54. Find the largest perfect square such that after erasing its last two digits, a perfect square is obtained again. (It is assumed that one of the erased digits is not zero.)
54. Answer: $41^{2}=1681$.

Let the number $n^{2}$ satisfy the condition of the problem, then $n^{2}=$ $=100 a^{2}+b$, where $0<10 a$ and therefore, $n \geqslant 10 a+1$. This means that $b=n^{2}-100 a^{2} \geqslant 20 a+1$, from which $20 a+141$, then $n^{2}-40^{2} \geqslant 42^{2}-40^{2}>100$.

$\nabla$ This topic is continued by problem 244.