Awesome Olympiad Problem For Maths Genius 😍| Aman Malik Sir

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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[abdulmujib]

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

[prakharsingh]

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

[tnsr-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.

[physicsolympiad-dt]

Sir please make an advanced illustration series of every chapter

[ishanarya]

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.

[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

[deadinlavapool]

x is four digit therefore 31<x<100
11(100a+b) insists that 11 is a factor of x^2 and therefore of x
So to get the answer one can just check the squares of 33, 44, 55, 66, 77, 88, 99 to get the answer

[p_bivan]

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"

[mathsbyiitians]

Gajab approach 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.

[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.

[aniruddhxiek]

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

[vibeduck]

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

[sparshsharma]

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

[iMvJ]

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

[prabhagupta]

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

[Aaravsrivastava]

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.

[mathskafunda]

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<X<100, X=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

[alscents]

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

[zen]

Another way sir (thoda lamba hein)
Assume number to be ab
Let a be variable and B be
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 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,