2002 AMC 10B Problem 8

1. A $31$-day month is $4$ full weeks ($28$ days) plus $3$ extra days. Each weekday shows up $4$ times in the full weeks, and the $3$ extra days, the $29$th, $30$th and $31$st, add a fifth copy of three weekdays. Those are the same weekdays as the $1$st, $2$nd and $3$rd.
2. July has five Mondays, so a Monday falls on July $1$, $2$ or $3$. That means July $1$ is a Monday, a Sunday or a Saturday.
3. August $1$ is $31$ days after July $1$. Moving $28$ days keeps the same weekday, so August $1$ is $3$ weekdays later than July $1$.
4. Make a table of the three cases:

1. "Must occur" means it has to happen in every case. Thursday is the only day in all three rows.

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

Weekday problems are about remainders when you divide by $7$. The length of a month minus the full weeks tells you how far the starting weekday moves and which days get a fifth visit. When a question says "must," list every possible case and keep only what they all share.

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