Awesome Olympiad Problem For Maths Genius 😍| Aman Malik Sir

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This question is for all maths genius and maths lovers.
If aabb=x^2, Find x
This is a very amazing question from Maths Olympiad which will surely test your mathematics skills and critical thinking.
You'll not be able to directly implement the direct mathematics fundamentals rather there will be critical points you need to think about from the core maths perspective.
Solving such questions will also train your mind to open up and go for in-depth critical thinking.
Maths Olympiad problems are surely a must-do thing for you if you want to develop such skills.
Stay tuned with ‪@BHANNATMATHS‬ for more such interesting questions.
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#maths #mathematics #olympiads #amansirmaths #jee #jeemains #iitjee #jeeadvanced #olympiadmath #mathsolympiad

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11 сен 2024

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Комментарии : 414

@abdulmujib1140 Год назад

Alternate Soln : aabb = x² 11(100a + b) = x² 100a +b = 11z Means aabb should be divisible by 11 Sum of odd digits = a+b = Sum of Even Digits = a+b = Difference of both = 0 So After dividing number is a0b So by divisibility of 11 Sum of odd digits = a+b Sum of even digits = 0 Difference of both = a+b So a+b should be = 11 By this, a = 7 b= 4

@Ajay_Vector Год назад

I done the same

@pranavshukla7778 Год назад

Initial idea was main part of the problem then we may continue in infinite ways

@DharmendraKumar-me2my Год назад

Can anyone explain this further more clearly 😅 After 100a+b=11z

@Ajay_Vector Год назад

@@DharmendraKumar-me2my We have 11(100a+b) = x² For this 100a+b must have a factor of 11 i.e. a0b which is a 3 digit no. Must divisible by 11 => (a+b)-(0) should be a multiple of 11 => because a,b can't be zero as "aabb" is a 4 digit no. And a,b ≤ 9 => a+b =11 Then we get different (a,b) Hence different a0b no. As 209,308,407,506,605,704,803,902 And 704 on dividing by 11 we a perfect square no. 64 => a= 7, b = 4 aabb = 7744 = 11.11.64 => x = 11.8 => x = 88

@tannusharma3966 Год назад

Sir apka online paid batch kha milega class 12 ka?????

@prakharsingh3243 Год назад

me before playing the video: aabb=x² a²b²=x² x=ab!! 😂😂😂

@dishaa_rawat 2 месяца назад

Still you did it wrong. This should be - x²= (ab)² *x = ±ab* → Ans.

@user-pe9lt3bb6h 2 месяца назад

@@dishaa_rawatcorrect But root(1) isnt -1

@SaurabhSingh-me1ci Год назад

9a+1=1 is also a perfect square, but since 'a' cannot be zero or 'b' cannot be greater than 9 hence we choose 64 as the perfect square for the rest of the solution.

@19-biswarooptalukdar99 Год назад

Why a can't be zero...

@19-biswarooptalukdar99 Год назад

The number than will reduced to bb....

@Sanvi565 Год назад

@@19-biswarooptalukdar99 b cant be 11 as in a number the face value of a digit cant exceed 9

@19-biswarooptalukdar99 Год назад

@@Sanvi565 thank you....understood

@user-jq4iq9be3p Год назад

Sir please make an advanced illustration series of every chapter

@ranvijaygoyal1765 Год назад

Yes sir please make this for advance

@priyabratapanda3264 Год назад

Yes sir

@unknown79884 Год назад

Yes sir please

@rkdrkd7362 Год назад

Yes sir

@samirghosh2856 Год назад

Yes sir please

@ishanarya16 Год назад

This is a question from NTSE stage-2. I am in 10th class and I have done this question myself without anyone else's help. It was a very proud moment for me as you made a video on this question. Love mathematics and your teaching too.

@sirak_s_nt Год назад

Exactly I'm also in 10th and solved it myself.. Btw are you NTSE aspirant? Any coaching?

@ishanarya16 Год назад

@@sirak_s_nt PW Vidyapeeth student. Maths is love Trying for PRMO but not getting enough resources

@vedants.vispute77 Год назад

Yes I am also the fellow, the last ones. Now unfortunately the exam is scrapped from 2022

@sirak_s_nt Год назад

@@vedants.vispute77 now u r in 11th? Which stream?

