Clocks, Calendars & Time
Clock angles, calendars, day-of-week, time-rate arithmetic
Clocks, Calendars & TimeAMC 10 Step by Step
Name: Date: October 28, 2026
Logic, Statistics & Misc · 12 problems have this as their main idea
12
primary-topic problems (0.9% of all)
5
more as a secondary topic
Where it appears
8
P1-10
1
P11-15
1
P16-20
2
P21-25
What you need to know
- How fast the hands move. The minute hand goes all the way around () in minutes: per minute. The hour hand goes one number () in minutes: per minute.
- Where the hands are. Measure clockwise from the . At hours and minutes, the hour hand is at degrees and the minute hand is at degrees. Example: at 4:20 the hour hand is at and the minute hand at , so they are apart.
- The angle between the hands is degrees. If that is more than , subtract it from to get the smaller angle. Example: at 10:10, , so the angle is .
- Weekdays repeat every days. So to move forward days, move forward by the remainder of divided by (this is " mod "). Example: if March 1 is a Monday, then May 1 is days later. Since , May 1 is days after Monday, a Saturday.
- Year to year. A common year has 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 , except century years, which must be divisible by . Example: and are leap years; and are not.
How AMC 10 tests it
- Clock-angle questions (positions 5–12): the angle at a given time, or the first time after :00 when the hands make a given angle.
- Day-of-week questions across several years (positions 10–20): count the days mod . 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 days apart, and that tells you whether a -day year lies between them.
Standard approaches
- For the hands, write both angles in terms of , the minutes past the hour. Set their difference equal to or , where is the target angle, and keep only answers with . Example: between 1:00 and 2:00 the hands are perpendicular twice: at (difference ) and at (difference ).
- For weekdays, count the days between the two dates exactly, find the remainder mod , and step forward or backward in the right direction.
- 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) (B) (C) (D) (E)
Let be the number of minutes past 7:00.
Place the hands. The hour hand starts at and moves per minute, so it is at degrees. The minute hand is at degrees. The hour hand is ahead by degrees.
Set up "perpendicular." The hands are perpendicular when this gap is or (minute hand behind or ahead).
- gives , so , about minutes.
- gives , so , about minutes.
Both are between and , so both happen before 8:00.
Subtract. minutes. The answer is .
Pitfalls
- Putting the hour hand right on the hour mark. At 3:30 it is halfway between and , so the angle is , not .
- Giving an angle bigger than when the smaller angle is wanted. At 10:10 the answer is , not .
- Counting " years later" as weekdays later without adding one more for each February 29 in between.
- Off-by-one errors when counting days. The 17th is days after the 10th (same weekday), not .
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, easiest first
Open any of them to try it here; the worked solution is on the problem's own page.