AMC 10 Step by Step
Printable problems

Clocks, Calendars & Time

Clock angles, calendars, day-of-week, time-rate arithmetic

What you need to know

  • How fast the hands move. The minute hand goes all the way around (360∘360^\circ) in 6060 minutes: 6∘6^\circ per minute. The hour hand goes one number (30∘30^\circ) in 6060 minutes: 0.5∘0.5^\circ per minute.
  • Where the hands are. Measure clockwise from the 1212. At hh hours and mm minutes, the hour hand is at 30h+0.5m30h+0.5m degrees and the minute hand is at 6m6m degrees. Example: at 4:20 the hour hand is at 120+10=130∘120+10=130^\circ and the minute hand at 120∘120^\circ, so they are 10∘10^\circ apart.
  • The angle between the hands is ∣30h−5.5m∣|30h-5.5m| degrees. If that is more than 180∘180^\circ, subtract it from 360∘360^\circ to get the smaller angle. Example: at 10:10, ∣300−55∣=245|300-55|=245, so the angle is 360−245=115∘360-245=115^\circ.
  • Weekdays repeat every 77 days. So to move forward nn days, move forward by the remainder of nn divided by 77 (this is "nn mod 77"). Example: if March 1 is a Monday, then May 1 is 31+30=6131+30=61 days later. Since 61=8⋅7+561=8\cdot7+5, May 1 is 55 days after Monday, a Saturday.
  • Year to year. A common year has 365=52⋅7+1365=52\cdot7+1 days. So a date moves one weekday later each year, or two if a February 29 comes in between. Example: January 1 is a Friday in 2027, a Saturday in 2028, and a Monday in 2029, because 2028 has a February 29.
  • Leap years are the years divisible by 44, except century years, which must be divisible by 400400. Example: 20242024 and 20002000 are leap years; 19001900 and 21002100 are not.

How AMC 10 tests it

  • Clock-angle questions (positions 5–12): the angle at a given time, or the first time after hh:00 when the hands make a given angle.
  • Day-of-week questions across several years (positions 10–20): count the days mod 77. Getting the leap days right is the whole point.
  • Working out which year was a leap year: two dates on the same weekday are a multiple of 77 days apart, and that tells you whether a 366366-day year lies between them.

Standard approaches

  1. For the hands, write both angles in terms of mm, the minutes past the hour. Set their difference equal to ±θ\pm\theta or ±(360∘−θ)\pm(360^\circ-\theta), where θ\theta is the target angle, and keep only answers with 0≤m<600\le m<60. Example: between 1:00 and 2:00 the hands are perpendicular twice: at m=24011m=\frac{240}{11} (difference −90∘-90^\circ) and at m=60011m=\frac{600}{11} (difference −270∘-270^\circ).
  2. For weekdays, count the days between the two dates exactly, find the remainder mod 77, and step forward or backward in the right direction.
  3. Decide which years are leap years before you count 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) 3030 (B) 3281132\frac{8}{11} (C) 32101132\frac{10}{11} (D) 3333 (E) 3636

Let mm be the number of minutes past 7:00.

Place the hands. The hour hand starts at 7⋅30=210∘7\cdot30=210^\circ and moves 0.5∘0.5^\circ per minute, so it is at 210+0.5m210+0.5m degrees. The minute hand is at 6m6m degrees. The hour hand is ahead by 210+0.5m−6m=210−5.5m210+0.5m-6m=210-5.5m degrees.

Set up "perpendicular." The hands are perpendicular when this gap is 90∘90^\circ or −90∘-90^\circ (minute hand behind or ahead).

  • 210−5.5m=90210-5.5m=90 gives 5.5m=1205.5m=120, so m=24011m=\dfrac{240}{11}, about 21.821.8 minutes.
  • 210−5.5m=−90210-5.5m=-90 gives 5.5m=3005.5m=300, so m=60011m=\dfrac{600}{11}, about 54.554.5 minutes.

Both are between 00 and 6060, so both happen before 8:00.

Subtract. 60011−24011=36011=32811\dfrac{600}{11}-\dfrac{240}{11}=\dfrac{360}{11}=32\dfrac{8}{11} minutes. The answer is (B) 32811\boxed{\textbf{(B)}\ 32\tfrac{8}{11}}.

Pitfalls

  • Putting the hour hand right on the hour mark. At 3:30 it is halfway between 33 and 44, so the angle is ∣90−165∣=75∘|90-165|=75^\circ, not 90∘90^\circ.
  • Giving an angle bigger than 180∘180^\circ when the smaller angle is wanted. At 10:10 the answer is 115∘115^\circ, not 245∘245^\circ.
  • Counting "nn years later" as nn weekdays later without adding one more for each February 29 in between.
  • Off-by-one errors when counting days. The 17th is 77 days after the 10th (same weekday), not 88.

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)

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.