Pingin jd orang kaya belajar berhitung broo ya...🤣🤣😁👍

This is not by Rama nojan. This is by Khayyam. Iranian mathematician

5:48 sigma

2:43 this step has a mistake (√x-√y)-(√x+√y) (√x-√y) =(√x-√y)(-1√x+√y)

Could've taken modulo 4 and then you could've easily guessed that m would be either 1 or 3 mod 4 , and n being an even integer.

Easy

Easy 😉 by this 1/2(a+-b)

X-9 Y-4

how much time should you deal with ths problem,I spend 10 minutes,my method was: we have the area A(big square)=9A(small square) (let's take C the lenght of the short square and a the lenght of the big one) imply : a^2=9*c^2 imply a= 3*c this is the first result, then we have A(ABCD) = (c+a)^2 =(3c+c)^2 = 16c^2 and we have that the A(ABCD)=25 imply 16c^2=25 imply c=5/4

We also slove this by using substitute methods

Just did it in around 3 minutes. My Solutions for integral roots for this simultaneous equations involves these (providing hints only): At the point, when you get (√ x - √ y) (√ x + √ y -1) = 4; choose 1 * 4 and not 2 * 2 as possible roots. Think why and reply in comments 🙂 Later on when you have to put a value of x - y = 5 => x = 5 + y in the simultaneous equation above make √ y = z and then solve equation (from (2)): 5 + y + √ y = 11 => 5 + z^2 + z = 11 => z^2 + z - 6 = 0; solve this quadratic for z to get z = 2 as real root => √ y = 2 => y = 4; Hence x = 5 + y = 9. That means, (x,y) = (9,4).

Solved by guessing and checking perfect squares less than 11, ...... there are exactly 3 and 9 is only 2 away from 11

I expected a problem by Ramanujan to be more interesting. of course, "integer solutions" should be stated right up front.

At 2:01 , let a+ib= re^i∅ where ∅ is theta, taking natural logarithm on both sides we get ln(a+ib) =ln(r)+i∅ Consider ln(i) has only an imaginary part so a=0 and b=1 then r= 1 and ∅= arc tan(1/0) = π/2 Therefore, ln(i)=1+iπ/2 We have x= ln4/(ln4+ 2ln(i) ) So x= ln4/(ln4+2(1+iπ/2 ) x= ln4/(ln4+2+πi) which is the only solution for (-4)^x=4 Is it right? Consider it please.😊

Ramanujan was an ordinary mathematician he gave a length equation.

No lengthy solution needed this is time waste just take 3^a = 5^b = k , K^1/a = 3, K^1/b = 5. Multiply these both you'll get K^1/a+1/b = 15 and put the value of 1/a+1/b = 2. K²= 15 K = √15 ans.

Wrong its Ln not log

If x and y are integers that means you just made a very simple problem too complex. There is only few ways to get 7 with one integer (y in this case) 0+7, 1+6, 2+5, 3+4 In the next equation to get an integer (11 ) as a sum of an integer (x) and an another nunber (√y), That another number (√y) should be an integer as well. For √y to be an integer y should be a perfect square. Above I showed all the possible ways to get 7. If you look at it, you'll see only 0 and 4 are perfect squares. y must be equal to 0 or 4. But if y=0, √x=7 which means x=49. As 49+2 (x + √y) ≠ 11, x≠49, √x≠7, y≠0. Only option you are left with is y=4. If y=4 for √x+y to be equal to 7 √x has to be equal to 3. So x=3²=9 y=4, x=9 This explaination is long because it is so simplified, not because it is complex.

Bro, where can I access WhatsApp or Telegram channels?

knowing that x and y are integers makes possible to solve this in mind without paper just using a few guesses

hey that is cheating not ab-initio solution

Just by simple logic it took hardly 10 sec 1. x,y = perfect sq 2. x,y < 11 That's it.... x = 9 , y = 4

At first how do you find a+b? I don’t understand it.

