80. Given that $\mathrm{n}$ is a positive integer, and the tens digit of $n^{2}$ is 7, find $n-100\left[\frac{n}{100}\right]$. (Note: $[x]$ denotes the integer part of $\mathrm{x}$)
Solution: $n-100\left[\frac{n}{100}\right]$ is to find the tens and units digit of $\mathrm{n}$. The tens digit of $n^{2}$ depends on the tens and units digit of $\mathrm{n}$. Let the tens digit of $\mathrm{n}$ be $\mathrm{a}$, and the units digit be $\mathrm{b}, (10 a+b)^{2}=100 a^{2}+20 a b+b^{2}$. If the tens digit of $b^{2}$ is 0 or even, then the tens digit of $n^{2}$ is even, so the units digit $\mathrm{b}$ of $\mathrm{n}$ can only be 4 or 6.

When $b=4$, $(10 a+4)^{2}=100 a^{2}+80 a+16=100 a^{2}+10(8 a+1)+6$, only when the units digit of $8 a+1$ is 7, the tens digit of $n^{2}$ is 7, so $\mathrm{a}=2$ or 7;

When $b=6$, $(10 a+6)^{2}=100 a^{2}+120 a+36=100\left(a^{2}+a\right)+10(2 a+3)+6$, only when the units digit of $2 a+3$ is 7, the tens digit of $n^{2}$ is 7, so $\mathrm{a}=2$ or 7.
In summary, $n-100\left[\frac{n}{100}\right]$ has four possible results, which are $24, 26, 74, 76$.