2. 100 We know that $12^{2}=144$ ends with two 4s, and $38^{2}=1444$ ends with three 4s. For a natural number not ending in zero, what is the maximum number of identical digits at the end of its square?
[Solution] We know that $n^{2} \equiv 0,4,9,6,5$ or $1(\bmod 10)$, and $(10 a+b)^{2} \equiv$ $20 a b+b^{2}(\bmod 100)$, and $20 a b \equiv 0,20,40,60$, or $80(\bmod 100)$.
(1) If $b=5$, then $b^{2}=25, 20 a b \equiv 0(\bmod 100)$, so there is no natural number whose square ends in 55;
(2) If $b=3$ or 7, then $20 a b+b^{2} \equiv 9,29,49,69$ or $89(\bmod 100)$, so there is no natural number whose square ends in 99.
(3) If $b=4$ or 6, then $20 a b+b^{2} \equiv 16,36,56,76$ or $96(\bmod 100)$, so there is no natural number whose square ends in 66.
(4) If $b=1$ or 9, then $20 a b+b^{2} \equiv 1,21,41,61$ or 81 (mod100), so there is no natural number whose square ends in 11.

In summary, the repeated last digits of the square of a natural number with a non-zero unit digit can only be 4.
Assume $x^{2}$ has at least 4 repeated 4s at the end, then by
$$
10000 \equiv 0 \quad (\bmod 16)
$$

we know $x^{2} \equiv 4\left(10^{3}+10^{2}+10+1\right) \equiv 12(\bmod 16)$.
But $\quad n^{2} \equiv 0,1,4$ or $9(\bmod 16)$,
so there is no natural number with a non-zero unit digit whose square has more than 3 repeated 4s at the end.

Thus, $38^{2}=1444$ is the natural number with a non-zero unit digit whose square has the most repeated last digits.