First of all, No condition was given!! There should be infinite solutions by plotting a curve!! And the first answer appeared in my mind was also (6,0)!!
Without parameters this is also a correct answer. Using his method of selecting the two numbers multiply together to equal 36, using 6 x 6 also fits his method. 0 is not a natural number but there is no rule expressly stated that one must use natural numbers or repeated numbers.
x^2 - y^2 = 36 x^2 - y^2 = 6^2 x^2 + 6^2 = y^2 according to Pythagoras a^2 + b^2 = c^2 The value of b is 2 times greater than the value of the Egyptian triple: 4^2 + 3^2 = 5^2 Accordingly 8^2 + 6^2 = 10^2 If you rearrange everything in its place: 10^2 - 8^2 = 6^2 100 - 64 = 36
I have also got this idea. The problem can be solved in just two steps. A triangle having sides with 3:4:5 ratio is referred to as Egyptian Triangle The thing that any math teacher must tell their students in Grade 5. .
X^2-y^2=36 X^2-y^2=6^2 X^2=6^2+y^2 We know 3,4,&5 are Pythagorean triplets. 6 can be converted to 3x2. so to eliminate 2 from 6, Divide both sides by 4 (X/2)^2= 3^2+(y/2)^2 It means x/2= 5 & y/2=4 => X=10 Y=8
In principle at least one of the factors must be an even number. But if one of the factors is even then (X,Y) is (even, even) or (odd, odd) and in both cases both factors would be even.
@@fernandoreescolar418 Or we could say that the difference between X+Y and X-Y is 2Y which is even. So if X+Y is even minus even is even, therefore X-Y is even
x X x = 36 + y X y so x = 6 if y= 0 when y is 1, 2, 3, 4,5,6,7 then we do not get a number which will have a root which is a whole number when y = 8 , then y X y = 64 x X x = 36 + 64 x X x = 100 so x = 10 , y= 8 or x = 6 , y= 0
I have no specific formula for solving the problem. But by common-sense inspection and intuition, 100 - 64 = 36. 100 happens to be 10 square, and 64 is 8 square... So "x" must therefore be 10, and "y" equals 8. Or, "x" could be 6, and "y" equals 0. How's that for a quick solution.
@@DespejalaxconJorge x=6 & y=0. It is called "trivial' solution. On further inspection, 4 (=2 x 2) is a factor too & indicates that both factors (x-y) & (x+y) are each even. x^2 - y^2 = 4 x 9 = (2 x 9) x (2 x 1); (x+y) (x-y)= (2 x 9) x (2 x 1). Hence, ½(x+y) = f odd ½(x-y) = g odd & the product ¼(x+y)(x-y) = ¼(36) = 9. 9 has factors (9, 1) or (3. 3). Factoring out 4 (from 36), gives 9, which is a product of 9 & 1, as 3 & 3 are ruled out being equal (it gives y=0, the trivial solution) - as we seek unequal factors. Thus f=9 & g=1. It can also be proved that swapping the pair of solutions (f=1 & g=9) is acceptable (or admissible) too. adding them gives, (f+g) even, since the sum of two odd numbers is even. Subtraction gives (f-g) even, since the difference of two numbers is even. ½(x+y) = 9 ½(x-y) =1 ___________ added ½(2x) = 10 x = 10. __________subtracted ½(2y) = 8 y = 8 Proof: x^2 - y^2 = 10^2 - 8^2 = 100 -64 x^2 - y^2 = 36
There is a problem with the set N. Older texts considered that 0 was not a natural number but almost all modern texts consider 0 to be a natural number so the solution using modern texts would be x=6 and y = 0.
I thought of it simply as Pythagorean triple. Move y2 to the other side, think of 36 as 6 squared. Then we can quickly see that the resulting expression is just twice the basic 3, 4 ,5 Pythagorean triple, namely 6, 8 and 10.
