in my paper i figured out the solution x=45 y=1 by hit and trial as 45 square was nearest to 2023 but i was not sure that it was the only solution so i left the question , great analysis by you sir
@@aayushjawalekar4169 No bro, if you can explain the analytical reason for your approach by hit and trial in subjective paper then also it's acceptable 😊
Sir I tried another method, if we make this equation like this: 2^y+2-2+2023 = x² =>2(2^(y-1) - 1)+2025=x² Now x will be: => √2(2^(y-1) - 1)+2025 So now as 2025 is a perfect square and its clearly visible that making y=1 will make the value 2(2^(y-1) - 1) =0 hence we can say x=45 and y=1 Thanks sir👍
Take modulo 4 assuming y>1 since x^2=-1 not possible since squares are either 0 or 1 mod 4 now this implies y=0 or 1,so you get the answer PS:This only works since x and y are naturals
take modulo 4 to thik hai par uske aage jo likha hai vo samaj nhi aa rha maine modulo 4 karke try kara and usse answer shi aa rha hai x^2=-1 ka kya meaning hai iss line mein y=0or 1 wali condition meri bhi aa rhi hai par x wali condition match nhi ho rhi
However, there is one step without any rationale : subtracting one. We need a structural, and a much more elegant solution. I have one, as follows: I did it in the following way: x^2 - 2023 = 2^y. 1. First, LHS > 0. So, x > sqrt(2023). This gives the minimum value of x = 45. 2. For x = 45, x^2 - 2023 = 2025 - 2023 = 2 3. Even number cannot be the answer because RHS is odd. 4. Consider next number: 47. 47^2 - 45^2 = (47+45)(47-45) = even * even, which contains 4 as a factor. Add 2 to it, and it definitely does not contain 4 as a factor. So, 2^y will never have 4 as a factor. For example, in this case, 47^2 - 2023 = 184 + 2 = 186, which is not divisible by 4. This applies to all upcoming numbers (since new number square is greater than that of existing number by a number divisible by 4) So, Y can only be (0, 1). 0 does not satisfy, so Y = 1
I solved this in my mind in 5 seconds, u see there is a trick for perfect squares of numbers with unit digit having 5 , so 5² =25, 15² = 1×2 ,25 = 225 (we are doing 1×2 since ten's place is occupied by 1 and then we multiplt it by successive number) , 25² = 2×3,25 =625 , 35² = 3×4,25=1225 , 45² = 4×5,25 = 2025 , there we go we got 2025 now in lhs , 2ⁿ + 2023 we need to find the value of the power of 2 to make it 2025 if u look carefully the power should be 1 to make it 2023+2=2025 , now x is 45 ,y is 1 therefore x+y= 46 simple and very easy no need to waste 6 minutes lol
The very instant I saw 2023 and x², I rewrote it as 2025-2 because 2025 is 45², rearranged it as (x-45)(x+45) = 2(2^(y-1) -1). Rest is just deducing lhs and rhs as even or odd
@@mukul9221 no use x²-2023 then take log then use log property tumhe kuch gadbad dikhheji kyuki ayega log(ײ-2025) ayega agr x=45 dal doge toh log define hi nhi hoga
Sir simple agar ham mod 4 kare to we know that every perfect square is 0,1 mod 4 and if we claim that y>2 then 2^y would be 0 mod 4 and 2023 is 3 mod 4 which is not possible and hence y
Sir 45 ka square 2025 hota hai. Agar hum x ki jagah 45 rakhe aur aur y ki jagah 1 tho dono barabar ho jaenge Answer - 45+1 = 46 😅 Hit and trial se kiya
@@夜神月-l8q I am sad that congruence modulo is no longer in number theory. - _ -. Jokes apart, I think you can try by divisibility of 8 ignoring modulos, provided even sufficient amount of time is there to develop the concept of modulo, by ourselves in exam hall. Lol😂😂😂
Sir there can be one more easy solution We can write it as 2^y=x²-2023 Here we can conclude that 2^y will always be even untill y is not 0 , So taking the case when y is not zero then we can say that 2^y will always be even and+ive and hence x²-2023 has to be even and first of all x² term has to be greater than 2023 for term to be +ive so simply x is greater than 44²(1936) as 45² is 2025 And if we just take a hit and trial in this we can get that at x=45 , 2^y is 2 which is even and +ive and no number below this exist there can be above that but 2^y can't be less than 2 ( taking case of y is not equal to 0 and y is real natural number as give in question ) So we have 2^y=2 and hence y =1 So this method can also be used I think I don't whether this method is convenient or not but I just tried becz you inspire me to try different approaches you are my real inspiration sir❤❤ thanks for all I learned about maths is bhannaaaattt😅❤❤❤❤
It obvious and trivial that x, y = 45,1 is a solution. In general if y≥3, taking mod 8 x²= 0,1, 4 mod 8 and 2^y +2023 = -1 = 7 mod 8. Thus the modular equation is contradicted. Therefore, y≤2. Taking y = 2 gives 2027 which is not a perfect square. Therefore 45,1 is the only solution set.
