1. Consider the set $A=\{100,101,102, \ldots, 200\}$.

a) Determine the number of elements in $A$ that are divisible by 10.

b) Determine the elements of $A$ that, when divided by 21 give a remainder of 12, and when divided by 28 give a remainder of 5.
a) A contains 11 numbers divisible by 10

b) If $n=21 k+12$ and $n=28 p+5, k, p \in \mathbb{N}$, then $4 n=84 k+48,3 n=84 p+15$ and $4 n-3 n=84(k-p)+33$, that is, $n=84 q+33,100 \leq n \leq 200$, thus $n=117$. $.4 \mathrm{p}$