Solutions are : S=[(x,y)=(6,0);(10,8)] X=6 or 10 and Y=0 or 8 36 is = 6*6 or 18*2. The solutions would be if two pairs numbers and the first number must be more or equal than the second to be valid X^2-Y^2 = 36 Case 1 : (X+Y)(X-Y) = 18*2 ===> X+Y = 18 X-Y = 2 === > X=10, Y=8 S1={(X,Y)=(10,8)} Case 2 (X+Y)(X-Y) = 6*6 ===> X+Y = 6 X-Y = 6 === > X=6, Y=0 S2={(X,Y)=(6,0)} S={(X,Y)=(10,8) and X,Y)=(6,0)}
Because you are dealing with squares, there are actually 4 solutions: x, y can equal 10, 8 10, -8 -10, 8 -10, -8 When the negative numbers are squared, the result becomes positive so any of the 4 sets yields 100 - 64 = 36.
36 = 36 x 1 = 6 x 6 = 12 x 3 = 18 x 2; X^2 _ Y^2 = (X+Y) (X-Y) If X+Y = 36 and X-Y = 1 will give nonwhole number. Same with 6×6, 12 x 3. If X+Y = 18 and X-Y = 2, you get x=10 y = 8. 100-64= 36. Qed.
Could you kindly provide the source of your claim that this question is from ‘Harvard University Admission Exam’? I am not sure whether they have a written format of entrance exam.
Just saw your problem on the screen and solved it within 2 minutes in my head only, I'm a 43 year old chef and this is not a Harvard math problem but primary school Pitagorian variant.
Another solution would be 6 and 0. However, since the Solution must be natural Numbers, 0 is rejected. I got the Solution 0 and 6 using derivatives, however 0 is not a natural Number and the 6,0 are not correct.
Introducing complex, time-consuming, and inefficient problem-solving procedures, often seen in corporations run by some Indian immigrants, can disrupt and undermine an otherwise efficient society.
It took me 15 seconds of logical mental gymnastics to obtain the solution: What perfect square numeral less another perfect square numeral = 36? No notes on paper required. Let’s keep it simple and work SMART and CONCISE 👍
Please graph X values that you rejected…. You only find one solution not moving solution … think like a circle and each degree different (r) descend or ascend in one loop of (timplace) If your math doesn’t describe a movement so your at constant level of knowledge Spin your equation
It's obviously 100-36. Duh. If you can see that you have a shot at Harvard. Your chances are better if your father went there. If you get it after some some wandering around, you're qualified to apply for your local community college.
Simply by inspection:
Which square minus which square gives 36?
100 - 64 = 36.
Therefore x = 10
And y = 8
X = 6 & y = 0 also work
@@paulhardy3746 x and y belongs to natural numbers so 0 cant exist
@@paulhardy3746 Dude, x and y belong to N (Natural number). 0 aint a natural number.
x=6 and y =0 is the simplest answer !
true....ye idea to youtuber ke dimaag me bhi nai aaya hoga..
No brother. 0 is not a natural number
First of all, No condition was given!! There should be infinite solutions by plotting a curve!! And the first answer appeared in my mind was also (6,0)!!
Without parameters this is also a correct answer. Using his method of selecting the two numbers multiply together to equal 36, using 6 x 6 also fits his method. 0 is not a natural number but there is no rule expressly stated that one must use natural numbers or repeated numbers.
That was my immediate thought when I saw the question.
You're making a very easy math problem complicated
It is immediately clear that both x^2 and y^2 should be sguare numbers , then we have 100- 64=36, so x=10, and y=8
Да слишком легко)
@paulhardy3746
7 hours ago
X = 6 & y = 0 also work
x^2 - y^2 = 36
x^2 - y^2 = 6^2
x^2 + 6^2 = y^2 according to Pythagoras a^2 + b^2 = c^2
The value of b is 2 times greater than the value of the Egyptian triple: 4^2 + 3^2 = 5^2
Accordingly 8^2 + 6^2 = 10^2
If you rearrange everything in its place:
10^2 - 8^2 = 6^2
100 - 64 = 36
Please elaborate please
If x^2 - y^2 = 6^2
then x^2 + 6^2 = y^2 ==> How?
Since 6٨2 + y٨2 = x٨2, you can list all the pythagorean triple containing 6 where 6 is not the hypotenuse.
Using Pythagoras law
In right angled triangle sides 6,8,10
x² = y²+ 6²
10² = 8² +6²
I have also got this idea. The problem can be solved in just two steps.