Ye to sabse basic question hai hit and trial se hogya Really ye sachme aya tha mains me Sidhe y=1 rakhoge to 2025 kiska square hai 45 ka x=45 hogya Dono natural number hai khtm😂😂
Are you actually that dumb What jf more pairs of x and y could exist Clearly you do not have the iq level to solve this in the exam room Question had asked all the possible pairs of x and y Everyone could only predict a single possibility of y=1 and x=45 What if there were more You cant take risks
First I want to answer before seeing solution if we put y=1 then rhs become 2025 which is sq of 45 so x=45 then X+y= 46 It hit me when I was looking at squares ending with 25 which are generally of numbers ending with 5 Sir and others fellow students pls verify
Though I'm in 9th class, this question felt too easy. I just first tried to find square root of 2023 which was around 44 and after that we have to add a factor of 2 to make it a perfect square and square of 45 is 2025 which is 2023+2 so y = 1 and x = 45 and as it is given that y and x both are natural numbers so no ±45=x. Therefore, x+y = 45+1 =46
@@divi7117 bhai tum naye log lag rhe ho for IITJEE maths. Teri baat galat nhi he obviously par tu isse ye conclude nhi kar sakta ki x and y ki aur ki possible value nhi he. And sir ne 6 minutes deeply samjhane me lagaye varna seconds me chalta he he dimag. Tujhe abhi bhi doubt lagrha he to check out the video of the author of black book usme dekhle. tumhari galti nhi he tum 9th me ho tumhe idea nhi he IIT waale kabhi bhi koi bhi twist daal dete he question me
Solution is on spot! I solved it in a different way, like y = log2(x^2 - 2023) now there are two cases, one is either x is very large and after subtracting to 2023 we'll get something again very large to make it power of 2 , but if we see graph of x^2 and 2^x we can see that slope of x^2 after y = 2023 is somewhatt nearly 88deg so if we have to make x^2 very large than 2023 it is not possible because compared to graph of 2^x , x^2 slope is mild at larger values so only possibility is that x^2 is slightly greater than 2023 and 2^x is not large then there slopes would be comparable. for eg: at x =2 slope of x^2 is 4 and slope of 2^x is 2.77
bhai tum naye log lag rhe ho for IITJEE maths. Teri baat galat nhi he obviously par tu isse ye conclude nhi kar sakta ki x and y ki aur ki possible value nhi he. And sir ne 6 minutes deeply samjhane me lagaye varna seconds me chalta he he dimag. Tujhe abhi bhi doubt lagrha he to check out the video of the author of black book usme dekhle.
Ok that's correct for this question because there is given x is a natural no. But u can't say even * even is div. By 4 or not even 2 because remember 0 is also an even no. That can cause confusion in another questions where it is not given ... Noone noticed it hoping it to be noticed by someone and this confusion should be clear so that maths can be bhannat ❤
my soln x^2 = 0, 1 (mod4) for y >1 2^y = 0 (mod 4) and 2023 = 3 (mod 4) therefore, 2^y + 2023 = 3 (mod 4) when y >1 but x^2 is 0 or 1 ( mod 4) which implies that y =1 for y=1 , x= 45 this is only soln your solution is also amazing sir
Solution:- We know that when a perfect square is divided by 4 it only gives remainder 0 and 1. So x² can only give remainder 0 or 1. On RHS 2023 gives remainder 3 and 2^y can give remainder 1,2 and 0. If it gives remainder 0 then RHS in total will give remainder 3 which is not possible for a perfect square so it must give remainder 1 or 2 as 1+3 will give total remainder 0 and 2+3 will give total remainder 1. That is only possible when y=0 and 1. If we put y=0 we get x=√2024 which is not a natural number. If we put y=1 we get x=45.
Ye jitne bhi log tukka laga rahe hai.. Unke liye 615 + x²= 2^y. (x, y are non zero integers /can be both negative or positive) Find maximum value of x+y. Lagao isme tukka
@@sirak_s_nt bro us ques me max nahi pucha to possibility hai ki ya to ek solution hai ya fir agar kai solution hoge to unka sum same hona padega so hit and trial is best way in that question
It took some concept I learned this year for Olympiad prep Any square number can be written is the form 4k and 4k+1 Here X² is odd, thus x²= 4k+1 We get 4k-2022=2^y As there exists only one +ve pair of value of x,y y=1, x²=2025 Thus (x,y)=(45,1) Hence ,the req ans is 46
Sir I have an alternate solution, As x is a natural number, so unit digit of x² can range from {0,1,4,9,6,5}. Coming to RHS, 2^x unit place can be {1,2,4,8,6} adding 3 of "2023" to unit place gives values ranging as {4,5,7,1,9}. Digits that get tallied on LHS and RHS are {4,5,9}. Which corresponds to {0,1,4} power of x on RHS, so values of y can be the same. For y=0, RHS=2024 y=1, RHS=2025 y=4, RHS=2039, Of the above cases. Only y=1 satifies the condition, so y=1, and from there we can get value of x as 45. So final answer is (45+1)=46.
If y is greater than or equal to 2 then x^2 is of the form 4k-1 which is not possible so ans=(2025)^1/2+1=46.because we know that every perfect square is of the form (2k+1)^2 or (2k)^2 or 4k+1 or 4k
Sir, alternate method can be, taking cases of y either y=1 or y>=2 for y=1, you get x=45 for y>=2 the RHS becomes a number in the form of 4k+3 where k is a natural no and this format can never be a perfect square. Therefore, x+y=46 . Exam mein question , find number of ordered pairs and summation of all ordered pairs tha.
