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Lunar New Year maths for CBSE Class 10

A ready-to-teach lesson for Lunar New Year, whenever it falls: the 12-year cycle of animals through Euclid’s division lemma, HCF and LCM and arithmetic progressions. A 5-minute starter, a 35-minute main activity, an extension and full worked answers. No calculator.

Level
CBSE Class 10 (Class 9 can try most of it)
Time
40 minutes, plus a 10-minute extension
Topics
Real numbers: Euclid’s division lemma, HCF and LCM; Arithmetic progressions; Remainders and repeating patterns
Equipment
No calculator. Every answer works out exactly.

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 minHCF and LCM at the festival
Main: task C12 minRemainders
Extension10 minFast finishers or homework

Starter (5 minutes)

Quick-fire mental maths.

  1. Write 2030 = 12q + r with 0 ≤ r < 12.
  2. Find the LCM of 10 and 12.
  3. Find the HCF of 144 and 180.
  4. Find the 10th term of the AP 8, 20, 32, …

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, for the lunar year that begins in calendar year Y: write Y − 4 = 12q + r; the animal number is r + 1.

  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 AP. Find the first term, last term and number of terms.
  3. Find the sum of the years in part (b).

Task B: HCF and LCM at the festival (11 min)

Use prime factorisation or Euclid’s division algorithm.

  1. Use Euclid’s division algorithm to find HCF(420, 660).
  2. Red lanterns hang every 4 m and gold lanterns every 6 m along a street, both starting at 0 m. At how many places from 0 m to 100 m are a red and a gold lantern together?
  3. 48 dancers and 36 drummers form the greatest possible number of identical groups. How many groups, and who is in each?

Task C: Remainders (12 min)

Euclid’s division lemma: every integer a can be written as a = bq + r, 0 ≤ r < b.

  1. Find the smallest positive number that leaves remainder 3 when divided by 12 and remainder 2 when divided by 5.
  2. Show that the square of any positive integer is of the form 3m or 3m + 1 for some integer m.
  3. A string of beads repeats a pattern of 12 colours. Which colour (1 to 12) is the 500th bead?

Extension (10 minutes)

For fast finishers, or as homework.

  1. Find the smallest number greater than 1 that leaves remainder 1 when divided by 7, 10 and 12.
  2. Show that n2 − 1 is divisible by 8 for every odd positive integer n.

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.
  • If students test the rule on their own birth year, remind them that the lunar year usually begins in January or February.
  • Task C (b) is the textbook style of proof using Euclid’s division lemma: a good model for board answers.

Starter

  1. q = 169, r = 2
    • 12 × 169 = 2028, and 2030 − 2028 = 2.
  2. 60
    • 22 × 3 × 5 = 60
  3. 36
    • 144 = 24 × 32, 180 = 22 × 32 × 5, so HCF = 22 × 32 = 36.
  4. 116
    • 8 + 9 × 12 = 116

Task A: The 12-year cycle

  1. 12
    • 2027 = 12 × 168 + 11, so r = 11 and the animal number is 12.
  2. 2008, 2092; 8 terms
    • r = 0 means Y = 12q + 4: 2008, 2020, …, 2092.
    • 2092 = 2008 + (n − 1) × 12 gives n = 8.
  3. 16 400
    • S8 = 8/2 × (2008 + 2092) = 4 × 4100 = 16 400

Task B: HCF and LCM at the festival

  1. 60
    • 660 = 420 × 1 + 240; 420 = 240 × 1 + 180; 240 = 180 × 1 + 60; 180 = 60 × 3 + 0
    • HCF = 60
  2. 9 places (every 12 m)
    • LCM(4, 6) = 12: 0, 12, 24, …, 96 is 9 places.
  3. 12 groups of 4 dancers and 3 drummers
    • HCF(48, 36) = 12; 48 ÷ 12 = 4 and 36 ÷ 12 = 3.

Task C: Remainders

  1. 27
    • 3, 15, 27, …; 27 = 5 × 5 + 2.
  2. Proof
    • Any integer is 3q, 3q + 1 or 3q + 2.
    • (3q)2 = 3(3q2); (3q + 1)2 = 3(3q2 + 2q) + 1; (3q + 2)2 = 3(3q2 + 4q + 1) + 1.
    • So the square is 3m or 3m + 1.
  3. Colour 8
    • 500 = 12 × 41 + 8

Extension

  1. 421
    • LCM(7, 10, 12) = 420, and 420 + 1 = 421.
  2. Proof
    • Write n = 2k + 1. Then n2 − 1 = 4k2 + 4k = 4k(k + 1).
    • k(k + 1) is a product of consecutive integers, so it is even. So 4k(k + 1) is a multiple of 8.

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

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