2002 AMC 10B Problem 8

P1-10~3 minPrint

Clocks, Calendars & Time Split into exhaustive cases and handle each

Suppose July of year $N$ has five Mondays. Which of the following must occur five times in the August of year $N$ ? (Note: Both months have $31$ days.)

Show solution without answering

Key insight

A 31-day month has three weekdays that appear five times: those of the 1st, 2nd and 3rd. Try each possible start for July and see which August day always appears.

Solution

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.
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.
August $1$ is $31$ days after July $1$ . Moving $28$ days keeps the same weekday, so August $1$ is $3$ weekdays later than July $1$ .
Make a table of the three cases:

July 1	August 1	five times in August
Monday	Thursday	Thu, Fri, Sat
Sunday	Wednesday	Wed, Thu, Fri
Saturday	Tuesday	Tue, Wed, Thu

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

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Prealgebra, Chapter 14: Counting (§14.3)