@vedants.vispute77 Год назад

@@sirak_s_nt I will give jee adv this june

@deadinlavapool7840 Год назад

x is four digit therefore 31

@mathskafunda4383 Год назад

We don't have to do that also. A perfect square can only end in 1, 4, 5, 6 or 9. Also, the mod 4 of any perfect square can only be 0 or 1. Thus aa11, aa55, aa66 and aa99 get eliminated instantly. Thus b=4 and aa44 is the only possibility. If a perfect square ends with 4, its square root must have unit digit either 2 or 8. As, aabb has to be divisible by 11, X must also be divisible by 11. Thus, the only two possibilities of X=22 or 88. As 31

@Rahulkumar-ft8jw 11 месяцев назад

@@mathskafunda4383mod 4 matlab??

@p_bivan11 Год назад

Easy question 1000a+100a+10b+b = x² 11(100a+b)=x² 100a+b should be formed like 11.( )² (100a+b) /11 = whole no. 99a/11 + a+b/11 = whole no. a+b = 11 If (a, b) = (2,9) Then, 100a+b = 209 209/11 = 19 (not a perfect square) If (a, b) = (3, 8) 308/11 = 28 By looking at pattern we will get no. like.... 19,28,37,46,55, (64) : perfect square 64*11= 704 We get (a, b) = 7,4 aabb = 7744 = 11².8² "x = 88"

@devanshdwivedi623 Год назад

Beautiful✨

@ARN48411 Год назад

I solved it myself by little bit another method.

@abhinavtiwari5585 Год назад

Put a=5 and b=4. ??? Also satisfy this equation...😅

@p_bivan11 Год назад

@@abhinavtiwari5585 it's a number not multiplied

@LUCKY_PRINCE_ 6 месяцев назад

Alternate solution without solving any equation: 11(100a+b)=x2 100a+b=11n, where n is a perfect square as the above quantity is greater than 100; using hit and trial. Substitute n= 16,25,36,64 we get n=64 and 100a+b=704, by comparison we get a=7 and b=4

@mathsbyiitians Год назад

Gajab approach sir..... Many people including me can't thought of this simple approach in the first place. But you again proved that best way to solve mathematics is to keep your basics up to date and in mind.

@thenameishitesh Год назад

Yes , my seniors used to advise me that IIT doesn't mean a lot , lot lot of hardwork ..... U have to first build the ground floor ( i,e clear ur basics ) to achieve a building

@aniruddhxie2k215 Год назад

6:30 Sir here a cannot be 0,1 as a+b = 11 so let's say a is 0 then b =11 but that's not possible as a and b are digits so they can only go from 0 to 9 Same logic for when a=1 So we can already reject 2 cases

@sparshsharma5270 Год назад

aabb = x^2 By Euclid's division lemma, c=dq+r c=x, d=10, r=(1,2,3,4,5,6,7,8,9) c^2=10q+r, r=(1,4,5,6,9) Since a symmetric number is divisible by 11, then By Euclid's proposition -: If a divides b and b divides c, then a divides c also - we have, aabb divides x^2 => x^2 divides 11 => x divides 11 Let x=11y, => x^2 = 121y^2 => aabb = 121*y^2 So, aabb/121 = y^2 y^2 = (9,16,25,36,49,64) => y = (3,4,5,6,7,8) => 121*y^2=aabb By the sqaure number rule, y=(4,5,6,8) By the symmetric number rule, y=8 So, x=11y=11*8=88 aabb=88^2=7744

@aniceguy6065 Год назад

Genius bro

@user-uz4gp6lh8n 10 дней назад

aabb=11(100a+b),for a =7,b=4,100a+b=704 and x^2=11*704=11*11*8*8 ,so x=11*8=88

@iMvJ27 Год назад

By just mere looking... Me screaming out of my lungs 88² =7744. x =88. Perks of ssc preparation ❤😂

@bhaskarkhandewal3257 Год назад

Same with me, solved it in head in three minutes

@The.Sigma. Год назад

But you have to prove in Olympiad

@siddharth8334 Год назад

who

@a_grimpo_khrel9650 Год назад

@@bhaskarkhandewal3257 .

@a_grimpo_khrel9650 Год назад

@@The.Sigma. .

