10,11

The number $N$ is a perfect square and does not end in zero. After erasing the last two digits of this number, another perfect square is obtained. Find the largest number $N$ with this property.
Let $a$ be a number such that the square of $a$ becomes the number $N$ after erasing its last two digits.

Then $N>100 a^{2}$, hence $\sqrt{N}>10 a$. By the condition, the number $\sqrt{N}$ is an integer. Therefore, $\sqrt{N} \geq 10 a+1$, which means $N \geq (10 a+1)^{2}=100 a^{2}+20 a+1$. On the other hand, $N<1700$, but the first two digits must form the number $a^{2}=16$. Thus, the largest possible value of $N$ is 1681.

## Answer

1681.00