Awesome Olympiad Problem For Maths Genius 😍| Aman Malik Sir
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- čas přidán 28. 05. 2023
- 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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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
I done the same
Initial idea was main part of the problem then we may continue in infinite ways
Can anyone explain this further more clearly 😅
After 100a+b=11z
@@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
Sir apka online paid batch kha milega class 12 ka?????
me before playing the video:
aabb=x²
a²b²=x²
x=ab!! 😂😂😂
Still you did it wrong. This should be -
x²= (ab)²
*x = ±ab* → Ans.
@@dishaa_rawatcorrect
But root(1) isnt -1
Sir please make an advanced illustration series of every chapter
Yes sir please make this for advance
Yes sir
Yes sir please
Yes sir
Yes sir please
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.
Why a can't be zero...
The number than will reduced to bb....
@@19-biswarooptalukdar99 b cant be 11 as in a number the face value of a digit cant exceed 9
@@Sanvi565 thank you....understood
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.
Exactly I'm also in 10th and solved it myself.. Btw are you NTSE aspirant? Any coaching?
@@sirak_s_nt PW Vidyapeeth student.
Maths is love
Trying for PRMO but not getting enough resources
Yes I am also the fellow, the last ones. Now unfortunately the exam is scrapped from 2022
@@vedants.vispute77 now u r in 11th? Which stream?
@@sirak_s_nt I will give jee adv this june
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.
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
x is four digit therefore 31
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
@@mathskafunda4383mod 4 matlab??
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.
Thanks sir, for solving problems for us😊😊😊
By just mere looking... Me screaming out of my lungs 88² =7744.
x =88.
Perks of ssc preparation ❤😂
Same with me, solved it in head in three minutes
But you have to prove in Olympiad
who
@@bhaskarkhandewal3257 .
@@The.Sigma. .
Behtareen Sir,
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"
Beautiful✨
I solved it myself by little bit another method.
Put a=5 and b=4. ???
Also satisfy this equation...😅
@@abhinavtiwari5585 it's a number not multiplied
You are a real hero of mathematics
Great sir
Best teacher of maths ever
aabb = x^2
x is divisible by 11
range of x is 32 - 99
square numers 33, 44, 55 etc
answer is 88
1 st one .... more such problems sir thank you so much
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
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
Wow sir .
Maja aa gaya
What an explanation!😊 Bahut interesting question h!! Aap book publish kro na apne collection of unique concept ko lekr!!!
@@musaifshaikh07yeah bro 😊 thanks!
@@musaifshaikh07 😊🙏. ,,radhe radhe🙏
Amazing 😍
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
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
Nice question sir ❤🎉🎉
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
Genius bro
You are great sir ❤❤
Sir please continue this series 🙏🙏🙏🙏
Interesting question
great thinking sir g 🙏👌🙏🙏👌🙏
thank you sir for this enormous question
Sir minor correction; x=88,-88
cannot be -88 as square of -88 gives 7744 only so basically square root of 7744 is modulus of 88 which only provides 88
bahtreen
me asf: “x = ab” 🗿
Same 😂
Same😂
Amazing question 😮😮
Sir Ji thank u very much
Sir plz bring more videos like that
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.
Gajab ka question
Thanks sir
thanks sir
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.
I like it, sir and i like you also. Thanks
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 ❤❤❤❤
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.
aabb=2233=x²
X=√(2233)≠perfect square but sir said that x is perfect square 👍
Excellent method
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
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
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
@@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
This is absolute art
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
Sir, I can solve this in different way.
Thanks sir.🎉
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
Right i also found it
Sir please make an advanced illustration series on every chapter
Sir please continue this Olympiad series
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
Second line samajh nhi aaya, please explain it
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
कतही जहर solution #bhannatmaths
sir please app bata dezeye ki ma per subject kitne question kru ek din mai for jee
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,
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
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)
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 " ,
142857
Big fan sir💟
Sir u r legend
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
If a, b, c are sides of a triangle and s be it's semi-perimeter, then prove the following 1
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.
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
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
Yes sir this is real math jisme koi faltu ka formula nhi yad krna bas apna logic use krna h
Great explanation ❤
Sir please aap mujhe ak baat bataiye ki aap alg hatke sochte kasa hai question ka bara ma please sir request❤
Sir please make adv illustration series like Physics Galaxy
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.😁😁
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
Sir, Please explain why 100a+b should be divisible by 11.
Wah
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
Hmm these type of question are very common from my daily ioqm practise question that I try from number theory
🙏🙏🙏🙏🙏🙏🙏🙏🙏👍👍👍👍👍👍👍sir you are a real hero of maths
Huge request to upload tough olympiad problems daily pLz.
7:20
9a + 1 at a = 0 is 1 which is also perft sqr
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?
That how I too did
I have solved one question similar for aba = x square and I got the answer 11 which was right and now also I got the right answer 88.
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 😅
Big fan sir ❤
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
💕LOVE FROM HINDUSTAN 💕
Yes, very good question 😊😊😊
Sir pw ke class 11 trigonometric function ka lect.11 samaj me nahi aya maximum and minimum value us me type 3 . Nahi aya h only
Easy Solution:
11(100a+b) => 100a+b = 11* x^2 put x =8 to get a= 7 and b= 4
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.
Sir apka online paid batch kha milega class 12 ka.....????
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,
yes, but 'a' can't be 0, otherwise "aabb" will not remain 4 digit no.
@@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.
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
After solving it in my try i am extremely happy
❤❤❤
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
this is how i did it
I got that by hit and trial...
Sir 1 bhi perfect square tho aap sirf 8² kyu liye
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.
❤ super ❤
Wow
(a-1)aa(b+1)bb=x squre then find x & that number (a-1)aa(b+1)bb.