A triangle having sides with 3:4:5 ratio is referred to as Egyptian Triangle
The thing that any math teacher must tell their students in Grade 5. .
Should be 10^2=8^2+6^2
Excellent work!
Solved the problem with easy method
The trick is you need to have a pair of even numbers in order to attain an outcome of natural numbers as the solution.
It's true
X^2-y^2=36
X^2-y^2=6^2
X^2=6^2+y^2
We know 3,4,&5 are Pythagorean triplets. 6 can be converted to 3x2.
so to eliminate 2 from 6, Divide both sides by 4
(X/2)^2= 3^2+(y/2)^2
It means x/2= 5 & y/2=4
=> X=10 Y=8
X=6 & Y=0
Suberb
You are doing a great work, keep it on.
Analyzing the expression (X+Y)(X-Y), we come to the conclusion that both terms can only be even numbers. So only the pair 18-2 can be a solution.
can you elaborate?
In principle at least one of the factors must be an even number. But if one of the factors is even then (X,Y) is (even, even) or (odd, odd) and in both cases both factors would be even.
Wrong sir @@fernandoreescolar418
@@fernandoreescolar418
X is 10
Y is 8
@@fernandoreescolar418 Or we could say that the difference between X+Y and X-Y is 2Y which is even. So if X+Y is even minus even is even, therefore X-Y is even
You left out 6x6 which would solve that (x,y) =(6,0).
However there seems to be a debate whether 0 is a natural number or not.
But given the requirement that x+y > x-y, 6x6 cannot be an option.
Then again, (x+y)(x-y) = 6x6 when 0 ∈ N, so I think you are correct.
@attiylanen The requirement should be that x+y >= x-y if 0 is a natural number
Okay but (6+6)(6-6) doesn't that just give you 0?
@ohmy9261 you misunderstood my response. What I said is that x=6, y=0 solves the equation if 0 is considered a natural number
x X x = 36 + y X y
so x = 6 if y= 0
when y is 1, 2, 3, 4,5,6,7 then we do not get a number
which will have a root which is a whole number
when y = 8 , then y X y = 64
x X x = 36 + 64
x X x = 100
so
x = 10 , y= 8
or
x = 6 , y= 0
6 x 0 = 0
x*x - y*y= 36
x*x = 36 + y*y
x*x = 36 + ( and here I ran basic times table to 8.... 8 squared is 64)
x*x = 36 + 64 = 100
x = 10
you're the genius! coz i did the same and just squared 1-10 coz its always one of the answers.
Yes partial parts of circle ⭕️ one central moving with different r =R which is effected from outside
I have no specific formula for solving the problem. But by common-sense inspection and intuition, 100 - 64 = 36. 100 happens to be 10 square, and 64 is 8 square... So "x" must therefore be 10, and "y" equals 8. Or, "x" could be 6, and "y" equals 0. How's that for a quick solution.
Call it guesswork
Perfect . Thank you for your good explanation and teaching 👌🏻👌🏻
The condition is that the sum of factors should be even (if x, y ∊ N).
Only (18, 2) satisfy this condition, that yields x=10 and y=8.
Para muchos estaría también la solución x=6 e y=0 debido a que la discusión sobre si el 0 es un número natural o no sigue abierta.
@@DespejalaxconJorge
x=6 & y=0. It is called "trivial' solution.
On further inspection, 4 (=2 x 2) is a factor too & indicates that both factors (x-y) & (x+y) are each even.
x^2 - y^2 = 4 x 9 = (2 x 9) x (2 x 1);
(x+y) (x-y)= (2 x 9) x (2 x 1).
Hence,
½(x+y) = f odd
½(x-y) = g odd
& the product ¼(x+y)(x-y) = ¼(36) = 9.
9 has factors (9, 1) or (3. 3).
Factoring out 4 (from 36), gives 9, which is a product of 9 & 1, as 3 & 3 are ruled out being equal (it gives y=0, the trivial solution) - as we seek unequal factors.
Thus f=9 & g=1. It can also be proved that swapping the pair of solutions (f=1 & g=9) is acceptable (or admissible) too.
adding them gives, (f+g) even, since the sum of two odd numbers is even.
