Australian Olympiad Problem (Question 8)

A three-digit number $n$ with the same congruence to $\operatorname{mod} 10^3$ when squared has one of the following sets of properties: $\begin{cases} n \equiv 1 (\operatorname{mod} 2^3) \\ n \equiv 0 (\operatorname{mod} 5^3) \end{cases} \begin{cases} n \equiv 0 (\operatorname{mod} 2^3) \\ n \equiv 1 (\operatorname{mod} 5^3) \end{cases}$ For each case, we get $625$ and $376$ respectively. Hence, the largest possible value for $n$ is $\boxed{625}$ .

There are actually three solutions , which are :

1

376

625.

625 is the highest with 625^2 = 390625. And this divided by a thousand has a remainder of 625.