@vibeduck17 Год назад

aabb = x^2 x is divisible by 11 range of x is 32 - 99 square numers 33, 44, 55 etc answer is 88

@dmc7325 Год назад

Alternate method: We can assume that 100a+b is of the form 11k^2 and find all its possible values from k=0 to 9. We have to search for the value whose tenth digit is 0 and that is possible for k=8. Thus we get a=7and b=4.

@prabhagupta6871 Год назад

7:26 1 is also a perfect square but if a=0 then b will be 11 which is not possible so a=7 only

@als2cents679 Месяц назад

Maine toh sirf 11 kah divisibility rule apply kiya. Difference in sum of alternate digits must be a multiple of 11. aabb = n^2 aabb = 11 * a0b since a and b are single digit numbers and a + b = 0 or 11, gives a + b = 0, gives a = b = 0, but then aabb is not a 4 digit number, hence a + b = 11, which means a0b = { 209, 308, 407, 506, 605, 704, 803, 902 } aabb = 11 * a0b aabb = 11 * 11 * (a0b / 11) a0b / 11 = { 19, 28, 37, 46, 55, 64, 73, 82 } Since (a0b / 11) needs to be a perfect square, the only answer is a0b = 64 aabb = 11^2 * (a0b / 11) = 11^2 * 64 = 11^2 * 8^2 = 88^2 = n^2 which gives n = 88

@zen9506 Год назад

Another way sir (thoda lamba hein) Assume number to be ab Let a be variable and B be 1......9 Case 1 B=1 A1*A1 is a square with unit digit 1 so the tens digit also should be 1 For tens digit a+a=(any number with unit digit 1) ie ___1 not possible so eliminate B=2 (no is a2*a2) Unit digit is 4 Tens digit is 4a=____4 A=6 satisfy (check through 4 table) So 64*64=3844 not possible B=3 (A3*A3) Unit digit is 9 9a=___9 a can't be 1 or 2 as any number from 1to 31 has square of 3 digits B=4 Unit digit is 6 but here 1 is carried so 8a=______5 as 1 carried should be added No case so emlinate B=5 unit digit is 5, 2 carry 10a=____3 eliminate B=6 Unit digit is 6 ,3 carry 12a=____3 Not possible B=7 Unit digit is 9,4 carry so 14a=____5 No case B=8 Unit digit is 4, 6 carry so 16a=___8 2 cases a=3 and a=8 A=3 square is 1444 A=8 Square is 7744 so x =88 Thank you,

@aadijaintkg Год назад

Alternate Solution aabb = x² aobo+aob = x² 11(aob) = x² Now, we can say that aob = 11 × kb where, k+b = some number which ends with 0 And then we can say that 11² kb = x² Now, kb should be a perfect square of any number from {4,5,...9} And by that we can say 8²= 64 and 6+4=10 Thus, kb = 8² Hence, 88²= x² Thus, x = 88

@anshika6689 Год назад

You are a real hero of mathematics

@mathskafunda4383 Год назад

A perfect square can only end in 1, 4, 5, 6 or 9. Also, the mod 4 of any perfect square can only be 0 or 1. Thus aa11, aa55, aa66 and aa99 get eliminated instantly. Thus b=4 and aa44 is the only possibility. If a perfect square ends with 4, its square root must have unit digit either 2 or 8. As, aabb has to be divisible by 11, X must also be divisible by 11. Thus, the only two possibilities of X=22 or 88. As 31

@toofaaniHINDU 4 месяца назад

Second line samajh nhi aaya, please explain it

@shyamaldevdarshan Год назад

What an explanation!😊 Bahut interesting question h!! Aap book publish kro na apne collection of unique concept ko lekr!!!

@shyamaldevdarshan Год назад

@@musaifshaikh07yeah bro 😊 thanks!

@shyamaldevdarshan Год назад

@@musaifshaikh07 😊🙏. ,,radhe radhe🙏

@dhruvmishra3859 Год назад

If a, b, c are sides of a triangle and s be it's semi-perimeter, then prove the following 1

@Aaravsrivastava117 Год назад

how i do this plz see - 11(100a+b) must be greater than 100 {as a ans b are lie btw 1-9} and 100a+b must contain 11 in its factor so to make a perfect square of multiple 11 and above 100 are (100a+b) - 44*4, 55*5, 66*6, 77*7, 88*8 , 99*9 now we can easily find no.

