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22 December

National Mathematics Day puzzles

22 December is National Mathematics Day in India, in honour of the mathematician Srinivasa Ramanujan. Celebrate with eight original puzzles: sums of cubes, partitions, nested square roots and a calendar puzzle. Every puzzle has two hints and a full solution, and none needs a calculator.

Teachers: open ‘For teachers’ to project a puzzle or add puzzles to a worksheet for your class. All free.

For teachers: project, add to a worksheet or set as homework

Press Project on any problem to show it full screen with a timer, the hints, the answer and the worked solution one step at a time (arrow keys move between problems; Space reveals the next step; F full screen; Esc closes). Switch on the ‘Add to worksheet’ buttons, pick problems, then print them from the worksheet builder or set them as homework for a class, with the full solutions as the mark scheme. Free problems are free for every class; problems marked ‘With a plan’ can be set by teachers with a plan or school licence. Ready-made sessions: maths club packs.

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Problem I166

Number theoryShort answer

1729 can be written as a sum of two positive cubes in two different ways: 1729 = a3 + b3 = c3 + d3. Find a + b + c + d.

Hint

List the cubes up to 1729: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728.

Second hint

For each cube, check whether 1729 minus it is also a cube.

Full worked solution

Answer: 32

  1. The cubes below 1729 are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331 and 1728.
  2. 1729 − 1728 = 1, so 1729 = 13 + 123.
  3. 1729 − 1000 = 729, so 1729 = 93 + 103. No other cube leaves a cube.
  4. a + b + c + d = 1 + 12 + 9 + 10 = 32.

Why this works: A short, organised list beats guessing: checking each cube once finds every way.

Where it leads: 1729 is the smallest number that is a sum of two positive cubes in two ways; such numbers are called taxicab numbers.

Strategy: Organised cases

Problem I167

CombinatoricsShort answer

A partition of 7 is a way of writing 7 as a sum of positive whole numbers, where order does not matter (so 3 + 4 and 4 + 3 are the same, and 7 on its own counts). How many partitions does 7 have?

Hint

Organise by the largest part.

Second hint

Largest part 7: 1 way; 6: 1; 5: 2; 4: 3; 3: 4; 2: 3; 1: 1.

Full worked solution

Answer: 15

  1. Largest part 7: 7. Largest 6: 6+1. Largest 5: 5+2, 5+1+1 (2 ways).
  2. Largest 4: 4+3, 4+2+1, 4+1+1+1 (3 ways).
  3. Largest 3: 3+3+1, 3+2+2, 3+2+1+1, 3+1+1+1+1 (4 ways). Largest 2: 2+2+2+1, 2+2+1+1+1, 2+1+1+1+1+1 (3 ways). Largest 1: 1 way.
  4. Total: 1 + 1 + 2 + 3 + 4 + 3 + 1 = 15.

Why this works: Fixing the largest part first means each partition is listed exactly once.

Where it leads: The partition numbers grow very fast (p(100) is over 190 million) and have surprising divisibility patterns, such as p(5k + 4) always being a multiple of 5.

Strategy: Organised cases

Problem I168

Number theoryShort answer

Write the date 22 December 2026 as the eight-digit number 22122026. What is the remainder when it is divided by 9?

Hint

A number and the sum of its digits leave the same remainder on division by 9.

Second hint

2 + 2 + 1 + 2 + 2 + 0 + 2 + 6 = 17.

Full worked solution

Answer: 8

  1. Because 10 ≡ 1 (mod 9), every power of 10 leaves remainder 1, so a number has the same remainder as its digit sum.
  2. Digit sum: 2 + 2 + 1 + 2 + 2 + 0 + 2 + 6 = 17.
  3. 17 = 9 + 8.
  4. The remainder is 8.

Why this works: Remainders mod 9 only see the digit sum, because 10 leaves remainder 1.

Where it leads: The same idea gives the divisibility tests for 3 and 11 and the check digits used on book numbers and bank cards.

Strategy: Spot the pattern and generalise

Problem I169

AlgebraShort answer

Evaluate √(1 + 2√(1 + 3√(1 + 4 × 6))).

Hint

Work from the inside out.

Second hint

1 + 4 × 6 = 25.

