2. Given a divisor (not 0) and a dividend, we can always find a quotient and a remainder that satisfy
dividend $=$ divisor $\times$ quotient + remainder
where, $0 \leq$ remainder < divisor. This is division with a remainder. When the remainder is 0, we also say the divisor divides the dividend evenly, or the divisor is a factor of the dividend (the dividend is a multiple of the divisor).
Please write down all natural numbers greater than 1 and not exceeding 88 that can be divided evenly by 6: $\qquad$
Solution: $6,12,18,24,30,36,42,48,54,60,66,72,78,84$. Numbers that can be divided by 6 must be multiples of 6, and they are required to be no more than 88. Therefore, we have $6,12,18,24,30,36,42,48,54,60,66,72,78,84$.