Question

A Question on Number theory.
I am clueless

Answer 1

If aabb is a 4 digit number , and is a perfect square , Find the number

Answer 2

@ikram002p

Answer 3

it's 88^2 , i'm trying but i've not found any way to prove it.

Answer 4

\[\large aabb = 1000a + 100a + 10b + b = x^2\]

Answer 5

\[\large 1100a + 11b = x^2\]

Answer 6

\[\large 11(100a + b) = x^2\]

Answer 7

since 11 is a prime, there must exist atleast one more 11 on the left hand side for it to be a perfect square :

100a + b = 11k

Answer 8

100a + b = 11k

99a + a + b = 11k

a+b = 11(k- 9a)

a+b = 11m

Answer 9

Sorry i was afk , wait let me read

Answer 10

next, notice that a and b are digits : 0<a<=9, 0<=b<=9

Answer 11

I didn't get this actually
since 11 is a prime, there must exist atleast one more 11 on the left hand side for it to be a perfect square

Answer 12

thats a good question to ask :)

Answer 13

2^2 * 3^2 = 36


see that 2 occurs two times and 3 occurs two times on left hand side

Answer 14

yes

Answer 15

For a perfect square, the prime number occurs in even powers

Answer 16

So, if 11 is a factor of x^2, then 11^2 mist also be a factor cos, x^2 is a perfect square, 11 cannot occur in odd powers : 11^1

Answer 17

*In the prime factorization of a perfect square, the prime number occurs in even powers

Answer 18

Yes yes i get it , let me read after that

Answer 19

okie good :)

Answer 20

You split that 100a into 99a +a was awesome

Answer 21

yes 0<a<=9, 0<=b<=9

Answer 22

yes :) thats because a digit can only between 0 and 9 (inclusive)

Answer 23

yess

Answer 24

a+b = 11m

solve in light of the previous info about range of a, b values

Answer 25

a cannot be 0 :)

Answer 26

again, a cannot be 0 cos its the left most digit of a four digit number

Answer 27

a+b = 11m

can m be 2 ?
can m be 3 ?

Answer 28

nope,

a+b = 11 is the only possible solution (why ?)

Answer 29

\[\large 11(100a + b) = x^2\]

\[\large 11(100a + 11-a) = x^2\]

\[\large 11(99a + 11) = x^2\]

\[\large 11^2(9a+1) = x^2\]

Answer 30

so the problem simplifies to this :
find \(a\) such that \(9a+1\) is a perfect square given that \(0\lt a\le 9\)

Answer 31

9 cases to work with, numbers are small... you can work it :)

Answer 32

Well let me work and tell you

Answer 33

a=7 ?

Answer 34

9(7) + 1

64

yep !

Answer 35

so the number is 77xx

Answer 36

b=4 !

Answer 37

i mean not 4!
4

Answer 38

Excellent ! i bet you're excited as i am :) actually there is another shortcut for this problem if you know the digital roots and the rules for last digit of perfect square

Answer 39

but that method hides many details, so don't bother about it :)

Answer 40

I like this concept Number theory , it stimulates logical thinking , Yes i am very much excited

Answer 41

Number theory = Queen of mathematics
you know who said it :)

Answer 42

---->RAMANUJAN<----

Answer 43

<3

Answer 44

Ultimately , Maths is about numbers

Answer 45

I'm really not very good at these kinds of things and always watch these but never feel like I get it.

My attempt was to establish an upper and lower bound by seeing that the square root of 9988 and 1100 would give me a range of about 33 to 100 as being the numbers to square.

Then assuming we could have 1-9 in the higher digits and 0-9 in the second digit without repeating this meant there were 81 different combinations I could make a four digit number like this lol.

Answer 46

You could try watching some youtube videos on Number theory , those are really good

Answer 47

I watch three videos a day

Answer 48

xxyy

81 combinations

yeah not too much off from 10 cases lol

im not sure if we can solve these kind of problems without cases...

Answer 49

@No.name whats your favorite problem in number theory so far

Answer 50

I really like your reasoning @ganeshie8 for why there must be another 11. That's really pretty nice!

Answer 51

there was one which i did not understand but watched it till the end like a idiot , because i was interested

Answer 52

@No.name if you ever find a good number theory video that you like, feel free to send me a link I would be very interested!

Answer 53

Yes sure :)

Answer 54

I will compile them and send to you together

Answer 55

Awesome =)