Ramanujan : The Mathemagician 🧙🏽‍♂️

Jaise RAMAnujan ko Mata Naamgiri Devi padati this waise hi mujhe bhi mere BAJRANGBALI MATH Padhate hai So I solved this in just 2 sec And Maine bhi ek equestion banaya hai √x+y+z=21 x+√y+z=21 x+y+√z=21 JAI SHREE RAM HAR HAR MAHADEV

Aap sab log sirf ek hi tarah ka soln post karte ho pls find another method to solve this equation 😅

And here i thought there were some amazing analitic solutions😢😢😢

m² + 1232 = 3ⁿ 3ⁿ is odd for all n ∈ ℕ 1232 is even Hence, m² is odd m is odd 1232 is a multiple of 4 Hence, 1232 ≡ 0 mod 4 When m is odd, The last digits of m are 1, 3, 5, 7 or 9 Last digit of 1² is 1 Last digit of 3² is 9 Last digit of 5² is 5 Last digit of 7² is 9 Last digit of 9² is 1 Hence, The last digits of m² are 1, 5 or 9 1 ≡ 1 mod 4 5 ≡ 1 mod 4 9 ≡ 1 mod 4 Hence, Odd m² ≡ 1 mod 4 Hence, When m² is odd, (m² + 1232) ≡ (0 + 1) mod 4 (m² + 1232) ≡ 1 mod 4 Therefore, 3ⁿ ≡ 1 mod 4 The values of 3ⁿ, starting from n = 1 and n ∈ ℕ in ascending order, are: 3, 9, 27, 81, 243, 729, 2187, 6561, ….. Hence, If n is odd, The last digit of 3ⁿ is either 3 or 7 3 ≡ 3 mod 4 7 ≡ 3 mod 4 Hence, 3ⁿ ≡ 3 mod 4 when n is odd and n ∈ ℕ If n is even, n ∈ ℕ, The last digit of 3ⁿ is either 9 or 1 9 ≡ 1 mod 4 1 ≡ 1 mod 4 Hence, 3ⁿ ≡ 1 mod 4 when n is even Therefore, n is even Hence, Let n = 2k, where k ∈ ℕ m² + 1232 = 3ⁿ m² + 1232 = 3²ᵏ 3²ᵏ - m² = 1232 (3ᵏ - m)(3ᵏ + m) = 1232 Since 3ᵏ and m are odd, (3ᵏ - m) is even, (3ᵏ + m) is also even Let a = 3ᵏ - m b = 3ᵏ + m Since m ∈ ℕ , 3ᵏ + m > 3ᵏ - m Hence, b > a b - a = 3ᵏ + m - (3ᵏ - m) = 3ᵏ + m - 3ᵏ + m = 2m Hence, Let’s find the even factors of 1232 from the prime factorisation of 1232 1232 = 2⁴ x 7 x 11 Express 1232 as a product of two even factors, with one of them being a multiple of 4, (A): 1232 = 2² x (2 x 7 x 11) = 4 x 308 (B): 1232 = 2³ x (2 x 7 x 11) = 8 x 154 (C): 1232 = 2⁴ x (7 x 11) = 16 x 77 (D): 1232 = (2⁴ x 7) x 11 = 112 x 11 (E): 1232 = (2⁴ x 11) x 7 = 176 x 7 For (A), a = 4 b = 308 b - a = 308 - 4 = 304 Hence, 2m = 304 m = 304/2 = 152 However, 152 is not odd For (B), a = 8 b = 154 b - a = 154 - 8 = 146 Hence, 2m = 146 m = 146/2 = 73 73 is odd Sub. m = 73 into given equation: 73² + 1232 = 3ⁿ 5329 + 1232 = 3ⁿ 3ⁿ = 6561 n = 8 For (C), a = 16, b = 77 b - a = 77 - 16 = 61 Hence, 2m = 61 However, 61 is not divisible by 2 Hence, m ∉ ℕ for 2m = 61 For (D), a = 112, b = 11 a - b = 112 - 11 = 101 Hence, 2m = 101 However, 101 is not divisible by 2 Hence, m ∉ ℕ for 2m = 101 For (E), a = 176, b = 7 b - a = 176 - 7 = 169 Hence, 2m = 169 However, 169 is not divisible by 2 Hence, m ∉ ℕ for 2m = 169 Hence, Ans: m = 73, n = 8