@@lamshywy8920 wo NTA ki problem nahi h.. Ye bohot hi basic baate h ki perfect squares 4k, 4k+3, 8k, 8k+1, 8k+4, 16k, 16k+4, 16k+8, 16k+1, 7k+2, 7k, 7k+1 7k+4 etc form h hote. Prime number 6k+5, 6k+1, 4k-1, 4k+1 form ke hote. Khud se bhi pta ki jaa skti h
We can also go by another approach which kinda involves some hit and trial but is effective since we know x+y will have one value only The approach goes like this... Since we know that y is a natural no. i.e. it would be 1,2,3,4..... so on. Hence, RHS will always be an odd number, therfore x must be an odd no. Since square of an odd natural no. is only odd. Now we know that x is an odd no. So now let's assume the range of x... Since x² lies in the range of 2000's hence x should be near 40 to 50 (just assuming, if no soln comes that means our assumption is wrong). Hence, now we know that x is an odd no. And lies between 40 to 50. Now let's take the mid of the range say 45 so that we can specify x. Therefore x² comes out to be 2025... Jackpot for some x belonging to our assumed range there exists a natural no. Y (i.e. 1) for which the given equation is satisfied.. Hence, x+y=46
Bro kya tum prayas se pdh rhe hi mai bhi start krne wali ho ashish sir ke diye hue ktk wagrah tumse solve ho jata hai agar koi aur tips ho to batanaa please reply
it is a easy question when you observe it you can see that x will be an odd number and you can just go with hit and trial by putting 45 47 ,, so x=45 and y=1. easy 2min question
Sir I literally found the solution 😨 So, I assume a number who's square was closer to 2023 and it was 45 who's square is 2025 so X^2 = 45^2 45^2= 2023+2 Therefore 2^y = 2 Therefore y=1 And X=45
Sir if we use trial and error method we can easily find the answer, it is very easy question for the 7th and 8th class students who knows the concept of squares and square roots.
Maine 2023 ke sabse pass wala square dekha jo hota hai 2025 square of 45 toh isko x ki jgh rakha aur 2025-2023 kiya Fir 2 bacha lhs me aur rhs me 2^y bacha equate krke y =1 aygya aur x = 45 phle se liya tha hence x+y = 46 Sir ye shi solution hai ?
Sir easiest solution ye hoga ki y cannot be greater than 2 kyunki phir RHS (4k +3) ke form ka ho jayega jo kabhi bhi perfect squae ni ho skta so y=1 aur x =45
Easy. Check parity. Upfront checking power of 2 we find x must be odd. Now use mod 4. If x is odd then x = 1 mod 4. But if y greater than 1 then rhs is congruent to 3 mod 4. Not possible. Hence y must be 1 so x = 45. Cheerio!!!😊
sir mere paas ek short trick hai☺. Aap pehele powers ko compare karlo jisse aapko 2=y+1 mil jaega aur aap further directly 2 second me solve kar sakte ho☺
Find the values of a, b, c and x, y, z with the value of ... (We let that...) k=x×b⁴-b³-b²+b-2(z×c+x×a)×b° Where we have an equality as, a×x+y+b×y-z+c×z = {b-[x×b⁴-b³-b²+b-2(z×c+x×a)×b°]}÷ (2×y) as well as we also have that, b²=a×c , y²=x×z , y²=b²/2 and.... 1). a, b, c belongs to natural numbers. as well as... 2). x, y, z also belongs to natural numbers. and... 3). k belongs to real numbers. Sir please answer and solve my question 🙏🙏 I am you're big fan ❤️
If y=1, then RHS becomes 2025, and x^2 = 2025 implies that x = +/- 45, if x is an integer. But, if x is a natural number, then x = -45 is not feasible. Then only solution is : x = 45.
Ise aese bhi Kar sakte hai na 2^y=x^2-2023 If y belongs to N so start by putting value y=1(minimum value of N) If y=1, 2=x^2-2023 x^2=2025 x=45,-45(-45 will be rejected) X=45 Hence,x+y=45+1=46 This way of solving is easier just you have to know square of 45.
I rarely comment on any video ,but i wil comment here because ,ye question ka infinite solution hai corresponding to natural number jiske liye dono x & y real ho & here x+y ka minimum value poocha jana chaiye jo 46 hoga.
x,y are natural numbers so if take root on both sides x=sqrt(2^y+2023) as x is natural RHS should be perfect square so I looked for the square near 2023 which I got of 45 Therefore x=45,y=1 Again this is hit and trial Thanks for correct explanation !!
I solved this under 5 seconds, without looking at the solution. Here's how, I first thought of numbers you can square to get a value close to 2023... I first tried 50 then got 45[which is 2025].... Therefore x equals 45 and y equals 1. Ans in 46.
Just looking at the question and hving the knowledge that 2+2023 = 45^2 one could have have answered as 46. The sheer fact that only one answer is being asked one ahiudk ahve answered...I am surprised so many students left it
Sir, Can we analyse like this?, that Since x²=2^y+2023 Therefore, 2^y+2023 must be a perfect square, and for this only possible value for 'y' is 1. And therefore x=45..
Sir I am in class 9th and solved it in first attempt.😊 Sir thank you a lot because your videos has helped me a lot and brought me in a position such that I am no longer afraid of competitive questions.😊😊
Bohot easy for a student of Vedic maths like me. We all know ki jiss number ke end meiñ 25 aata hai and start ke digits do consecutive numbers ka product ho toh voh ek perfect square hota hai. Here 20 is the product of 4 and 5. So clearly add 2 to 2023 and you will get 2025 which is aperfect square of 45. Done. Answer x=45 and y=1.
I saw the problem and instantly the following solution came to me, which is much easier too: The RHS is odd, so must the LHS be. Any odd square is 1(mod 8). 2023 is 7(mod 8). So we just need another 2(mod 8) as contribution from 2^y. For y>=3, 2^y=0(mod 8). And y^2=4(mod 8). So y=1 is the only solution.
One Liner Solution: If y > 1, then by checking modulo 4 on RHS we get x^2 = 3 (mod 4), impossible as a square is either 0,1 (mod 4), so y = 1 only possibility and x = 45.
