Just unfortunate this was so easy by inspection
Exactly what I said Y is less then seven so it only can be 4 or 1
this guy loves being obscure with logic to drag out his videos so that they contain more ads. stop watching his videos because hes not being genuine.
Observe that both y
Surely this is trivial. The question states x,y are integers! Simple inspection gives the solution.
9 and 4. Under 30 seconds.
4:35 integet factoring is completely arbitrary and incorrect assumption, limiting the number of solutions...
I respect your explanation but just to say if we are looking for integers then for y we already have only 6 possibilities so we can check them all and thats it since you used thus method at the end of your sophisticated solution
Because the resulting term is always a positive integer, both x and y must be perfect squares.
Both X and Y need to be perfect squares and Y
9^¹'²+4=7
9+4¹'²=11
Do us a favor and please stop with the cheesy pictures of Einstein. I'm sure you can pick another brilliant person in history, especially one that is known for math, their are many of them. Thank you.
Well said
Do us a favor and please use the correct form of their, they're and there
@@rotreal9863Beat me to it. Thank you.
Ramanujan or even Leonardo Da Vinci
Actually, you should notice at begining, that both sqrt must be integers.
Othewise, factorisation at 5:50, is not so simple - both part may be real number.
Hardest? Hardly. A few observations quickly narrow down the possibilities. This can then be solved in one minute.
• integers x,y - so for both LHS's to be integers, x & y must both be squares; non-zero ones at that
• √x + y = 7 - so y must be 1 or 4, because √x can't be < 0
y = 1? Then x = 36; x + √y > 36. Nope
y = 4? Then x = 9; and x + √y = 9 + 2 = 11. Solved!
Fred
Recommend in future, look for multiple ways to solve problems, then pick the easiest to do and/or the clearest/simplest.
After some trial and error its x=9 and y=4 that fulfill both equations. I didnt bother to watch. Maybe for bigger Numbers a systematic approach would help. Il there is one.
X=9. Y=4
x+y=7 (3)+(4)=7 (y ➖ 4x+3). x+y=11 (1)+(10)=11 (y ➖ 10x+1) .
Cases for x or y equals to 0 are impossible, so x,y are not 0. From first equation If y is integer then square root of x must be integer so x must be perfect square and positive, also square root of x is positive and equals to 7-y, then square of root of x must be also smaller than 7. Similarly by second equation y must be possitive, and perfect square. Cases are x=1, 4 or 9, easily 1 and 4 are discarded, and the only integer solution is x=9. First equation easily leads to y=4.
BS, it is the easiest question
if we end up to analyze 4, then we can analyze 7 and 11 from the begging and solve in a second....?!
aka The Ramanujan problem
At a quick glance, x = 9 and y = 4 .
Hardest exam question? It took me about 2 seconds to solve it by inspection!
Very complex solution. There are better solutions in the comments!
let Vx=u , u^2=x , u^4=x^2 , --> , y=(11-x^2)^2 , y=121-22x+x^2 , x^2-22x+Vx+114=0 , u^4-22u^2+u+114=0 ,
add 3u^3 , -3u^3 , we get , (u-3)(u^3+3u^2-13u-38)=0 , u=3 , Vx=3 , x=9 , x+Vy=11 , 9+2=11 , Vy=2 , y=4 , solu. , x=9 , y= 4 ,
test , V9+4=3+4 , 3+4=7 , OK , 9+V4=9+2 , 9+2=11 , OK ,