Formula sheetClasses 8 to 12
IOQM and RMO Toolkit: Theorems, Formulas and Practice
Everything the Indian Mathematical Olympiad route leans on, from IOQM to RMO and INMO: the pattern of each paper, the theorems in number theory, algebra, combinatorics and geometry, and ten problems to try.
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10 practice questions at the end. Pick an answer for each question, then press Check my answers: your score and an explanation for every question appear at once. Nothing you choose leaves this page.
01The route and the papers
| Stage | 2026–27 | The paper |
|---|---|---|
| IOQM | 6 September 2026 | Classes 8 to 12. 30 questions with answers from 00 to 99 in 3 hours: questions 1–10 carry 2 marks, 11–20 carry 3 and 21–30 carry 5. No negative marking |
| RMO | 15 November 2026 | For IOQM qualifiers: 6 problems with full written proofs, in 3 hours |
| INMO | 17 January 2027 | For RMO qualifiers: 6 proof problems in 4.5 hours |
| IMOTC | After INMO | A training camp at HBCSE, Mumbai, which picks India's team for the International Mathematical Olympiad |
02Number theory
- Divisibility by 3 or 9 if the digit sum is; by 11 if the alternating digit sum is; by 8 if the last three digits are
- HCF and LCM gcd(a, b) × lcm(a, b) = ab; gcd(a, b) = gcd(b, a mod b)
- Number of divisors n = paqbrc has (a + 1)(b + 1)(c + 1) divisors
- Sum of divisors multiply (pa+1 − 1)/(p − 1) over the prime powers
- Fermat's little theorem ap−1 ≡ 1 (mod p) when p is prime and does not divide a
- Euler's theorem aφ(n) ≡ 1 (mod n) when gcd(a, n) = 1; φ(n) = n × ∏(1 − 1/p)
- Wilson's theorem (p − 1)! ≡ −1 (mod p) for every prime p
- Zeros at the end of n! n ÷ 5 + n ÷ 25 + n ÷ 125 + …, keeping only whole parts
- Squares leave 0 or 1 when divided by 4, and 0, 1 or 4 when divided by 8
- Chinese remainder theorem remainders for coprime divisors fix the number modulo their product
- Last digits work modulo 10 or 100; powers repeat in short cycles
03Algebra and inequalities
- Quadratic ax2 + bx + c roots add to −b/a and multiply to c/a
- Cubic x3 + px2 + qx + r roots add to −p, pairs multiply and add to q, all three multiply to −r
- A sum of squares p2 + q2 = (p + q)2 − 2pq
- Factorising a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)
- Sophie Germain a4 + 4b4 = (a2 + 2b2 + 2ab)(a2 + 2b2 − 2ab)
- AM ≥ GM (a + b)/2 ≥ √(ab) for a, b ≥ 0, with equality when a = b
- Cauchy–Schwarz (a1b1 + a2b2 + …)2 ≤ (a12 + a22 + …)(b12 + b22 + …)
- Telescoping 1/(k(k + 1)) = 1/k − 1/(k + 1)
- Sums 1 + 2 + … + n = n(n + 1)/2; squares add to n(n + 1)(2n + 1)/6; cubes to (n(n + 1)/2)2
04Counting
- Choosing C(n, r) = n!/(r!(n − r)!); arranging r of n: n!/(n − r)!
- Stars and bars x1 + … + xk = n in whole numbers ≥ 0 has C(n + k − 1, k − 1) solutions
- Inclusion–exclusion |A ∪ B| = |A| + |B| − |A ∩ B|, and so on for three sets
- Pigeonhole n + 1 objects in n boxes put two in one box
- Round a table n different people sit in (n − 1)! ways
- Grid paths from (0, 0) to (m, n), moving right or up, in C(m + n, m) ways
- Diagonals a polygon with n sides has n(n − 3)/2
- Catalan numbers C(2n, n)/(n + 1): 1, 1, 2, 5, 14, 42
05Geometry
- Angle bisector the bisector from A cuts BC in the ratio AB : AC
- Ceva AD, BE and CF meet at one point exactly when (BD/DC)(CE/EA)(AF/FB) = 1
- Power of a point PA × PB = PC × PD for two chords through P, and = PT2 for a tangent PT
- Cyclic quadrilateral opposite angles add to 180°; Ptolemy: AC × BD = AB × CD + AD × BC
- Area of a triangle ½ab sin C; Heron: √(s(s − a)(s − b)(s − c)); rs; abc/(4R)
- Sine rule a/sin A = b/sin B = c/sin C = 2R
- Cosine rule c2 = a2 + b2 − 2ab cos C
- Centroid divides each median in the ratio 2 : 1
- Right triangle the circumcentre is the midpoint of the hypotenuse
06How to work the IOQM paper
- Small cases first try n = 1, 2, 3, 4 and look for the pattern
- Answers run from 00 to 99 a huge answer usually means you missed a remainder or a digit sum
- Pace about 6 minutes a question: faster on 2-markers, slower on 5-markers
- No negative marking give an answer to every question, after checking
- Old papers solve past IOQM and PRMO papers in 3 hours, then study the solutions
07Practice problems1 mark each
1Find the remainder when 2100 is divided by 7.
2How many positive divisors does 360 have?
3How many zeros does 125! end with?
4p and q are the roots of x2 − 7x + 3 = 0. Find p2 + q2.
5Find the least value of x + 16/x for x > 0.
6In how many ways can 10 identical sweets be shared among 3 children so that each child gets at least one?
7A right triangle has legs 20 and 21. Find its hypotenuse.
8How many diagonals does a 12-sided polygon have?
9Chords AB and CD of a circle meet at P inside the circle. PA = 4, PB = 9 and PC = 6. Find PD.
10Find the last two digits of 220.
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Answers and explanations
- 10223 = 8 ≡ 1 (mod 7) and 100 = 3 × 33 + 1, so 2100 ≡ 2.
- 224360 = 23 × 32 × 5, so it has (3 + 1)(2 + 1)(1 + 1) = 24 divisors.
- 331Count the factors of 5: 125 ÷ 5 = 25, 125 ÷ 25 = 5 and 125 ÷ 125 = 1, so 25 + 5 + 1 = 31.
- 443p + q = 7 and pq = 3, so p2 + q2 = (p + q)2 − 2pq = 49 − 6 = 43.
- 508By AM ≥ GM, x + 16/x ≥ 2√16 = 8, with equality at x = 4.
- 636Give each child one sweet first, then share the other 7 freely: C(9, 2) = 36.
- 729202 + 212 = 400 + 441 = 841 = 292.
- 854n(n − 3)/2 = 12 × 9 ÷ 2 = 54.
- 906Power of a point: PA × PB = PC × PD, so 36 = 6 × PD and PD = 6.
- 1076210 = 1024 ends in 24, and 242 = 576 ends in 76, so 220 ends in 76.
How to use this sheet
- Check the dates and rules each year on the HBCSE Olympiad pages.
- Write full solutions to every practice problem: RMO marks the proof, not the answer.
- Keep a notebook of tricks you meet, sorted into the four topics.
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