Sir 2^y (even) + 2023 (odd) will be odd so x^2 is odd and hence x has to be odd. However to prove fhat y can be only 1, one needs this factorization after subtracting 1 on both sides. Fantastic solution Sir.❤
One more way to think is that if x is established to be odd then the form of x is 2n+1 so x^2 will be 4n^2 + 4n +1 and therefore when divided by 4 will give a remainder of 1. Now 2023 mod 4 is 3 so the only option of 2^y is 2 (y=1)which will give a remainder of 1 when divided by 4 all other higher values of y will give a reminder of 3 as higher values of y (y>=2) will give a reminder of 0 when 2^y is divided by 4 and therefore 2^y + 2023 will give a reminder of 3.
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Kya solution h 🥶🥵
My respect for aman malik sir 📈📈📈📈
Bhai ye common problem hai
Yeh amazing soln nhi hai
There are better methods
tum ek method bata do
@@pratyakshYT95.5
@@pratyakshYT95.5 btao better method...mujhe to ye wala mast laga
in my paper i figured out the solution x=45 y=1 by hit and trial as 45 square was nearest to 2023 but i was not sure that it was the only solution so i left the question , great analysis by you sir
Mene bhi same yahi hit Kia but 30 seconds me hi question skip karke aage badh gya tha
Mujhe toh yaad hi nhi aisa koi question tha paper mein? Meri bhi same shift thi
I think udhar x and y belongs to n nhi tha
Aur isliye ek aur possible case bnega -45 wala
question different tha kuch complex numbers se related tha jitna yaad hai
🛐🛐🛐🛐
What a mind blowing solution!
This channel is very underrated...
Aman sir, aapne aise solutions le aate ho saamne ki dekh kr maza aa jaata hai, aap jaisa teacher maine shayad hi dekha hai
x=45, y=1 (by hit and trial).
Frrr same i did
NAH, IF it's mcq then only it can happen
@@aayushjawalekar4169 Why would it not work lmao it's literally an answer and if it's wrong then the question can't be right
@@aayushjawalekar4169 No bro, if you can explain the analytical reason for your approach by hit and trial in subjective paper then also it's acceptable 😊
Sir I tried another method, if we make this equation like this: 2^y+2-2+2023 = x²
=>2(2^(y-1) - 1)+2025=x²
Now x will be:
=> √2(2^(y-1) - 1)+2025
So now as 2025 is a perfect square and its clearly visible that making y=1 will make the value 2(2^(y-1) - 1) =0 hence we can say x=45 and y=1
Thanks sir👍
Nicee
🤌✨✨❤
This solution only proves y=1 is a solution. It does not prove that y=1 is the only solution.
It's giving u single value for it; But not a complete proof!
Same ways I solved it😊.
What a way of solving this, since I'm in my graduation but still I like to solve these JEE sums. Great explanation sir.
I always love the solution when there is no loss of generality. Great solution sir.
MATHS IS NOT A SUBJECT IT IS FEELING
Ha bhai mera bhi favourite subject hai
❤❤
My favourite subjects are maths and physics ❤❤❤❤
It's love bro
If you treat it like subject then you cannot understand it.😊
Mathematics is a language of God ❤😊
Take modulo 4 assuming y>1 since x^2=-1 not possible since squares are either 0 or 1 mod 4 now this implies y=0 or 1,so you get the answer
PS:This only works since x and y are naturals
take modulo 4 to thik hai
par uske aage jo likha hai vo samaj nhi aa rha
maine modulo 4 karke try kara and usse answer shi aa rha hai
x^2=-1 ka kya meaning hai iss line mein
y=0or 1 wali condition meri bhi aa rhi hai
par x wali condition match nhi ho rhi
@@krrishrohilla2945 take modulo 4 on both sides 2023 leaves a remainder of -1 or 3 from 4 but that is not possible for a perfect square so y=1
\\
Mene bhi exactly same thought se Kiya ❤
this question don't demand any concept from any chapter of the syllabus just pure common sense, indeed a good question
It was a basic level olympiad question from prmo diophantine equation
@@thevibetree1 Seriously
@@KUMAR_ALOK_JEE2024kbhi olympiads nhi diya hai kya? 11th jee se tough maths 9th ioqm wagera me hota hai
Kya solution hai sir🔥🔥😯 ekdum bhannat🔥🔥
However, there is one step without any rationale : subtracting one. We need a structural, and a much more elegant solution. I have one, as follows:
I did it in the following way:
x^2 - 2023 = 2^y.
1. First, LHS > 0. So, x > sqrt(2023). This gives the minimum value of x = 45.
2. For x = 45, x^2 - 2023 = 2025 - 2023 = 2
3. Even number cannot be the answer because RHS is odd.
4. Consider next number: 47. 47^2 - 45^2 = (47+45)(47-45) = even * even, which contains 4 as a factor. Add 2 to it, and it definitely does not contain 4 as a factor. So, 2^y will never have 4 as a factor. For example, in this case, 47^2 - 2023 = 184 + 2 = 186, which is not divisible by 4. This applies to all upcoming numbers (since new number square is greater than that of existing number by a number divisible by 4) So, Y can only be (0, 1). 0 does not satisfy, so Y = 1
I solved this in my mind in 5 seconds, u see there is a trick for perfect squares of numbers with unit digit having 5 , so 5² =25, 15² = 1×2 ,25 = 225 (we are doing 1×2 since ten's place is occupied by 1 and then we multiplt it by successive number) , 25² = 2×3,25 =625 , 35² = 3×4,25=1225 , 45² = 4×5,25 = 2025 , there we go we got 2025 now in lhs , 2ⁿ + 2023 we need to find the value of the power of 2 to make it 2025 if u look carefully the power should be 1 to make it 2023+2=2025 , now x is 45 ,y is 1 therefore x+y= 46 simple and very easy no need to waste 6 minutes lol
Yes bro 😂😂😂same thinking.What a coincidence.Class 9th mind same
Us Bhai Us
Agar subjective me ata to?
