Some solutions resorted to the use of calculators but the problem did say that this was not necessary. Correct solutions were obtained from Andrei of School 205 Bucharest and Mary of Birchwood High School. The idea is to use the properties of numbers to eliminate non squares so for example:

No square number ends in a 7 so 846 348 767 is not a perfect square

Similalry no perfect square ends in 15, so 987 465 315 is not a perfect square.

Andrei Lazanu explains as follows:

For a perfect square to end in 5, the number from which it is obtained must also end in 5. If you raise a number ending in 5 to the second power:

Therefore the number must end in 25.

Andrei continued as follows:

993 762 576 – I do not observe a contradiction.

1 175 624 800 – this number is multiple of 102. For it to be a perfect square 11756258 must also be a perfect square, which is not possible, because the square numbers don't end in 8.

2 345 832 406 – this number can't be a square number. There isn't a square number that ends in 06.

The idea to find the last two digits of a perfect square is to raise a number, let it be a three digit number abc to the second power:

(100a + 10b + c) 2 = 104a2 + 100b2 + c2 + 2 * 103 ab + 200ac + 20bc = 100(100a2 + b2 + 20ab + 2ac) + 20bc + c2

So, for the last 2 digits of the square number, only the last two digits of the initial number are significant.

I also observe that the last two digits could be obtained by considering only b = 0, 1, 2, 3 and 4 because:

20 (b + 5)c + c2 = 100c + 20(bc + c2)

The possibilities are:

The only 22 possibilities are therefore 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89.

I observe that 76 is on the list of possibilities.

So the only remaining number is 993 762 576, which is 315242.