Subtraction gives (f-g) even, since the difference of two numbers is even.
½(x+y) = 9
½(x-y) =1
___________ added
½(2x) = 10
x = 10.
__________subtracted
½(2y) = 8
y = 8
Proof:
x^2 - y^2 = 10^2 - 8^2 = 100 -64
x^2 - y^2 = 36
Let X=Y+ n, plug into the equation, you can quick immediately get the answers
Amazing 😍
This guy is insane, period!
X=10, Y=8. 100 minus 64 = 36. Simples
X^2-Y^2=36 X=±10, Y=± 8 X=± 6, Y = 0 (X,Y)=(10,8)
There is a problem with the set N. Older texts considered that 0 was not a natural number but almost all modern texts consider 0 to be a natural number so the solution using modern texts would be x=6 and y = 0.
He explained the math so well that I'm only 11 but I can still understand the math.👍
(X x X) - (Y x Y) =36
(10 x10) - (8 x 8) = 36
(100) - (64) = 36
X = 10 and Y=8
I thought of it simply as Pythagorean triple. Move y2 to the other side, think of 36 as 6 squared. Then we can quickly see that the resulting expression is just twice the basic 3, 4 ,5 Pythagorean triple, namely 6, 8 and 10.
İ am from Azerbaydjan.Veru useful.Thanks
XX - YY = 36
.
*. 36 = 10x10 - 8x8 = 100 - 64
**. X = 10, Y = 8.
X=10 Y=8 10x10=100 8x8=64 100-64=36 (6x6)
(X-Y).(X+Y)=36 , (10-8).(10+8)=36
Solutions are : S=[(x,y)=(6,0);(10,8)]
X=6 or 10 and Y=0 or 8
36 is = 6*6 or 18*2.
The solutions would be if two pairs numbers and the first number must be more or equal than the second to be valid
X^2-Y^2 = 36
Case 1 :
(X+Y)(X-Y) = 18*2 ===>
X+Y = 18
X-Y = 2 === > X=10, Y=8 S1={(X,Y)=(10,8)}
Case 2
(X+Y)(X-Y) = 6*6 ===>
X+Y = 6
X-Y = 6 === > X=6, Y=0 S2={(X,Y)=(6,0)}
S={(X,Y)=(10,8) and X,Y)=(6,0)}
X=10 y=8. Or just in a second x=6 and Y 0. Can do it.
(10x10=100) - (8x8=64) = 100-64 =36
And where exactly is the "solution" here? Simply permutation of the possibilities which are not that much and can be done w/o need to write at all.
you are missing ONE obvious solution : x=6 and y=0 !!
Zero is not a natural number. x and y are natural numbers
@@nagalakshmidevi7022 there is no such restriction imposed in the question.So anyhting that logically satisfies the equation , will do the JOB !
x2 = 36 + y^2 => x = SQRT(36+y2) Now if y2 = 64 => y = 8 since y belongs to N => x = 10
What about 6 x 6 = 36?
If x+y=x-y,y=0
Thank you 🎉❤
To eliminate cases, in the condition that result shall be natural number, then one should look for factors that when added shall be even
Because you are dealing with squares, there are actually 4 solutions: x, y can equal
10, 8
10, -8
-10, 8
-10, -8
When the negative numbers are squared, the result becomes positive so any of the 4 sets yields 100 - 64 = 36.
X=13/2. Y=5/2
X*x-y*y= 100 - 64
X= +-100 , y= +-6
Or, x=+-6, y=0
36 = 36 x 1 = 6 x 6 = 12 x 3 = 18 x 2; X^2 _ Y^2 = (X+Y) (X-Y)
If X+Y = 36 and X-Y = 1 will give nonwhole number. Same with 6×6, 12 x 3.
If X+Y = 18 and X-Y = 2, you get x=10 y = 8. 100-64= 36. Qed.
10 squared - 8 squared = 36 (100-64)
x=10 y=8 is the answer
X² - y² =10² -8²
100 - 64 = 36.
Could you kindly provide the source of your claim that this question is from ‘Harvard University Admission Exam’? I am not sure whether they have a written format of entrance exam.