@@Wanderer-dd4mj it will never come , there's no chapter which has these types of question in ncert. Ncert is either too childish , or too hard
Ong same bhai
The very instant I saw 2023 and x², I rewrote it as 2025-2 because 2025 is 45², rearranged it as (x-45)(x+45) = 2(2^(y-1) -1). Rest is just deducing lhs and rhs as even or odd
Gave JEE in 2011 but man what a solution to this beautiful question. Just randomly youtube recommended me your video
X= 45 and Y = 1
Hit and trial...
Better for ioqm beginner question
jo log IOQM ki thodi bhi taiyari kiye hey unke liye bahut hi jyada asaan tha
Pata nahi jee coaching Wale number theory kyon nahi padhate
Diophantine eqn. Right
@@prashant34049 star batches me number theory ke basics padhate h and geometry me RMO ke pyq and Pathfinder krwate..
@@prashant34049 mostly nobody teach
Bhai tune kiya kaise diophantine se
Kidhar kagaya modulo
Elsborate krega mera to aa hi nai rha
@@monujhembrom9279
🤯🤯🤯🤯🤯 mind blowing sol.
This is why i tell to my freinds that you are the one who is giving me ideas.
Use both side log you can see directly
@@SKYLORD6905please provide solution using log
@@SKYLORD6905 Log 2023?
@@mukul9221 no use x²-2023 then take log then use log property tumhe kuch gadbad dikhheji kyuki ayega log(ײ-2025) ayega agr x=45 dal doge toh log define hi nhi hoga
Mera shift
Ques me summation of all possible x and y puchatha. Thank you sir legendary soln...
29th shift 2 me tha yeh?
@@utkarshpandey01 yess
@@soumabhopal9901 kon sa question tha
Yaha par x, y € N tha ya R
Kyunki R mein toh ans 2 ho jayega
Sir simple agar ham mod 4 kare to we know that every perfect square is 0,1 mod 4 and if we claim that y>2 then 2^y would be 0 mod 4 and 2023 is 3 mod 4 which is not possible and hence y
Exactly, simple ioqm examples lol
which chapter is it from? number theory?
Sir 45 ka square 2025 hota hai.
Agar hum x ki jagah 45 rakhe aur aur y ki jagah 1 tho dono barabar ho jaenge
Answer - 45+1 = 46
😅 Hit and trial se kiya
Normal Application of Number Theory in Math Olympiad(Divisibility Theory)
One Liner, if y>2 then mod 8 gives x^2=-1(mod 8), which gives no solution to x. Then only possible values are 0
Congruence modulo is not in jee syllabus -_-. But number theory can be also used .
@@夜神月-l8q I am sad that congruence modulo is no longer in number theory. - _ -. Jokes apart, I think you can try by divisibility of 8 ignoring modulos, provided even sufficient amount of time is there to develop the concept of modulo, by ourselves in exam hall. Lol😂😂😂
Sir there can be one more easy solution
We can write it as 2^y=x²-2023
Here we can conclude that 2^y will always be even untill y is not 0 , So taking the case when y is not zero then we can say that 2^y will always be even and+ive and hence x²-2023 has to be even and first of all x² term has to be greater than 2023 for term to be +ive so simply x is greater than 44²(1936) as 45² is 2025
And if we just take a hit and trial in this we can get that at x=45 , 2^y is 2 which is even and +ive and no number below this exist there can be above that but 2^y can't be less than 2 ( taking case of y is not equal to 0 and y is real natural number as give in question ) So we have 2^y=2 and hence y =1
So this method can also be used I think I don't whether this method is convenient or not but I just tried becz you inspire me to try different approaches you are my real inspiration sir❤❤ thanks for all I learned about maths is bhannaaaattt😅❤❤❤❤
What a beautiful solution sir and the beauty of maths damnnn !!!!
It obvious and trivial that x, y = 45,1 is a solution. In general if y≥3, taking mod 8
x²= 0,1, 4 mod 8 and 2^y +2023 = -1 = 7 mod 8. Thus the modular equation is contradicted. Therefore, y≤2. Taking y = 2 gives 2027 which is not a perfect square. Therefore 45,1 is the only solution set.
wrong h
45,1 is the true solution daal ke check krlo
Galat hai 45 and 1
@@HoldOnItwillGetBetter are ha ik galti se type ho gya mene & y = 2 me 2027 value ko already soln me reject kiya hua h
Whi itna easy q kaise aagya but shyd exam temperament me bachha confuse hosakta hai.
Felt in love with maths again sir. Koti koti naman Aman sir❤
Ye to sabse basic question hai hit and trial se hogya
Really ye sachme aya tha mains me
Sidhe y=1 rakhoge to 2025 kiska square hai 45 ka x=45 hogya
Dono natural number hai khtm😂😂
Ha ye ese question hote h ki dekh ke phli bari me click kra toh ho gya
... Vrna exam pressure me Kbhi kbar soch ni pate
Are you actually that dumb
What jf more pairs of x and y could exist
Clearly you do not have the iq level to solve this in the exam room
Question had asked all the possible pairs of x and y
Everyone could only predict a single possibility of y=1 and x=45
What if there were more
You cant take risks
Matrix Wale ka kya tha ans ......
