Awesome Olympiad Problem For Maths Genius 😍| Aman Malik Sir

[@abdulmujib1140]

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]

I done the same

[@pranavshukla7778]

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

[@DharmendraKumar-me2my]

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

[@Ajay_Vector]

@@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]

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

[@prakharsingh3243]

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

[@dishaa_rawat]

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

[@Digital_aman007]

Same

[@prakharsingh3243]

@@dishaa_rawat "☝️🤓"

[@t3ns0r-ss]

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]

Why a can't be zero...

[@19-biswarooptalukdar99]

The number than will reduced to bb....

[@Sanvi565]

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

[@19-biswarooptalukdar99]

@@Sanvi565 thank you....understood

[@physicsolympiad-d9t]

Sir please make an advanced illustration series of every chapter

[@ranvijaygoyal1765]

Yes sir please make this for advance

[@priyabratapanda3264]

Yes sir

[@unknown79884]

Yes sir please

[@rkdrkd7362]

Yes sir

[@samirghosh2856]

Yes sir please

[@LUCKY_PRINCE_]

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

[@ishanarya16]

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]

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

[@ishanarya16]

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

[@vedants.vispute77]

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

[@sirak_s_nt]

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

[@vedants.vispute77]

@@sirak_s_nt I will give jee adv this june

[@thatfabana07]

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

[@p_bivan11]

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]

Beautiful✨

[@ARN48411]

I solved it myself by little bit another method.

[@abhinavtiwari5585]

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

[@p_bivan11]

@@abhinavtiwari5585 it's a number not multiplied

[@mathsbyiitians]

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.

[@prabhagupta6871]

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

[@sparshsharma5270]

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]

Genius bro

[@dmc7325]

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.

[@gidskdsfjiafjifdifjdif]

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.

[@anshika6689]

You are a real hero of mathematics

[@mathskafunda4383]

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]

Second line samajh nhi aaya, please explain it

[@deadinlavapool7840]

x is four digit therefore 31

[@mathskafunda4383]

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]

​@@mathskafunda4383mod 4 matlab??

[@vibeduck17]

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

[@shyamaldevdarshan]

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

[@shyamaldevdarshan]

@@musaifshaikh07yeah bro 😊 thanks!

[@shyamaldevdarshan]

@@musaifshaikh07 😊🙏. ,,radhe radhe🙏

[@iMvJ27]

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

[@bhaskarkhandewal3257]

Same with me, solved it in head in three minutes

[@The.Sigma.]

But you have to prove in Olympiad

[@siddharth8334]

who

[@a_grimpo_khrel9650]

@@bhaskarkhandewal3257 .

[@a_grimpo_khrel9650]

@@The.Sigma. .

[@als2cents679]

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

[@aadijaintkg]

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

[@als2cents679]

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)

[@gaurabhraaz]

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

[@ATB25659]

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]

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

[@himeshpatel1139]

Great sir Best teacher of maths ever

[@GurpreetSinghMadaan]

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

[@zen9506]

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,

[@ragedgamer2850]

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]

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

[@ragedgamer2850]

@@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.

[@Zerotoinfinityroad]

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

[@Anmol_Sinha]

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]

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]

@@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

[@Aaravsrivastava117]

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.

[@AlstonDsouza-jl7ow]

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

[@dhruvmishra3859]

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

[@amit-jx5lh]

You are great sir ❤❤

[@googleverify9772]

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

[@localtry]

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.

[@Aaditya7447]

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

[@tanmaykumarkeshari4642]

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

[@24rohitanand9c7]

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

[@globalolympiadsacademy4116]

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

[@SarveshSingh-kw8xb]

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

[@ritheshstidie]

@ 7:23 there are two perfect squares sir (1 and 64). Why only you took 64 not 1

[@19-biswarooptalukdar99]

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]

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

[@lsytz]

Thanks sir, for solving problems for us😊😊😊

[@RahulSingh-rn9mm]

Wow sir . Maja aa gaya

[@ankush.0369]

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

[@vinaygodara1111]

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

[@honestadministrator]

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]

this is how i did it

[@rescvbhvvnnvvn]

Behtareen Sir,

[@impresent2005]

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]

💕LOVE FROM HINDUSTAN 💕

[@AdeshBenipal]

Sir minor correction; x=88,-88

[@girindrapandita5168]

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

[@laxmanprasadnarwariya1789]

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

[@rajivgorai5381]

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

[@gamingupam2256]

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

[@scifo7826]

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

[@Insight.crime111]

Sir please continue this series 🙏🙏🙏🙏

[@ajayagar84]

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

[@Chad_maker]

2:47 esa kese likh sakte hai ???

[@omsd8088]

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

[@diptidutta5503]

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

[@neeldobariyavii400]

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 ❤❤❤❤

[@ashwanibeohar8172]

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]

142857

[@ankitupadhyay646]

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

[@SVijaypratap]

कतही जहर solution #bhannatmaths

[@karuOP]

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

[@Surendra.2805]

Sir please continue this Olympiad series

[@surendrabishnoi2929]

sir bhut fyada hota h jb hamko 100 tk ke square yasd ho ye questions toh 5 sec m khtm thanks 🙏👍👍😊😊😊

[@unknown79884]

Sir please make adv illustration series like Physics Galaxy

[@aroo999yt2]

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

[@brainbusters4423]

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.🎉

[@piyushkumar13ok]

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]

That how I too did

[@abinashkaushik8014]

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

[@bollyboodgoldsilverscreen8612]

Sir ek abbb Bala solutions bi bataiye Answer hai 1444 = (38)2 but ey muje yad hai ye kese nikla muje nahi pata aap bataiye

[@Ajay_Vector]

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

[@131raghav]

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.

[@soumyadipsardar1233]

2:41 Sir yah Kaise likha aapane.. samajh Nahi Aaya

[@rishikeshkumawat9249]

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

[@kavyanshtyagi2563]

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

[@krishnaats7141]

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

[@ashwanibeohar8172]

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

[@p.msaini7515]

Sir aap ko vapas pw kab jao ge this year or jana ho aap ko plz sir m class 11 me hu or aap ke pw ke old video me syllabus sayad pura syllabus nahi h

[@hemamrutia201]

Huge request to upload tough olympiad problems daily pLz.

[@abhijnanmallick9849]

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

[@dhipin9590]

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

[@Vipyograj123]

Sir please aap mujhe ak baat bataiye ki aap alg hatke sochte kasa hai question ka bara ma please sir request❤

[@chiragwadhawan7033]

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

[@Amol-Bhāratvāsi]

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 😅

[@fantacycricketbull7658]

For what natural numbers n is the fraction (3n + 4)/5 an integer. Sir is question ko maine bohot try kiya par solve nahi ho raha hai. Aur iska koi sahi solution nahi hai

[@aakarshmanishdubey5087]

General solution: {2+10n,7+10n : n belongs to whole numbers } ... Or {2,7,12,17,22,27...........}

[@kashyaptandel5212]

me asf: “x = ab” 🗿

[@Puzzlarium1]

Same 😂

[@MDDilshan-oq5or]

Same😂

[@Math-625]

Nice question sir ❤🎉🎉

[@harshitmathpal4015]

thank you sir for this enormous question

[@orangesite7625]

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

[@avinash6740]

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]

what do we have to find? a and b?

[@avinash6740]

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

[@narendradubey8015]

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

[@rajpal2453]

Gajab ka question

[@ytmircalnoob9924]

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😸😸

[@lokeshhari]

x=6 , a=2 b=3

[@lokeshhari]

x=6, a=2, b=3

[@tannusharma3966]

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

[@sanjeevdhurwey4704]

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]

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