Hello George,

A number is a perfect square if it can be written as the product of some integer with itself. For instance the number 9 is a perfect square because it can be written as the product of 3 with itself.(9 = 3*3) Well, 40...006 = 4*(10...001.5) = 2*2*(10...001.5) which implies that our original number is a perfect square if and only if the number 10...001.5 is a perfect square. Can 10...001.5 be written as the product of an integer with itself? Notice that this number is NOT an integer. Can two integers multiply to produce something which is not an integer?

Tyler

George,

Here is another simple approach to your type of problem. Just look for patterns. Call your number N. First ask yourself what the last digit of N must be must be for its square to end in 6. (For example, the square of any number ending in 3 or in 7 will end in 9: 13² = 169, 23² = 529, ... 17² = 289...) You will quickly see that N must end in 6.
Now ask what the final two digits of N must be for N² to end in ...06. If you know algebra, you can finish the problem with a simple equation (Let d be the digit of N that comes before the 6, so that the final two digits form the number 10*d+6. Now square 10*d+6 and see if for any d the last two digits can be 06.) If that argument sounds complicated to you, a more elementary method is to square the ten possibilities (6, 16, 26, ... 96), and see whether any of the ten resulting squares can end with ...06. (You will note that no matter how many digits N has, its final two digits will determine the final two digits of N².)

Chris