@@vanshgupta6454 tumhara shift konsa tha aur exactly kya question tha wo batao fir ans bata dunga Mera 27shift 2 tha
Haa isse ho jata hai aise competitive setting mein
But it doesn't hurt to know the full beauty of the question
This is also called solution by parity. Olympiads preparers would have found this extremely easy
First I want to answer before seeing solution if we put y=1 then rhs become 2025 which is sq of 45 so x=45 then
X+y= 46
It hit me when I was looking at squares ending with 25 which are generally of numbers ending with 5
Sir and others fellow students pls verify
Nice concept
What an analysis ... really fantastic.. Great Aman ji
This question is so amazing. Solving it with this method is even more amazing
Sir aap great ho kasam se itna accha explanation kisi ka nahi dekha hoga maine
It can be done easily
x²-2023=2^y
Take log base 2 both side ,as x and y are natural number
There is only one possibility that is x=45 and y=1
Though I'm in 9th class, this question felt too easy. I just first tried to find square root of 2023 which was around 44 and after that we have to add a factor of 2 to make it a perfect square and square of 45 is 2025 which is 2023+2 so y = 1 and x = 45 and as it is given that y and x both are natural numbers so no ±45=x. Therefore, x+y = 45+1 =46
for real this was more of a logical one
@@divi7117 bhai tum naye log lag rhe ho for IITJEE maths. Teri baat galat nhi he obviously par tu isse ye conclude nhi kar sakta ki x and y ki aur ki possible value nhi he. And sir ne 6 minutes deeply samjhane me lagaye varna seconds me chalta he he dimag. Tujhe abhi bhi doubt lagrha he to check out the video of the author of black book usme dekhle. tumhari galti nhi he tum 9th me ho tumhe idea nhi he IIT waale kabhi bhi koi bhi twist daal dete he question me
Physics is love ❤
But math me everything 🔥🔥
Sir, Apne aisa samjhaya ki mai class 9th ka student isko samajh gaya.
Its a very very easy question, but maybe not in the exam!!!
First when i saw this qiestion i took the values of perfect square greater that 2023 which is 2025 =45²
i.e x=45 and y=1
Solution is on spot! I solved it in a different way, like y = log2(x^2 - 2023) now there are two cases, one is either x is very large and after subtracting to 2023 we'll get something again very large to make it power of 2 , but if we see graph of x^2 and 2^x we can see that slope of x^2 after y = 2023 is somewhatt nearly 88deg so if we have to make x^2 very large than 2023 it is not possible because compared to graph of 2^x , x^2 slope is mild at larger values so only possibility is that x^2 is slightly greater than 2023 and 2^x is not large then there slopes would be comparable. for eg: at x =2 slope of x^2 is 4 and slope of 2^x is 2.77
The amount of illogical argument in this answer is too damn high!
Bhai neend puri krle apni
wow what a problem solving, so helpful to know the actual way to solve now i can use these ideas in other different situations too , thank you ! sir!
Sir I want to conclude one thing that you are a person from different planet.❤
Maths ko dimag se nhi dil se padha jata h kyuki yeh feeling h❤
X^2 = 2^y +2025-2
X*2-2*y=45*2-2*1
By comparing x =45
Y = 1
bhai tum naye log lag rhe ho for IITJEE maths. Teri baat galat nhi he obviously par tu isse ye conclude nhi kar sakta ki x and y ki aur ki possible value nhi he. And sir ne 6 minutes deeply samjhane me lagaye varna seconds me chalta he he dimag. Tujhe abhi bhi doubt lagrha he to check out the video of the author of black book usme dekhle.
Ok that's correct for this question because there is given x is a natural no. But u can't say even * even is div. By 4 or not even 2 because remember 0 is also an even no. That can cause confusion in another questions where it is not given ... Noone noticed it hoping it to be noticed by someone and this confusion should be clear so that maths can be bhannat ❤
but you can see that x will be an odd number as 2^y will be an even and by adding it with 2023 will result in odd .
0 is divisible by every number
my soln
x^2 = 0, 1 (mod4)
for y >1 2^y = 0 (mod 4)
and 2023 = 3 (mod 4)
therefore, 2^y + 2023 = 3 (mod 4) when y >1
but x^2 is 0 or 1 ( mod 4)
which implies that y =1
for y=1 , x= 45
this is only soln
your solution is also amazing sir
Olympiad wala lag rahe ho.. lol
Solution:-
We know that when a perfect square is divided by 4 it only gives remainder 0 and 1. So x² can only give remainder 0 or 1. On RHS 2023 gives remainder 3 and 2^y can give remainder 1,2 and 0. If it gives remainder 0 then RHS in total will give remainder 3 which is not possible for a perfect square so it must give remainder 1 or 2 as 1+3 will give total remainder 0 and 2+3 will give total remainder 1. That is only possible when y=0 and 1. If we put y=0 we get x=√2024 which is not a natural number. If we put y=1 we get x=45.
x+(2^y/2)=289
x-(2^y/2)=7
So,by adding,x=148 and 2^y/2=141
Although I am in class eight presently but I have solved this problem by using class eight concepts
Can you tell me which concepts?
Sir -44 bhi a skta h
Kyuki x^2 =2025
So x =-/+ 45
And -45+1=-44
X belongs to N h bhai
Ye jitne bhi log tukka laga rahe hai.. Unke liye
615 + x²= 2^y. (x, y are non zero integers /can be both negative or positive) Find maximum value of x+y. Lagao isme tukka
ban gaya cool?
@@bruhyou197 mere se toh bana hi hua tabhi toh diya h 😅
@AkshunChauhan-bt6ey x, y = (59, 12) & (-59, 12) so maximum sum is 71, minimum sum is -47
@@sirak_s_nt bro us ques me max nahi pucha to possibility hai ki ya to ek solution hai ya fir agar kai solution hoge to unka sum same hona padega so hit and trial is best way in that question
@@HarishSharma-s7ba/c to situation
It took some concept I learned this year for Olympiad prep
Any square number can be written is the form 4k and 4k+1
Here X² is odd, thus x²= 4k+1
We get 4k-2022=2^y
As there exists only one +ve pair of value of x,y
y=1, x²=2025
Thus (x,y)=(45,1)
Hence ,the req ans is 46
What a great thinking....maza hi aa gya...tukke se to Mera bhi ho gya..par te analysis dekh ke dil khush ho gya....