@als2cents679 Месяц назад

a = 1 bhi perfect square hain, lekin woh value use nahin kar sakate kyonkay a + b = 11 with a = 1 deta hain b = 10 (not single digit number)

@globalolympiadsacademy4116 Год назад

In the last stage , a >1 as a+b = 11 and both are digits so we can avoid testing for a =0 and 1. Also if 9a +1 is y^2 then y°2-1 = 9a or (y+1)*(y-1) = 9a as a

@GurpreetSinghMadaan 11 месяцев назад

Aabb is obviously an 11 multiple. If it is equal to x^2, then x^2= 11^2 x N^2 =121xN^2. The four digits aabb min max are (1000 to 9999), so N^2 can have values from 8 to 82. The values being (9, 16, 25,36,49,64,81) for integer values on N. 64 solves for 121x64= 7744

@ATB25659 Год назад

If a=2 and b=3 then the x=6 If a=2 and b=4 then the x=8 I think above these two situations also satisfy this aabb=x² equation. Please tell me more about regarding this question.

@harjassingh1385 Год назад

aabb=2233=x² X=√(2233)≠perfect square but sir said that x is perfect square 👍

@gidskdsfjiafjifdifjdif 11 месяцев назад

SIMPLEST ANSWER WOULD BE 0 since it is not given anywhere that A not equal to B therefore a=b therfore aaaa type no. and a = then x sq. =0 and x=0 ans.

@impresent2005 Год назад

Sir my way of thinking : (sir thode detail me explain kiye hai mene plz ek bar padna jarur) I had first read the number carefully and I'm pretty sure that these number aabb must be divisible by 11 . As we divide these number by 11 we have a0b x 11 = aabb And as aabb is a perfect square a0b must be again divisible by 11 Till now we had factories aabb = 11 x 11 x (something)..... Now these something must be a "square" Becoz if it's not a square then no. aabb will not be a perfect square. And if we multiply a perfect square with 11 we get a number in a0b form. So, 11*16 = 11*25 = . . . 11*64 = 704 Hence, Factors of aabb is 11*11*64 Hence x = 8

@impresent2005 Год назад

💕LOVE FROM HINDUSTAN 💕

@AdeshBenipal Год назад

Sir minor correction; x=88,-88

@girindrapandita5168 Год назад

cannot be -88 as square of -88 gives 7744 only so basically square root of 7744 is modulus of 88 which only provides 88

@kashyaptandel5212 Год назад

me asf: “x = ab” 🗿

@Puzzlarium1 Год назад

Same 😂

@MDDilshan-oq5or 11 месяцев назад

Same😂

@Anmol_Sinha Год назад

I did it by long division method. aabb / 11 = a0b. As this is divisible by 11, a+b=11 by divisibility. 9a+1 is a square number say m² 9a = (m+1)(m-1), as a is not 11, 9 = m+1 and a is thus 7. b=11-a=4 Ans is thus 7744

@aadijaintkg Год назад

This may become wrong in some case like if you have an equation that 8×9= (m+1)(m-1) Then as per you way of solving the m will comes out with 7, 10 but by solving 72= m² - 1 m will be square root of 73

@Anmol_Sinha Год назад

@@aadijaintkg as far as I understand, the problem is that I assumed that if 9a = (m+1)(m-1), then any 1 of those factors must be 9 even though it may not be depending on a. I didn't want to brute force and I realized that if I got a solution using it(luck) then I wouldn't have to do so much work lol

@localtry Год назад

Sir you have a challenge 👇 can you solve this integral?? ∫(x/tanx)dx , where limit is 0 to π/2. Answer is (π/2)ln2.