Full worked solution

Answer: 3

  1. Innermost: 1 + 4 × 6 = 25, and √25 = 5.
  2. Next: 1 + 3 × 5 = 16, and √16 = 4.
  3. Next: 1 + 2 × 4 = 9, and √9 = 3.
  4. The value is 3.

Why this works: Each layer has the form 1 + n(n + 2) = (n + 1)2, so every square root comes out exactly.

Where it leads: Continuing the pattern forever gives the famous infinite nested radical √(1 + 2√(1 + 3√(1 + …))) = 3.

Strategy: Working backwards, Spot the pattern and generalise

Problem I170

Number theoryShort answer

1729 = 7 × 13 × 19. Find the sum of all the positive divisors of 1729 (including 1 and 1729).

Hint

Every divisor is a product of some of the primes 7, 13 and 19.

Second hint

Expand (1 + 7)(1 + 13)(1 + 19).

Full worked solution

Answer: 2240

  1. Each divisor uses each of 7, 13, 19 either 0 or 1 times: 8 divisors.
  2. Expanding (1 + 7)(1 + 13)(1 + 19) produces each divisor exactly once.
  3. 8 × 14 × 20 = 2240.
  4. The sum is 2240.

Why this works: The sum of divisors factorises over the primes, so a product of small brackets replaces listing all eight divisors.

Where it leads: The divisor-sum function σ(n) defines perfect numbers (σ(n) = 2n) and is one of the central functions of number theory.

Strategy: Spot the pattern and generalise

Problem I171

CombinatoricsShort answer

In how many ways can 10 be written as a sum of different positive whole numbers, where order does not matter (10 on its own counts)?

Hint

Sort by how many parts there are.

Second hint

One part: 1 way. Two parts: 4. Three parts: 4. Four parts: 1.

Full worked solution

Answer: 10

  1. One part: 10.
  2. Two different parts: 9+1, 8+2, 7+3, 6+4 (4 ways).
  3. Three different parts: 7+2+1, 6+3+1, 5+4+1, 5+3+2 (4 ways). Four: 4+3+2+1 (1 way); five different parts need at least 15.
  4. Total: 1 + 4 + 4 + 1 = 10.

Why this works: Counting by the number of parts keeps the list short and complete; the smallest possible sum rules out large numbers of parts.

Where it leads: Euler proved that partitions into different parts are as many as partitions into odd parts: check that 10 has 10 of those too.

Strategy: Organised cases

Problem I172

Number theoryShort answer

How many whole numbers from 1 to 2026 are multiples of 22 or of 12 (or both)?

Hint

Add the two counts and subtract the numbers counted twice.

Second hint

A multiple of both 22 and 12 is a multiple of their LCM, 132.

Full worked solution

Answer: 245

  1. Multiples of 22: ⌊2026/22⌋ = 92.
  2. Multiples of 12: ⌊2026/12⌋ = 168.
  3. Multiples of both = multiples of LCM(22, 12) = 132: ⌊2026/132⌋ = 15.
  4. Total: 92 + 168 − 15 = 245.

Why this works: Inclusion–exclusion: numbers in both lists are counted twice, so subtract them once; and ‘both’ means the LCM, not the product.

Where it leads: With more conditions the alternating sum continues; it is the same principle that counts derangements.

Strategy: Count the opposite

Problem I173

LogicShort answer

22 December 2026 is a Tuesday. On which day of the week will 22 December 2030 fall?

Hint

A normal year moves the weekday on by 1 (365 = 52 × 7 + 1); a year containing 29 February moves it by 2.

Second hint

Which of the four year-long gaps from 22 December 2026 contains a 29 February?

Full worked solution

Answer: Sunday

  1. From 22 Dec 2026 to 22 Dec 2027: 365 days, +1.
  2. To 22 Dec 2028: contains 29 Feb 2028, 366 days, +2. To 22 Dec 2029: +1. To 22 Dec 2030: +1.
  3. Total shift: 1 + 2 + 1 + 1 = 5 days after Tuesday.
  4. Wednesday, Thursday, Friday, Saturday, Sunday.

Why this works: Only the remainder of the number of days on division by 7 matters, so each year contributes 1 or 2.

Where it leads: The same modular counting gives Zeller’s congruence for the weekday of any date.

Strategy: Spot the pattern and generalise

About Srinivasa Ramanujan

Read about his life and work from these sources:

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