The correct option is (c).

A common year has 365 days and a leap year has 366. From the division of 365 by 7, we get $365=52 \times 7+1$ and from the division of 366 by 7 we get $366=52 \times 7+2$. Therefore,

$$
\begin{aligned}
\text { common year } & =52 \text { weeks }+1 \text { day } \\
\text { leap year } & =52 \text { weeks }+2 \text { days }
\end{aligned}
$$

Thus, a common or leap year has, at a minimum, 52 and, at a maximum, 53 Sundays (one Sunday for each of the 52 weeks and, perhaps, one additional Sunday for the day or the two days that complete the year).

Each of the 12 months of the year has, at a minimum, 28 days and, at a maximum, 31 days, therefore, it has, at a minimum, 4 Sundays and, at a maximum, 5 Sundays. Considering that $12 \times 4=48$, we conclude that

i) In a year with 52 Sundays, there are still $52-48=4$ Sundays left. Each of these will be in a different month, because no month can have six Sundays; thus, we have four months with 5 Sundays.

ii) Analogously, in a year with 53 Sundays, there are 5 Sundays left, which will be in different months. Therefore, in this case, we will have five months with 5 Sundays.
In a year, at most how many months have five Sundays?
(a) 3
(b) 4
(c) 5
(d) 6
(e) 7