@himeshpatel1139 Год назад

Great sir Best teacher of maths ever

@tanmaykumarkeshari4642 Год назад

Easy Solution: 11(100a+b) => 100a+b = 11* x^2 put x =8 to get a= 7 and b= 4

@AlstonDsouza-jl7ow Год назад

One more useful information square numbers end with 1,4,5,6,9 for b u can eliminate the rest and substitute in a+b=11

@Aaditya7447 Год назад

If we can rational number also then answer should this also If a be - 1 by root 2 b be root 1 Then x will be 1 which is a perfect square

@honestadministrator Год назад

x^2 = 1000a + 100a + 10b + b = 11 ( 100 a + b) = (11) ^2 x square number. Hereby one needs to check whether either of the following numbers are of the form a a b b : 121*9, 121*16, 121*25, 121*36, 121*49, 121*64, 121*81. Only 121*64 =7 7 4 4 is of this form

@TheHellBoy05 Год назад

this is how i did it

@dhipin9590 7 месяцев назад

Sir you won’t believe it is my first advanced or Olympiad level question I could solve on my own that too by not this method it took me a long solution but It increased my confidence a lot

@ankush.0369 Год назад

Le me: aabb= x² a²b²=x² => X=ab Solved...🤣🤣

@rajivgorai5381 Год назад

A perfect square number has four digits, none of which is zero. The digits from left to right have values that are: even, even, odd, even. Find the number

@xpscorp Год назад

Thanks sir, for solving problems for us😊😊😊

@mananjaykumartiwari6329 Год назад

Sir 1 bhi perfect square tho aap sirf 8² kyu liye

@Anmol_Sinha Год назад

If we take 1, then a=0 As b=11-a as he proved earlier, we will get b =11 which is not possible as b is a single digit number.

@laxmanprasadnarwariya1789 Год назад

(a-1)aa(b+1)bb=x squre then find x & that number (a-1)aa(b+1)bb.

@Ajay_Vector Год назад

Sir intro me hint mil jaati hai ques solve karne ki Phir khud soch nhi paate Isiliye if possible remove intro in upcoming vides

@unknown79884 Год назад

Sir please make adv illustration series like Physics Galaxy

@googleverify9772 Год назад

Yes sir this is real math jisme koi faltu ka formula nhi yad krna bas apna logic use krna h

@gaurabhraaz 10 месяцев назад

Alternate solution:- No. is of 4-digit, so it must be lies between 1000-9999 Let's start squaring with smallest possible no. 22×22= 484 (3-digit no.) 33×33= 1089 (smallest 4-digit perfect square no.) 100×100= 10000 (5-digit no.) 99×99= 9801 (largest 4-digit perfect square no.) Now the no. lies between, 33²(1089) & 99²(9801) Now, calculating squares from 33 to 99, like 44², 55², 66² .....and so on.. At the end you will find that 88²= 7744 Give it a thumbs up 👍 if you got it

@ajayagar84 Год назад

Sir aapke solution me jab aap second time perfect square khoj rahe the tab, 1 bhi ek perfect square tha. We need to reject that option because a cannot be 0 given a+b=11 and a and b are digits

@thealphagamer1292 Год назад

After solving it in my try i am extremely happy

@Zerotoinfinityroad 3 месяца назад

7:20 9a + 1 at a = 0 is 1 which is also perft sqr

@ankitupadhyay646 Год назад

Critical thinking use krne k liye best questions CAT mein ate hai, maths , reasoning , DI , solve kro dimag pura khul jayega

@omsd8088 Год назад

This ques is simple if a:2 And b:4 OR a:4 b:2 So x would be :8 X:8

@piyushkumar13ok Год назад

I used hit& trial method. Firstly we know that last digit of perfect square can never be (2,3,7,8) then b={014569} and a can not equal to 0. So I got x= 88. Can my solution is valid?

@kamyahaloi1724 Год назад

That how I too did

@19-biswarooptalukdar99 Год назад

Sir in 6:39 minutes in this video, when putting the value of a as 0, the value of whole expression is coming 1 which is also a perfect square....so why don't you take that as a solution.....???

@vishalgupta2547 Год назад

Because for that value a is zero and thus aabb will not be a 4-digit number

@amit-jx5lh Год назад

You are great sir ❤❤

@24rohitanand9c7 4 месяца назад

sir one small doubt/correction a cannot be 0 because aabb is 4 digit number 6:41 / 9:01

@131raghav Год назад

I tried this question before watching the solution, i eventually solved it but it took me a lot of time , so that how i solved it First of all find how many possibilities can be for aabb , it is 9 * 10 = 90 , but a perfect square ends with (0 , 1 , 4 , 5 , 6 ,9) so the possibilities goes down by 9 * 6 = 54 , now comes the tricky part , i observed that perfect square whose last two digits are same always have the same last two digits 44 , so the possibilities comes down to only 9 , (1144 , 2244 , 3344 .... 9944) then i checked with short tricks that out of these 9 which all could possibly be perfect square that came down to 2 that were 5544 and 7744 , now simply checked which of it was a perfect square by nornal method.

