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Lunar New Year maths for IB Maths SL and HL

A ready-to-teach lesson for Lunar New Year, whenever it falls: the 12-year cycle of animals as remainders. A 5-minute starter, a 35-minute main activity on cycles, arithmetic sequences and remainders, an extension and full worked answers.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Arithmetic sequences and series; Remainders and cycles (modular arithmetic); LCM and repeating patterns; Proof (extension)
Equipment
No GDC needed.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minThe 12-year cycle
Main: task B11 minTwo cycles together
Main: task C12 minClock and calendar arithmetic
Extension10 minFast finishers or homework

Starter (5 minutes)

No GDC.

  1. Find the remainder when 2030 is divided by 12.
  2. Find the lowest common multiple of 10 and 12.
  3. Find the 10th term of the arithmetic sequence 3, 15, 27, …
  4. Today is a Wednesday. What day of the week will it be in 100 days?

Main activity (35 minutes)

Task A: The 12-year cycle (12 min)

Number the 12 animals of the cycle 1 to 12 in the traditional order (Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig). In this activity, the lunar year that begins in calendar year Y has animal number ((Y − 4) mod 12) + 1, where ‘mod 12’ means the remainder on dividing by 12.

  1. Find the animal number for the lunar year that begins in 2031.
  2. The calendar years from 2001 to 2100 with animal number 1 form an arithmetic sequence. Find its first term, last term and number of terms.
  3. Find the sum of those years.

Task B: Two cycles together (11 min)

A 10-year cycle (labels 1 to 10) runs alongside the 12-year cycle (animals 1 to 12). In this model, both are at 1 in year 0, and each moves on by one every year.

  1. After how many years does the pair (label 1, animal 1) first come back?
  2. How many different (label, animal) pairs occur in one full cycle?
  3. There are 10 × 12 = 120 possible pairs, but only 60 occur. Show that (label 1, animal 2) never occurs by counting the years in one cycle that give it.

Task C: Clock and calendar arithmetic (12 min)

Work with remainders.

  1. Find the remainder when 72026 is divided by 12.
  2. Find the remainder when 1 + 2 + 3 + … + 100 is divided by 12.
  3. Find the smallest positive number that leaves remainder 3 when divided by 12 and remainder 2 when divided by 5.

Extension (10 minutes)

For fast finishers.

  1. Find the smallest number greater than 1 that leaves remainder 1 when divided by 7, by 10 and by 12.
  2. Prove that if n is not divisible by 2 or 3, then n2 leaves remainder 1 when divided by 12.

For teachers

Teacher notes and full worked answers
  • Lunar New Year falls on a different date each year: use this pack in the weeks around it.
  • Students may want to check the rule against their own year of birth: remind them that the lunar year usually begins in January or February, so a birthday early in the calendar year may belong to the previous lunar year.
  • Task B (c) is the key idea: the two cycles can only meet in pairs with the same parity.

Starter

  1. 2
    • 12 × 169 = 2028, and 2030 − 2028 = 2.
  2. 60
    • 10 = 2 × 5, 12 = 22 × 3, so LCM = 22 × 3 × 5 = 60.
  3. 111
    • 3 + 9 × 12 = 111
  4. Friday
    • 100 = 7 × 14 + 2, so 2 days after Wednesday.

Task A: The 12-year cycle

  1. 12
    • 2031 − 4 = 2027 = 12 × 168 + 11, so the remainder is 11.
    • 11 + 1 = 12
  2. 2008, 2092; 8 terms
    • Animal number 1 needs (Y − 4) mod 12 = 0, that is Y = 4 + 12k.
    • The first after 2000 is 2008 (= 4 + 12 × 167); the last up to 2100 is 2092.
    • (2092 − 2008) ÷ 12 + 1 = 8 terms
  3. 16 400
    • S = 8/2 × (2008 + 2092) = 4 × 4100 = 16 400

Task B: Two cycles together

  1. 60 years
    • We need a multiple of 10 that is also a multiple of 12: LCM(10, 12) = 60.
  2. 60
    • The pattern repeats every 60 years, and no pair repeats inside one cycle, so 60 pairs occur.
  3. 0 years
    • Label 1 needs the year to be a multiple of 10, which is even. Animal 2 needs the year to leave remainder 1 when divided by 12, which is odd.
    • A year cannot be both even and odd, so there are 0 such years.

Task C: Clock and calendar arithmetic

  1. 1
    • 72 = 49 = 4 × 12 + 1, so 72 leaves remainder 1.
    • 72026 = (72)1013 leaves remainder 11013 = 1.
  2. 10
    • 1 + 2 + … + 100 = 5050.
    • 5050 = 12 × 420 + 10
  3. 27
    • Remainder 3 on dividing by 12: 3, 15, 27, 39, …
    • 27 = 5 × 5 + 2, so 27.

Extension

  1. 421
    • It is 1 more than a common multiple of 7, 10 and 12. LCM(7, 10, 12) = 420.
    • 420 + 1 = 421
  2. Proof
    • Such an n leaves remainder 1, 5, 7 or 11 when divided by 12.
    • 12 = 1, 52 = 25 = 2 × 12 + 1, 72 = 49 = 4 × 12 + 1, 112 = 121 = 10 × 12 + 1.
    • In every case the remainder is 1.

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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