10.124 $* *$ Find the largest square number, such that after removing its last two digits, you still get a square number. Assume the two removed digits are not both 0.
Let $n^{2}$ satisfy the condition. Let $n^{2}=100 a^{2}+b, 010 a$, i.e., $n \geqslant 10 a+1$. Thus, $100>b=n^{2}-100 a^{2} \geqslant 20 a+1$. Therefore, $a \leqslant 4$.

When $a=4$, only $n=41$ satisfies the condition (if $n>41$, then $n^{2}-40^{2} \geqslant 42^{2}-40^{2}>100$). Therefore, the largest square number that satisfies the condition of this problem is $41^{2}=1681$.