@AMGMineCrick 6 месяцев назад

Sir, mindblowing answer to a simple, very simple looking question 🤯 I spent almost 2 hours trying to solve it and did not get the point rhat it is a perfect square so the number should be divisible by 11 again 😅

@brainbusters4423 7 месяцев назад

This was a easy question. I am 10th grade and i could do this very easily. I did it in a different way, first i wrote all 81 possibilities and checked if theyre perfect squares. It took me 5-10 min but i feel very happy after solving it.🎉

@gamingupam2256 Год назад

a=7 , b=4 and x=88 , yeh to bahut easy approach tha 🤔

@prajjawaltiwari9566 Год назад

I got that by hit and trial...

@karuOP Год назад

Alt digits ka sum same hai, toh number dekhke hi we can conclude 11 se divisible hoga

@ragedgamer2850 9 месяцев назад

5:11 a has possible values from 0-9 but as b is the last digit of a square number it can only be 0,1,4,5,6,9,

@toofaaniHINDU 4 месяца назад

yes, but 'a' can't be 0, otherwise "aabb" will not remain 4 digit no.

@ragedgamer2850 4 месяца назад

@@toofaaniHINDU you're right,there's another logic for this, as a+b must be 11, b or a can't be 0 as it will make the other one's value to be 11 which is not possible in this case.

@aroo999yt2 Год назад

x^2=aabb =a^2b^2 X=ab

@RupeshKumar-rp5qi Год назад

Agar mujhe kisi bhi halat me iske ans tak pahunchna hota to mai 32 se 99 no. Ka square nikalta q ki 4 digit tak inhi ka square pahunchta h phir ans match katwa deta..... But abhi maine google se dekha Sare no. Ka sq. Aur 88 ka sq. 7744 mujhe mil gya

@neeldobariyavii400 Год назад

Oh sir please tell me how you build up this much good thinking skills in maths sir I also want this type of thinking all I love maths very much but I can't solve hard questions ❤❤❤❤

@narendradubey8015 11 месяцев назад

sir solution mai a ko aapne 0 bhi consider kiya hai but a 0 toh possible hi nhi hai kyunki usse vo 2 digit no. ban jayega

@sanjeevdhurwey4704 Год назад

I did it in a unconventional way by looking at the number it was clear that aabb=a0b×11(by basic division) , now the challenge was to make aob=y×11 , such that y is a perfect square since only the square of one digit number can make a 3 digit number by multipliying with 11 , I started my trial and error by dividing 11 by 901 hoping for a quotient of 81 and eventually ended up at 704 which gave quotient of 64 the square of 8 . It might sound ridiculous to some , but this os what I could think using basic division and some intuition.😁😁

@aadijaintkg Год назад

So you can write a0b equals to kb × 11 where "k" belongs to +ve integer and "b" is same as question but while writing this there must be one condition and then is k+b = some number whose unit place is zero

@ashwanibeohar8172 Год назад

If " abcdef " is a 6 digit number such that on multiplying it by any digit from 1 to 6, there is no change in digits, no change in their sequence, only digits rotate from left to right eg " bcdefa", defabc " ,

@vishalgupta2547 Год назад

142857

@ashwanibeohar8172 Год назад

X can not be one digit (as square will be Max or 3 digit number), and x can't be 3digit ( as square wil be 5digit, or 6 digit). So x should be two digit Now by trial and error Do The square of 11,22,33,44,55,66, --- In the sequence, when you square 88 You get 88*88 = 7744 And that is the soln 8

@krishnaats7141 Год назад

Answer is 88. I have not yet watched the video but I remember doing this problem for my IIT preparation 15 years ago.

