2010 AMC 10B Problem 5

Since $31 = 4 \cdot 7 + 3$, every weekday occurs four times, and the weekdays of days $29, 30, 31$, which are the same as days $1, 2, 3$, occur a fifth time. So the first day and the two days after it appear five times; the other four weekdays appear four times.

Monday and Wednesday have equal counts exactly when the three-day block $\{d, d+1, d+2\}$ (with $d$ the starting weekday) contains both of them or neither.

• Both: the block must be Mon, Tue, Wed, so $d =$ Monday.
• Neither: the block must avoid Monday and Wednesday. Checking each start, Thu-Fri-Sat and Fri-Sat-Sun work; Sat-Sun-Mon, Sun-Mon-Tue, Tue-Wed-Thu, and Wed-Thu-Fri each contain one of the two days.

That gives $3$ possible starting days: Monday, Thursday, Friday.

The answer is $\boxed{\textbf{(B)}\ 3}$.

A month's weekday counts depend only on the remainder of its length mod $7$: the leftover days are the ones that get an extra occurrence. Turning "same number of Mondays and Wednesdays" into a statement about a three-day window makes the check a quick list of seven cases.

• Only finding the start day Monday (both days occur five times) and forgetting the starts where neither Monday nor Wednesday gets a fifth occurrence.
• Counting Saturday as a valid start: the block Sat-Sun-Mon gives Monday five occurrences but Wednesday only four.