Sir I have an alternate solution,
As x is a natural number, so unit digit of x² can range from {0,1,4,9,6,5}.
Coming to RHS, 2^x unit place can be {1,2,4,8,6} adding 3 of "2023" to unit place gives values ranging as {4,5,7,1,9}.
Digits that get tallied on LHS and RHS are {4,5,9}. Which corresponds to {0,1,4} power of x on RHS, so values of y can be the same.
For y=0, RHS=2024
y=1, RHS=2025
y=4, RHS=2039,
Of the above cases. Only y=1 satifies the condition, so y=1, and from there we can get value of x as 45.
So final answer is (45+1)=46.
my solution is similar to sir's.. and you were lucky to get the right answer as you never proved that there can't be more than one solution...
op
@@AaravSingh-qv6cp I proved that there are no more solution than this one, look closely
If y is greater than or equal to 2 then x^2 is of the form 4k-1 which is not possible so ans=(2025)^1/2+1=46.because we know that every perfect square is of the form (2k+1)^2 or (2k)^2 or 4k+1 or 4k
More good solutions are possible
Vote for olympiad sreries
Sir, alternate method can be,
taking cases of y
either y=1 or y>=2
for y=1,
you get x=45
for y>=2
the RHS becomes a number in the form of 4k+3 where k is a natural no and this format can never be a perfect square.
Therefore, x+y=46 .
Exam mein question , find number of ordered pairs and summation of all ordered pairs tha.
4k+3 perfect square nahi hai yeh kaisa pata chala?
@@_ehh (x^2)-3 isn't divisible by four
@@_ehh class 10 NCERT🙌
@@sirak_s_nt Iss sal class 10 NCERT se ye sari cheez remove ho gai hai 😭
@@lamshywy8920 wo NTA ki problem nahi h.. Ye bohot hi basic baate h ki perfect squares 4k, 4k+3, 8k, 8k+1, 8k+4, 16k, 16k+4, 16k+8, 16k+1, 7k+2, 7k, 7k+1 7k+4 etc form h hote. Prime number 6k+5, 6k+1, 4k-1, 4k+1 form ke hote. Khud se bhi pta ki jaa skti h
😮 Wow what a solution sir
Full respect to you sir
It was easiest question ever for all my friends and me
We can also go by another approach which kinda involves some hit and trial but is effective since we know x+y will have one value only
The approach goes like this...
Since we know that y is a natural no. i.e. it would be 1,2,3,4..... so on. Hence, RHS will always be an odd number, therfore x must be an odd no. Since square of an odd natural no. is only odd. Now we know that x is an odd no. So now let's assume the range of x... Since x² lies in the range of 2000's hence x should be near 40 to 50 (just assuming, if no soln comes that means our assumption is wrong). Hence, now we know that x is an odd no. And lies between 40 to 50. Now let's take the mid of the range say 45 so that we can specify x. Therefore x² comes out to be 2025... Jackpot for some x belonging to our assumed range there exists a natural no. Y (i.e. 1) for which the given equation is satisfied.. Hence, x+y=46
Ashish sir also solved this problem in parayas jee 1.0 2025 by kind of similar but more logical way 😊😊 he is the best maths teacher
Bro kya tum prayas se pdh rhe hi mai bhi start krne wali ho ashish sir ke diye hue ktk wagrah tumse solve ho jata hai agar koi aur tips ho to batanaa please reply
it is a easy question when you observe it you can see that x will be an odd number and you can just go with hit and trial by putting 45 47 ,,
so x=45 and y=1.
easy 2min question
Sir I literally found the solution 😨
So, I assume a number who's square was closer to 2023 and it was 45 who's square is 2025 so
X^2 = 45^2
45^2= 2023+2
Therefore 2^y = 2
Therefore y=1
And X=45
Brilliant problem and a nice solution strategy
This solution was just amazing like mannnn❤
Sir if we use trial and error method we can easily find the answer, it is very easy question for the 7th and 8th class students who knows the concept of squares and square roots.
sir i get many learnings about maths by watching your videos ,thankyou so much
Maine 2023 ke sabse pass wala square dekha jo hota hai 2025 square of 45 toh isko x ki jgh rakha aur 2025-2023 kiya
Fir 2 bacha lhs me aur rhs me 2^y bacha equate krke y =1 aygya aur x = 45 phle se liya tha hence x+y = 46
Sir ye shi solution hai ?
Yeh bahut aasan problem hai. Aur Aman sir ka soln koi bahut khatarnaaak nhi hai. Any IOQM student will nail it faster.
Sir easiest solution ye hoga ki y cannot be greater than 2 kyunki phir RHS (4k +3) ke form ka ho jayega jo kabhi bhi perfect squae ni ho skta so y=1 aur x =45
Hit and trial se karliya under 30 second😅😂
Bas closest perfect square nearby 2023 check kiya which is 2025
So x=45
y=1
Easy. Check parity. Upfront checking power of 2 we find x must be odd. Now use mod 4. If x is odd then x = 1 mod 4. But if y greater than 1 then rhs is congruent to 3 mod 4. Not possible. Hence y must be 1 so x = 45. Cheerio!!!😊
sir mere paas ek short trick hai☺. Aap pehele powers ko compare karlo jisse aapko 2=y+1 mil jaega aur aap further directly 2 second me solve kar sakte ho☺
Find the values of a, b, c and x, y, z with the value of ...
