Clocks & Calendars

1How fast each hand moves

A clock face is a circle of 360 degrees. The minute hand runs all the way around that circle once every 60 minutes, so it moves 6° per minute (360 / 60). The hour hand is slower: it runs all the way around once every 12 hours, which is 720 minutes, so it moves 0.5° per minute — or 30° in a full hour (360 / 12). Those two speeds, 6 and 0.5, are the only rates this topic needs; every later formula is built from them.

The running time is 3:40. At 3:00 exactly the minute hand sits on 12 (0° from the 12-mark) and the hour hand sits on 3 (90° from the 12-mark). Forty minutes later the minute hand has moved 6 \times 40 = 240° and points at 8. The hour hand has not stayed on 3: forty minutes at 0.5° each have dragged it 20° toward 4, so it now sits at 90 + 20 = 110°. The common trap is treating the hour hand as stuck on the hour mark. At 3:40 the hour hand is not at 3.

The relative speed is just the difference: the minute hand gains on the hour hand at 6 - 0.5 = 5.5° per minute. That 5.5 is what spaces coincidences and right angles later in this topic. Until you believe the hour hand moves with the minutes, every angle you compute will be 20° off at 3:40 — and off by 0.5M at any other time. The animation on this card is that drift: the minute hand sweeps toward 8 and the hour hand leaves the 3.

Build the rates

1. Minute hand360° / 60 = 6° per minute from the 12 o'clock mark. At 3:40 that is 6 \times 40 = 240°, which is the 8.
2. Hour hand360° / 12 = 30° per hour, so 30° / 60 = 0.5° per minute. At 3:40 the hour hand sits at 30 \times 3 + 0.5 \times 40 = 110°, not on the 3.
3. Relative gainMinute hand gains on hour hand at 6 - 0.5 = 5.5° per minute — the number behind coincide and right-angle timings.

2Angle between the hands

Once both positions are measured from 12, the angle between the hands is the absolute difference of those positions. There is a one-line formula that already includes the hour-hand drift: angle = |30H - 5.5M| degrees. H is the hour as shown (0 through 11); M is minutes past that hour. The 30H term is the hour hand sitting on the hour-mark — at 3:40 that is 30 \times 3 = 90. The 5.5M term packs minute-hand travel minus the hour-hand's extra drift: 6M - 0.5M = 5.5M, so 5.5 \times 40 = 220. Subtract: |90 - 220| = 130.

Two angles sit on every face: 130 and 360 - 130 = 230. Exam answers almost always want the smaller angle on the face, so if the formula returns more than 180° you replace it with 360° minus that value. 130 is already under 180, so 130 stands. Watching the hands at 3:40 makes the same 130 visible as the wedge from the hour hand at 110° to the minute hand at 240° — |110 - 240| is 130, matching the formula. You never need to sketch the face to get the number; plug H and M straight into the absolute-value formula. The hour hand's 0.5M term is already inside the 5.5.

How it works

1. Identify H and MH is the hour as shown (0–11); M is minutes past the hour. For the running 3:40, H = 3 and M = 40.
2. Compute |30H - 5.5M|One subtraction; keep the absolute value. |30 \times 3 - 5.5 \times 40| = |90 - 220| = 130°.
3. Take the smaller angleIf the result is over 180°, replace it with 360° minus that result. 130 is already the smaller angle, so it stands; 230 is the reflex you throw away.

3How often hands coincide or form a right angle

Hands coincide when the angle between them is 0° — they point at the same mark. Start from noon, when they already overlap. The minute hand then has to gain a full 360° on the slower hour hand to overlap again. At the relative speed 5.5° per minute that takes 360 / 5.5 = 720/11 \approx 65\tfrac{5}{11} minutes, a little after 1:05. The next overlap is the same gap later, a little after 2:10, and so on. Count the meetings in twelve hours: noon, then one after 1, 2, …, 10 — that is eleven. They coincide 11 times, not 12, because the meeting that would have been 'around 11' is the same meeting as noon/midnight. The 11-to-1 window merges with that shared endpoint.

Right angles are the same idea at 90°. The minute hand must gain 90° on the hour hand, which at 5.5° per minute takes 90 / 5.5 = 180/11 minutes, and it can do that on either side of the hour hand, so you get two right angles in most hour-ish windows. The same kind of shortfall that killed the twelfth overlap kills two of those windows (near 3 and near 9), leaving 22 right angles in twelve hours, not 24. 'How many times between…' questions are counting these meetings, not recomputing a face angle. The relative speed 5.5°/min is what spaces the meetings; the counts themselves are the facts to keep.

Order of attack

1. Name the eventCoincide means angle 0°; right angle means 90° (either side of the hour hand).
2. Use the standard count12 hours → 11 coincides, 22 right angles. Do not invent 12 and 24 — those are the naive once-per-hour guesses the options plant.
3. Space with relative speedMinutes between consecutive coincides: 360 / 5.5 = 720/11. The same 5.5 from the 3:40 card is doing the spacing.

4Odd days — the weekday remainder

Weekdays repeat every 7 days. Tuesday plus 7 days is Tuesday again; Tuesday plus 8 days is Wednesday. So only the remainder after dividing a span of days by 7 matters. That remainder is the number of odd days — the leftover 0, 1, 2, …, 6 that actually shift the weekday from a known anchor. A span of 14 days has 0 odd days and does not move the weekday at all; a span of 15 days has 1 odd day and moves it by one.