Congratulation! Solution is very well expressed
x=10, y=8 6, 8 and 10 are pyrhagorean triplets.
Вот именно что там решать если знаешь квадраты чисел от двух до десяти
Just saw your problem on the screen and solved it within 2 minutes in my head only, I'm a 43 year old chef and this is not a Harvard math problem but primary school Pitagorian variant.
Since x,y are naturals integers
Why you forgot the first case
X=6 and Y= 0
(6+0)*(6-0)= 36
Analyzing the expression (X+Y)(X-Y),
6 X 6 = 36
X =6
Y = 0
Another solution would be 6 and 0. However, since the Solution must be natural Numbers, 0 is rejected.
I got the Solution 0 and 6 using derivatives, however 0 is not a natural Number and the 6,0 are not correct.
• x = 10, y = 8
• x = 8, y = 10
• x = -10, y = -8
• x = -8, y = -10
No need to check every case. (x+y)+(x-y) = 2x must be an even number. SUm of 2 factor must be even so only (18,2) works.
6x6 is also 36 you need to present it and omitting it as x-y will be zero, but 6 and 0 is also a solution for x and y respectively
x=6, y=0 gives the right solution
x,y=(10,8)
I think a simple way was elongated to confuse, if you do math like this everything is going like a big problem.
X=13/2 Y=5/2
x=10, Y = 8
You can get 10 and 8 in your head through substitution in less than a minute.
You didn't apply the rule of multiple two negative numbers that are positive and you didn't put it in the probabilities
X=10; y=8
x + y = 18
x- y = 2
x = 10
y = 8
x = 15/2 =7.5
y = 12 - 7.5 =4.5
7.5² - 4.5² = 56.25 - 20.25
= 36
Why do you reject these numbers ?
X=10
Y=8
(x+y)(x-y) = 36 = 18*2
so x+y = 18 and x-y = 2
so 2x=20
so x=10 and y = 8
Check
100-64=36
Ramanujan would first list these solutions: x=6,y=0 and x=-6, y=0
Introducing complex, time-consuming, and inefficient problem-solving procedures, often seen in corporations run by some Indian immigrants, can disrupt and undermine an otherwise efficient society.
x =10, y=8
A common error in reasoning is that if we make the assumption at the beginning that we are considering only natural numbers, then the cases x
X=8
y=5
(6x6) - (0×0)
It took me 15 seconds of logical mental gymnastics to obtain the solution: What perfect square numeral less another perfect square numeral = 36? No notes on paper required. Let’s keep it simple and work SMART and CONCISE 👍
X= 6, Y= 0.
практически данное решение, методом подбора
Where is 6 x 6?
بما ان الاعداد الثلاث صحيحة
نكتب المربعات المحتملة وهي
من 1 ، 4 الى 400
ونحصل على مربعين هما 100 و 64 تحققان الحل
Please do not solve this question like that. There are no conditions saying x and y are integer.
Please graph X values that you rejected…. You only find one solution not moving solution … think like a circle and each degree different (r) descend or ascend in one loop of (timplace)
If your math doesn’t describe a movement so your at constant level of knowledge
Spin your equation
X = 10 and Y = 8, resulte in 5 seconds or less, why 9,30 minutes?
It is Pythagoras. 25=16+9, 100=64+36, 225=144+81, 400=256+144...
X=10 and y=8
Binomial. Then you can do it in your head.
How did you conclude that x+y is greater than x-y???
10. And. 8
36=6×6
Très bien
Deux solutions possibles:
✓ x=6 & y=0
✓ x=10 & y=8
? x=10 Y= 8.
It's obviously 100-36. Duh.
If you can see that you have a shot at Harvard. Your chances are better if your father went there.
If you get it after some some wandering around, you're qualified to apply for your local community college.
X=10 &y=8
If both negative?
great
How about 6x6=36
6x6
What took you so long?
(x+y) (x-y) = 9*4 = 36*1
If this is the best Harvard can come up with, God help us!! No one should need 9 minutes and/or a pencil to solve this one.
What happened to 6 x 6?
6²-6²=0≠36