Clocks, Calendars & Time

What you need to know

• The minute hand moves $6^\circ$ per minute; the hour hand moves $0.5^\circ$ per minute. At $h$ hours $m$ minutes the hour hand is at $30h + 0.5m$ degrees and the minute hand at $6m$ degrees.
• The angle between the hands is $|30h - 5.5m|$, or $360^\circ$ minus that if it exceeds $180^\circ$.
• Weekdays cycle with period $7$. A common year has $365 = 52 \cdot 7 + 1$ days, so a date advances one weekday per year, two if a February 29 lies between.
• Leap years are divisible by $4$, except century years, which need divisibility by $400$.

How AMC 10 tests it

• Clock-angle questions (positions 5–12): the angle at a given time, or when the hands first make a given angle after $h$:00.
• Day-of-week questions spanning years (positions 10–20): count days mod $7$; leap days are the whole point.
• "Which year is leap" deductions: two dates with the same weekday differ by a multiple of $7$ days, revealing whether a $366$-day year lies between.

Standard approaches

1. For hands, write both angles in terms of $m$ minutes past the hour and set the difference to $\pm$ the target.
2. For weekdays, count the days between the dates exactly, reduce mod $7$, and step in the correct direction.
3. Settle leap-year status before counting anything else.

Worked example

Between 7:00 and 8:00 the hands of a clock are perpendicular at exactly two moments. How many minutes apart are these moments?

(A) $30$ (B) $32\frac{8}{11}$ (C) $32\frac{10}{11}$ (D) $33$ (E) $36$

Solution. At $m$ minutes past 7:00 the hour hand is at $210 + 0.5m$ degrees and the minute hand at $6m$ degrees, so the minute hand trails by $210 - 5.5m$ degrees. Perpendicular means this equals $90$ or $-90$: $5.5m = 120$ gives $m = \dfrac{240}{11}$, and $5.5m = 300$ gives $m = \dfrac{600}{11}$. Their difference is $\dfrac{360}{11} = 32\dfrac{8}{11}$ minutes. The answer is $\boxed{\textbf{(B)}\ 32\tfrac{8}{11}}$.

Pitfalls

• Placing the hour hand on the hour mark: at 3:30 the angle is $75^\circ$, not $90^\circ$.
• Reporting an angle above $180^\circ$ when the smaller angle is wanted.
• Counting "$n$ years later" as $n$ weekdays without adding one per leap day in the interval.
• Off-by-one errors: the 10th to the 17th is $7$ days, not $8$.

Traps that recur

• Sliding the date by one: forgetting February 2003 has 28 days, or counting the tenth full day as the answer instead of the day the remainder falls on. (2003 AMC 10B #22)
• Counting 5 days forward from Tuesday (getting Sunday) when the target date is earlier, or ignoring leap years entirely. (2000 AMC 10 #25)
• Counting 50 leap years by forgetting that 1900 is not a leap year, or counting Feb 29, 1812 (which falls after the birthday) incorrectly. (2012 AMC 10A #12)
• Counting 120 days forward from the Monday (landing on Tuesday) instead of 119, because day 1 itself is the Monday. (2016 AMC 10B #4)