2010 AMC 10B Problem 5

P1-10~2 minPrint

Clocks, Calendars & Time Organized listing / direct enumeration

A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?

Show solution without answering

Key insight

31 days is 4 weeks plus 3 days, so the month's first three weekdays appear 5 times. Monday and Wednesday must both be among those three, or both not.

Solution

First figure out which weekdays appear more often than the others.

$31 = 28 + 3$ , and $28$ days is exactly $4$ weeks. In those $4$ weeks every weekday appears $4$ times.
The last $3$ days of the month (days $29, 30, 31$ ) fall on the same weekdays as days $1, 2, 3$ . Those three weekdays get a fifth appearance. Every other weekday appears only $4$ times.
So Monday and Wednesday have the same count exactly when the first three days of the month include both of them or neither of them.

Now check all seven possible first days:

First day	Days 1, 2, 3	Mondays	Wednesdays	Equal?
Sunday	Sun, Mon, Tue	5	4	no
Monday	Mon, Tue, Wed	5	5	yes
Tuesday	Tue, Wed, Thu	4	5	no
Wednesday	Wed, Thu, Fri	4	5	no
Thursday	Thu, Fri, Sat	4	4	yes
Friday	Fri, Sat, Sun	4	4	yes
Saturday	Sat, Sun, Mon	5	4	no

Three first days work: Monday, Thursday and Friday.

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

Why this works

How often each weekday shows up in a month depends only on the days left over after the full weeks. Here the $3$ leftover days get an extra appearance, so the question becomes a simple check of seven short lists. Remember that "the same number" can happen in two ways: both days get the extra appearance, or neither does.

Concise solution

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

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 $\boxed{\textbf{(B)}\ 3}$ .

Review in the AoPS books

Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)