How many four digit numbers for which the first two and the last two digits are the same are perfect squares?

Let number be 1000x+100x+10y+y It would be 11(100x+y). So right now number is a multiple of 11, So to be a perfect square it should be a multiple of 11^2

So, 11^2 * 2^6 =7744.

(When exponents of the factors of the number are even then it is a perfect square)

Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares? CAT 2007

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Consider four-digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

Answer

Since first two and last two digits are equal, let the four-digit number be XXYY

This number can be expressed as:

1000X + 100X + 10Y + Y = 1100X + 11Y = 11(100X + Y) = k2 (perfect square)

In order for this to be true, 100X + Y must be the product of 11 and a perfect square, and looks like X0Y.

11 x 16 = 176; 11 x 25 = 275; 11 x 36 = 396; 11 x 49 = 593; 11 x 64 = 704; 11 x 81 = 891

The only one that fits is 704. This means there is only one four-digit number that works, and it is 7744.

The correct option is A.

Answer:- 1

Explanation:-

Solution:

Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11*(11a+b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . . Now, let the required number be aabb. Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4.
Hence, 7744 is a perfect square.

you have given the wrong form for 4 digit no.
the correct form for 4 digit no. whose first 2 digits are same and last 2 digits are same should be 11*(100a+b)

It's easy to solve the problem with brute force, but here is a handmade solution. So, we are trying to find integers $a,b,x$ so that $1100a+11b=x^2$, $x>0$, $1\leq a\leq9$ and $0\leq b\leq9$. (In fact, we must have $30<x<100$.)

Considering the equation modulo 11, we get $x^2\equiv0\pmod{11}$. It is easy to check (there are only 11 cases, or one could use the fact that 11 is square free) that this implies $x\equiv0\pmod{11}$. Therefore $x=11y$ for some $y\geq1$. Plugging this into the equation gives $100a+b=11y^2$. Since we must have $x<100$, we must have $y\leq9$. Now there are only nine cases to check, and it should be easy enough to find all solutions by hand.

Even this amount of brute force can be avoided. From our new equation $100a+b=11y^2$ we get $b\equiv 11y^2\pmod{100}$. Therefore $b\equiv y^2\pmod{10}$ and the only squares modulo 10 are easy to find, so $b\in\{0,1,4,5,6,9\}$. We also get $b\equiv -y^2\pmod{4}$, and the only squares modulo 10 are 0 and 1, so $b\in\{0,3,4,7,8\}$. Combining these two conditions gives $b\in\{0,4\}$. If $b=0$, then from the condition $1100a=x^2$ we get that $x=10z$ for some integer $z$ which must satisfy $z^2=11$. There is no such integer $z$, so necessarily $b=4$.

Since $b=4$, $y$ must be even. Therefore $y=2w$ and we get $25a+1=w^2$ and for an integer $w$ with $0<w<5$. Since $w^2\equiv1\pmod5$, we have $w=1$ or $w=4$. If $w=1$, then $x=22$, which is too small, so $w=4$ and $x=88$ is the only option. And a little calculation shows that $x=88$ indeed gives a solution: $88^2=7744$.

(There are many ways to go about this problem. For example, I never used the fact that $100a+b=11y^2$ implies $a+b\equiv0\pmod{11}$ and therefore $a+b=11$.)