@Awesome.Rahul2005 Год назад

Sir please continue this series 🙏🙏🙏🙏

@rishikeshkumawat9249 Год назад

🙏🙏🙏🙏🙏🙏🙏🙏🙏👍👍👍👍👍👍👍sir you are a real hero of maths

@vinaygodara1111 11 месяцев назад

Why (9a+1) must be a perfect square????..at 6:15

@orangesite7625 Год назад

My method/ how i solved aabb = 1100a+ 11b = 11.(100a+b) = x² Now we can observe and assume (100a+b) = 11 t² Possible values of a = 1 to 9 (first digit ≠0 because it's 4 digit n.o) Possible values of b= 0,1,4,5,6,9 100a+b = a0b => sum of n.o in odd places - even places = a+b-0 = a+b {From possible values min(a+b) = 1+0 = 1 Max(a+b) = 9+9 =18} N.o divisible by 11 between (1,18) is only 11 so a+b =11 { Min(100a+b)=100+0=100;max= 909} Check (100a+b) = 11t² put t=3 we get 99100 ✓ 11 t² 5+2≠10 So directly without calculating we can say all other choices except t=8 are correct So t=8 and 11(100a+b) = 11. 11t²= (11t)² =x² So so t=8 so x=88 and {a0b is 704 so a=7,b=4} original number is 7744

@ytmircalnoob9924 Год назад

Sir iska answer 88 hoga without any derivation Hamare class 9 aur 10 ke mensulatin ke question mai √7744 ka bohot baar istamal huaa hai to ya number humko yaad ho gaya tha😸😸

@technopanipuri3054 3 месяца назад

As a person who already knew that 88 squared is 7744 i am proud to be able to solve it in just 8 seconds

@NoobPerson-xp7nn 6 месяцев назад

Well the longest strategy i can think if is putting 1=a b=2 and keep putting numbers and squate rooting them i did this in calculator ans is 7744 which is the square of 88. Of course its unethical

@SarveshSingh-kw8xb 3 месяца назад

5:20 pe doubt hain kaise a+b ka yog 22 aur 33 nhi hoga

@chiragwadhawan7033 Год назад

We can also do it by base method for ex a perfect square can have only 44,00 same digit at the end 00 is not possible because it does not lead to same digit of aa then we choose 44 as bb then after 44 is unit digit come with taking base as 12 the.then we make combination like 38,62,88 and the number is multiple of 11 so 88 is the answer then square it 7744 will be the answer

@abhijnanmallick9849 Год назад

I am very happy to solve it without anyone's help....... Very nice problem...

@kavyanshtyagi2563 Год назад

1 st one .... more such problems sir thank you so much

@KabeerHasanSKS Год назад

Not only 64, 1 is also a perfect square

@Introgant Год назад

Then a will become 0 and a+b = 11 and a and b are digits so b can't be 11 .

@babitah3007 Год назад

Sir I have a doubt X^2(x^2--3)=4 X^2=4 (why?) X^2-3=4(how?) Please reply sir

@Digital_aman007 Год назад

3×3×2×2. 9×4=6²

@abinashkaushik8014 11 месяцев назад

Sir, Please explain why 100a+b should be divisible by 11.

@lokeshhari Год назад

x=6, a=2, b=3

@Math-625 Год назад

Nice question sir ❤🎉🎉

@mananjaykumartiwari6329 Год назад

But sir 1 is also a perfect square

@scifo7826 Год назад

sir please app bata dezeye ki ma per subject kitne question kru ek din mai for jee

@diptidutta5503 7 месяцев назад

Sir, I can solve this in different way. Thanks sir.🎉

@Surendra.2805 Год назад

Sir please continue this Olympiad series

@lokeshhari Год назад

x=6 , a=2 b=3

@wasudeoumredkar1526 Год назад

Very much at Home in mathematics+very intelligent.

@avinash6740 Год назад

Easy question 88 is answer. One awesome problem from my side 1/a +1/b=3/2018.Make solution of it. Students will learn alot from concept of this question.

@mathgeek1823 Год назад

what do we have to find? a and b?

@avinash6740 Год назад

@@mathgeek1823 yes find possible pairs of a and b.

@stardomnights9612 Год назад

Sir aap Roz jee adv related 1 Q laya Karo na ...please

@nazishali9746 Год назад

Solving without formula Possibility of aabb can be 81 so I have started doing root of those numbers and then when I reached on Possibility number 59 I have found the number that is perfect root and it was 7744 the root of 88.hence I have found X.funny enough but found the answer.