(We let that...)
k=x×b⁴-b³-b²+b-2(z×c+x×a)×b°
Where we have an equality as,
a×x+y+b×y-z+c×z = {b-[x×b⁴-b³-b²+b-2(z×c+x×a)×b°]}÷ (2×y)
as well as we also have that,
b²=a×c , y²=x×z , y²=b²/2
and....
1). a, b, c belongs to natural numbers.
as well as...
2). x, y, z also belongs to natural numbers.
and...
3). k belongs to real numbers.
Sir please answer and solve my question 🙏🙏
I am you're big fan ❤️
2025 (x^2)) is the perfect a sqare of 45 (x) next to 2023. Implies that value of 2^y=2 ,meaning y=1. Therefore x+y= 46
but in student should you following approch
1) x,y both are natural numbers
2) find perfect squares near 2023 which is 2025 so y=1 and x=45.
If y=1, then RHS becomes 2025, and x^2 = 2025 implies that x = +/- 45, if x is an integer.
But, if x is a natural number, then x = -45 is not feasible.
Then only solution is : x = 45.
Ise aese bhi Kar sakte hai na
2^y=x^2-2023
If y belongs to N so start by putting value y=1(minimum value of N)
If y=1,
2=x^2-2023
x^2=2025
x=45,-45(-45 will be rejected)
X=45
Hence,x+y=45+1=46
This way of solving is easier just you have to know square of 45.
my solution is similar to sir's.. and you were lucky to get the right answer as you never proved that there can't be more than one solution...
I found that tha in RHS, 2025 is complete square of 45, if we put y = 1, then we get x = 45
X+y = 46
I rarely comment on any video ,but i wil comment here because ,ye question ka infinite solution hai corresponding to natural number jiske liye dono x & y real ho & here x+y ka minimum value poocha jana chaiye jo 46 hoga.
Then give some other values for y and x please!
I am Sachin sir student and i solve this question by only seeing thumbnail.❤❤❤❤😂 But thank you sir for complete explanation ❤😊😊
x,y are natural numbers so
if take root on both sides
x=sqrt(2^y+2023)
as x is natural RHS should be perfect square so I looked for the square near 2023 which I got of 45
Therefore x=45,y=1 Again this is hit and trial
Thanks for correct explanation !!
He is the GOAT MATHEMATICAN
What a solution sir,you are best, salute 🫡🫡 to you sir.
I solved this under 5 seconds, without looking at the solution. Here's how, I first thought of numbers you can square to get a value close to 2023... I first tried 50 then got 45[which is 2025].... Therefore x equals 45 and y equals 1. Ans in 46.
my solution is similar to sir's.. and you were lucky to get the right answer as you never proved that there can't be more than one solution...
Nice. I would have never have thought of this solution under that much pressure😅. Good luck.
Such questions are rare in JEE more kind of math Olympiad question
Btw good reasoning by sir 👍 👌
Can hit and try method also be a right method to solve this question
Sir🙏🙏..
Sir आप ऐसे ऐसे question दिखा कर मजा ला देते हैं सच कह रहा हूँ।
आपके तारिफ में और क्या कहूँ शब्द ही कम पड रहे हैं।
Just looking at the question and hving the knowledge that 2+2023 = 45^2 one could have have answered as 46. The sheer fact that only one answer is being asked one ahiudk ahve answered...I am surprised so many students left it
Sir, Can we analyse like this?, that
Since x²=2^y+2023
Therefore, 2^y+2023 must be a perfect square, and for this only possible value for 'y' is 1. And therefore x=45..
Sir I am in class 9th and solved it in first attempt.😊 Sir thank you a lot because your videos has helped me a lot and brought me in a position such that I am no longer afraid of competitive questions.😊😊
Just take mod 4 we get y
Bohot easy for a student of Vedic maths like me. We all know ki jiss number ke end meiñ 25 aata hai and start ke digits do consecutive numbers ka product ho toh voh ek perfect square hota hai. Here 20 is the product of 4 and 5. So clearly add 2 to 2023 and you will get 2025 which is aperfect square of 45. Done. Answer x=45 and y=1.
I solved this thing mentally . x=45 , y=1
Most easiest solution would be for y=1 it's x=45
And for higher values of y RHS is of form 4k+3 which can't be a perfect square by class 10 EDL 😅
I saw the problem and instantly the following solution came to me, which is much easier too:
The RHS is odd, so must the LHS be. Any odd square is 1(mod 8). 2023 is 7(mod 8). So we just need another 2(mod 8) as contribution from 2^y. For y>=3, 2^y=0(mod 8). And y^2=4(mod 8). So y=1 is the only solution.
One Liner Solution:
If y > 1, then by checking modulo 4 on RHS we get x^2 = 3 (mod 4), impossible as a square is either 0,1 (mod 4), so y = 1 only possibility and x = 45.
could you please tell which chapter is it from? kya yeh number theory se hai? usme bhi konse sub-topic se?
Yes this is from number theory itself, I don't think it has any such sub topics, this is the core idea of number theory (modular arithmetic).
Pranam Charan sparsh peri pagdi apke pairo me
Kuchh v likhu Kam hai gr8 teacher
Sir 2^y (even) + 2023 (odd) will be odd so x^2 is odd and hence x has to be odd. However to prove fhat y can be only 1, one needs this factorization after subtracting 1 on both sides. Fantastic solution Sir.❤
One more way to think is that if x is established to be odd then the form of x is 2n+1 so x^2 will be 4n^2 + 4n +1 and therefore when divided by 4 will give a remainder of 1. Now 2023 mod 4 is 3 so the only option of 2^y is 2 (y=1)which will give a remainder of 1 when divided by 4 all other higher values of y will give a reminder of 3 as higher values of y (y>=2) will give a reminder of 0 when 2^y is divided by 4 and therefore 2^y + 2023 will give a reminder of 3.