## Problem 1.

a) How many multiples of 6 are less than or equal to 610?

b) Specify which of these are multiples of 96 and determine their number.
Solution. a) A multiple of 6 is of the form $6 \cdot t$, where $t$ is a natural number. From $0 \leq 6 \cdot t \leq 610$ it follows that $0 \leq t \leq 101$. In total, there are 102 such numbers.

b) A multiple of 96 is of the form $96 \cdot k$, where $k$ is a natural number. Since $96=6 \cdot 16$, any multiple of 96 is a multiple of 6. Therefore, we seek the values of $k$ such that $6 \cdot 0 \leq 6 \cdot 16 k \leq 6 \cdot 101$. From $0 \leq 16 k \leq 101$, it follows that $0 \leq k \leq 6$. Thus, 7 numbers are obtained, namely: $0,96,96 \cdot 2, \ldots, 96 \cdot 6$.