Perfect square

[sommukh]

Consider 4 digit numbers with equal first two digits and equal last two digits (like aabb) ... how many such numbers are perfect squares?

1. 0
2. 1
3. 2
4. 3

[clock60]

i found only one number 88^2=7744
but i don`t know how to prove or exploit any approach....

[CyberSpy]

A square number can only end with digits 00,1,4,6,9, or 25 in base 10, as follows:

1. If the last digit of a number is 0, its square ends in 00 and the preceding digits must also form a square.
2. If the last digit of a number is 1 or 9, its square ends in 1 and the number formed by its preceding digits must be divisible by four.
3. If the last digit of a number is 2 or 8, its square ends in 4 and the preceding digit must be even.
4. If the last digit of a number is 3 or 7, its square ends in 9 and the number formed by its preceding digits must be divisible by four.
5. If the last digit of a number is 4 or 6, its square ends in 6 and the preceding digit must be odd.
6. If the last digit of a number is 5, its square ends in 25 and the preceding digits must be 0, 2, 06, or 56.

Now analysing all the above possibilities.

1 -> can be eliminated because the preding nos should be of the form (11, 22..99) none of these are perfect squares.

2 -> From this last two digits are 11.. Numbers preceding 1 must be divisible by 4.. none of the nos 111, 221,331... are divisible by 4. Eliminate this possibility.

3 -> the nos should be of the form 1144, 2244 etc analysing these by prime factorization we find only 7744 is prime.

4-> nos should be of the form 1199, 2299 and so on.. but none of the nos 119, 229... are divisible by 4. Eliminate this possibility.

5 ->when square no ends in 6 preceding digit should be odd. But in our problem the last two digits are equal.. so this possibility can be eliminated.

6 -> Number ending in 25 is out of scope since last two digits are not equal.


So we are left with only 7744. Answer is 1.