An ordinary year has 365 days. 365 divided by 7 is 52 weeks with 1 day left over, so 1 odd day. A leap year has 366 days: 52 weeks plus 2 leftover days, so 2 odd days. That is the whole calendar half of this topic in one sentence: convert the span to a leftover, then walk the weekday that many steps. Counting full date arithmetic when only the remainder is needed is wasted work and a common source of off-by-one errors. The running calendar uses this immediately: 2024 is a leap year, so the span from 1 January 2024 to 1 January 2025 carries 2 odd days.

Figure. Weekdays repeat every 7 days, so only the leftover after dividing the span by 7 shifts the weekday. An ordinary year (365) leaves 1; a leap year (366) leaves 2; a week (7) leaves 0.

How it works

1. Fix an anchor weekdayYou need one known date → day pair; everything else is a shift from there. Here the anchor is 1 January 2024 = Monday.
2. Count odd days in the spanConvert years, months or day counts to a total, then reduce mod 7. A leap year is 366 ≡ 2 (mod 7); an ordinary year is 365 ≡ 1.
3. Add mod 7Advance the anchor by that many weekdays (0 stays put, 1 is the next day, …). Monday + 2 = Wednesday.

5Which years are leap years

A year is a leap year if it is divisible by 4 — except century years, which must be divisible by 400. So 2000 is a leap year (2000 / 400 = 5) and 1900 is not (1900 / 400 is not a whole number, even though 1900 / 4 is). That single exception changes February's length — 29 days instead of 28 — and therefore the odd-day count for the year: leap years contribute 2 odd days, ordinary years contribute 1. February 29 is the extra leftover day.

The running calendar needs this rule before the weekday can be trusted. 2024 is not a century, and 2024 / 4 = 506 exactly, so 2024 is a leap year and carries 2 odd days. Getting the century rule wrong on a year that does end in 00 flips the odd-day contribution from 2 to 1 (or the reverse) and shifts every later weekday in the problem by one. 1900 and 2100 are the classic traps: both look leap if you only test ÷4, and neither is. A span that starts on 1 March of a leap year has already left February 29 behind, so that remaining year behaves like an ordinary year — a trap the 1 January running example does not hit.

Figure. Leap if divisible by 4, except century years, which must be divisible by 400. 2016 is leap; 2017 is not. 1900 is a century and fails ÷400; 2000 is a century and passes.

Classify the year

1. Century?If the year ends in 00, demand divisibility by 400 — not merely by 4. 2000 yes; 1900 no.
2. OtherwiseLeap iff divisible by 4. 2024 / 4 = 506, so 2024 is leap; 2023 is not.
3. Apply to odd daysLeap → 2 odd days for the year; ordinary → 1. February 29 sits inside that leap count.

6Day of the week from odd days

Given an anchor date and its weekday, advance by the odd days in the intervening span. For a full year from 1 January to the next 1 January, that span is just the year's own odd-day count: 1 if ordinary, 2 if leap. The running calendar is exactly this shape. 1 January 2024 is a Monday. 2024 is a leap year (divisible by 4, not a century), so it contributes 2 odd days. Monday plus 2 weekdays is Wednesday, and that is 1 January 2025.

Do not reopen the calendar and count every month unless the span is irregular — a date that is not the same calendar day a year later, or a span that starts after 28 February in a leap year (because the extra day then sits inside the span) versus one that finishes before it (the extra day sits outside). Year-length mod 7 is enough when both dates are the same calendar day a year apart. An ordinary year from a Sunday 1 January lands on Monday; answering Tuesday is the leap-year shift applied to a year that did not earn it.

Figure. 1 January 2024 is a Monday. 2024 is a leap year, so the span is 366 days and 366 mod 7 = 2. Monday plus 2 weekdays is Wednesday: 1 January 2025.

How it works

1. Confirm leap or ordinaryApply the century rule to the year that contains the span between the two identical calendar dates. 2024 / 4 works and 2024 is not a century, so leap.
2. Read odd daysLeap → 2; ordinary → 1. 366 mod 7 = 2 is the same fact as 'leap year → 2 odd days'.
3. Shift the weekdayAdd that many days to the anchor weekday, wrapping past Sunday. Monday + 2 = Wednesday, so 1 January 2025 is Wednesday.

7Odd days in 100 and 400 years

Longer calendar spans use packed remainders so you do not list every year. One hundred years with a non-leap century contain 24 leap years and 76 ordinary years: 24 \times 2 + 76 \times 1 = 124 odd days, and 124 \bmod 7 = 5. So 100 years contribute 5 odd days. Four hundred years contain one extra leap century (the one divisible by 400), which is 97 leap years and 303 ordinary years: 97 \times 2 + 303 = 497, and 497 \bmod 7 = 0. So 400 years contribute 0 odd days — a full cycle that lands on the same weekday. Those two facts sit on top of the ordinary/leap year counts and the century leap rule.

When a problem jumps across centuries, reduce with these codes instead of listing every year. Pull out as many 400-year blocks as you can and drop them (each adds 0). Each leftover 100-year block adds 5, then reduce mod 7. Finish with the remaining individual years at 1 or 2 odd days each. A 400-year block can be dropped entirely because it adds 0 odd days. 5 is the 100-year code; do not confuse it with a single ordinary year's 1.

Figure. Packed remainders: an ordinary year adds 1 odd day, a leap year adds 2, 100 years add 5, and 400 years add 0. Pull out 400-year blocks first, then add 5 per leftover century.

Using the codes

1. Split the spanPull out as many 400-year blocks as you can — each adds 0, because 497 odd days is a multiple of 7.
2. Pack centuriesEach leftover 100-year block adds 5 odd days (then reduce mod 7). 124 mod 7 = 5 is that code, not a guess.
3. Finish with yearsAdd 1 or 2 per remaining year by the leap rule, then reduce mod 7 once at the end.