2002 AMC 10B Problem 8

A $31$-day month is four full weeks plus three extra days, so exactly three weekdays occur five times: the weekdays of the $1$st, $2$nd and $3$rd of the month.

July has five Mondays, so Monday is one of those first three days. That gives three cases for July $1$: Monday, Sunday, or Saturday.

August $1$ comes $31$ days after July $1$, and $31 \equiv 3 \pmod 7$, so August $1$ is three weekdays later than July $1$:

• July 1 = Monday: August 1 = Thursday; five Thursdays, Fridays, Saturdays.
• July 1 = Sunday: August 1 = Wednesday; five Wednesdays, Thursdays, Fridays.
• July 1 = Saturday: August 1 = Tuesday; five Tuesdays, Wednesdays, Thursdays.

The only weekday on every list is Thursday.

The answer is $\boxed{\textbf{(D)}\ \text{Thursday}}$.

Calendar problems reduce to arithmetic mod $7$: a month's length mod $7$ tells you how far the weekday of the $1$st shifts, and the remainder tells you which weekdays get the extra occurrence. "Must occur" means every case has to agree, so list all cases and intersect. The middle case (July 1 = Sunday) is the one people forget.

• Assuming July 1 must be a Monday, which makes Monday, Tuesday and Wednesday all appear five times in August and leads to a wrong pick.
• Shifting by $30$ days ($2$ weekdays) or by $28$ days (no shift) instead of $31$ days, which moves every list by one day.