How many perfect squares(0,3,6,7,8)?

All square numbers end with either $0, 1, 4, 5, 6$ or $9$

Therefore, the only number that is a candidate is $36$ , which is the square of $6$

$\surd 36 = 6$ .

There's no other square numbers.

Therefore, the answer is $\fbox 1$ .

Who will write a correct solution first? Some help:

• The last digit of a perfect square can't be $2,3,7,8$
• If $n^2$ is divisible by $a$ , where $a$ isn't a square number, then $n^2$ is divisible by $a^2$
• If $\overline{abc\cdots}$ is divisible by 3 or 9, then $a+b+c+\cdots$ is divisible by 3 or 9 too
• If $\overline{abc\cdots}$ is divisible by $2^n$ , then the last $n$ digit is divisible by $2^n$

Can somebody solve that without brute-force? I can :)