A number containing just two digits 0,6 (like:60,660,606.......)prove that the number will not be a perfect square.
Let the number(N, say) consists of digits D_{0},D_{1},D_{2},...,D_{n} where each D_{i} is from {0,6} and D_{0} must be 6 to be a (n+1) digited number. Now observe that both 0 and 6 are evenly divisible by 2.
Consider 2 cases:
1. D_{n }= 6
2. D_{n }= 0
In case 1 if the last digit of N is 6 i.e. D_{n}=6 then dividing it by 2 gives us a number (N/2) with last digit 3 and hence cannot be a perfect square since any further division of (N/2) will not yield any 2 from it i.e. N consists of a single 2 and hence it is a non-square number.
In case 2 if the last digit of N is 0 i.e. D_{n}=0 then let the least consecutive sub-sequence of 0s from D_{n} backwards be upto D_{i} where 1 <= i <= n,since D_{0} must be 6 to be a (n+1) digited number.
Therefore,from D_{0} to D_{i-1} the number consists of mixes of 0s and 6s and from D_{i} to D_{n} the number consists of all 0s. Pictorially the view is as follows:
Now separate D_{0} to D_{i-1} as a string of 0s and 6s and D_{i} to D_{n} as a string of only 0s.
In other words the number is as follows: N = D_{0}...D_{i-1} X 10^{(n-i+1)} = 6...6 X 10^{(n-i+1)}
Now by previous argument we know that D_{0}...D_{i-1} can't be a perfect square and it also consists of a factor 6=2x3. So if the other part i.e "power of 10 is a perfect square[i.e if (n-i+1) is even]" then overall the number N can't be a perfect square.
If (n-i+1) is odd then 10^{(n-i+1)} can be factored as 2^{n-i+1} X 5^{n-i+1} where we see that the part 5^{n-i+1} still can't be a perfect square( as n-i+1 is odd and no other factors of 5 are there in the whole number ).
So in that case also this required result is proved.
Easy solution: Assume that the number ends in 6. We can do this because if it ends in -60 it cannot be a perfect square (all numbers in the form must be multiples of 100) and if it ends in -600 (two or more zeros) we can cross off pairs of zeros, resulting in a number ending in 60 or 6.
If the number ends in 6, it must end in either 06 or 66. Both of these numbers are 2 modulo 4, so a perfect square cannot end with these two digits.