19. Given that the difference between two 2-digit numbers is 58 and these last two digits of the squares of these two numbers are the same, find the smaller number.
19. Answer: 21
Let the numbers be $n$ and $m$ where $n>m$.
$$
n-m=58
$$
$n^{2}-m^{2}=(n-m)(n+m)-58(n+m)$ is a multiple of 100 , that is,
$58(n+m)=100 k$ for some integer $k$. This simplifies to
$$
29(n+m)=50 k
$$

Since $\operatorname{gcd}(29,50)=1$, we must have $n+m=50 p$ for some positive integer $p$.
Solving the above simultaneously with $n-m-58$ and bearing in mind that both $m$ and $n$ are two digit numbers, only $n=79$ and $m=21$ satisfy the question.
Thus the smaller number is 21 .