Got it fast but didnt find it mathematically x=9, y=4 root 9 + 4 = 7 9 + root 4 = 11

X=9, Y=4

Not necessarily you have to take only integer values as it is not specified but the case where sqrtx+ sqrty =2 will be neglected because these numbers and the other factor differ by an odd number so 5/2 basically does not satisfy but anyways thanks for the question

Y = 4 & X= 9. Clue is perfect squares

I think you complicated it a lot more then necessary

as a child, i solved this in 20 seconds 😁😁😁😁😁😁😁😁😁😁😁😁

Hit and trail...>>>>>

Great video and well explained. Thanks for sharing

I somehow solved it mentally without using this method (I guessed it)(I thought both variable should be a perfect square in order to get a perfect whole number root)

Bro which software do you use

It took 20 seconds. √x+y=11 X+√y=7 Put value of y to be less than 11 but higher than 7 for the guess and x to be small but both have to be perfect square in that range so i choose Y=9 saw that it fitted if x=4 and that ans

kzbin.info/www/bejne/pGSTc3t7nKqjl9ksi=OpFkmWC2AgDr42Xk

X= 9 Y=4

3:50 Why not -2, -2??

Sir the third stage simplification wasn't clear

you should've said early on that the solution is based on guessing. It would have given me the chance to tune out earlier

I reached half way but hesitated to try arbitrary values since I thought there may be a more subtle method. Anyway good puzzle. 👍

x = 9 y = 4

By observation I got the x=9, y=4 I haven't seen the solution yet now I will see the whole soln 😊😊😊?

(x)^(1/2) +y=7---------1st equation ; (y)^(1/2) +x=11------------2nd equation ; my solution is (x)^(1/2) =7 -y; we squaring on both sides we get (x)=49+(y)^2-14y put in 2nd equation we obtain (y)^(1/2)+49+(y)^2-14y=11 now we arranging and transforming process in y variable then we have (y)^(2)+14y-(y)^(1/2)-38=0 ; ((y)^(1/2))^(4)+14(((y)^(1/2))^2)-(y)^(1/2)-38=0 compare with a z^4 + b z^3 + c z^2 + d z + e = 0 this is fourth order(4) with respect to minimum one half order (1/2) of y (y^(1/2)) equation so only one valid value of x and y will exist that is x==21.0659574433 and y =2.41023339991 that is well satisfied and valid value you can check LHS=RHS when you will put value of x and y in equation only 1st but similarly processed as above arrange and transform in x term then ((x)^(1/2))^(4)-22(((x)^(1/2))^2)+(x)^(1/2)+114=0 compare with a z^4 + b z^3 + c z^2 + d z + e = 0 this is fourth order(4) with respect to minimum one half order (1/2) of y (x^(1/2)) equation so only one valid value of x and y will exist that is x=9.00 and y =4 that is well satisfied and valid value you can check LHS=RHS when you will put value of x and y in both equation 1st and 2nd equation.

This is 7th grade maths question. Don't spoil the name of Ramanujan. And you too need to improve your maths skill - so many unwanted steps!!

√x+y=7 x+√y=11 1) If we represent the graphs in the Cartesian plane we will realize that it has a unique solution. 2) Both solutions in R are positive. 3) In case x=0, the maximum value of y=7 4) In case y=0, the maximum value of x=11 5) In the event that x does not have an exact square root, y must be such that its infinite decimals complement those of √x to make it an integer. It is possible, but difficult 6) In the case that y does not have an exact square root, x must be such that its infinite decimals complement those of √y to make it an integer. It is possible, but difficult 7) That 5 and 6 are fulfilled at the same time is something that can be almost impossible 8) Forgetting about 5 and 6 because of what was mentioned in 7, we can assume that x and y are numbers with exact roots 9) Possible cases x={4,9}, y={1,4}, with x>y 10) Test values with the conditions of 9 and conclude that x=9, y=4