How to show that there is no perfect square whose decimal representation consists entirely of digits 6 and 0?
2026-03-29 04:48:08.1774759688
There does not exist a perfect square with all decimal digits 0 or 6
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Outline: Well, I can think of one, namely $0$. So let us assume there is at least one $6$.
The number would have to end in an even number of $0$'s, possibly none. Remove them. The resulting number must be divisible by $4$. But a number that ends in $06$ or $66$ cannot be divisible by $4$.