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Prepare for the OBMEP Brazilian Math Olympiad by mastering creative arithmetic reasoning, prime factorization, modular arithmetic, plane and spatial geometry, algebraic systems, systematic combinatorial counting, and parity/invariant deduction principles across 100 practice questions.

Sample OBMEP Practice Questions

Try these sample questions to test your OBMEP exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1The five-digit number $3a54b$ is divisible by $36$, where $a$ and $b$ are single digits with $a > b$. What is the value of $a + b$?
A.4
B.5
C.6
D.8
Explanation: A number is divisible by 36 if and only if it is divisible by both 4 and 9. Divisibility by 4 requires the two-digit number formed by its last two digits, $4b$, to be divisible by 4, so $b \in \{0, 4, 8\}$. Divisibility by 9 requires the sum of all digits, $3 + a + 5 + 4 + b = 12 + a + b$, to be a multiple of 9. If $b = 0$, then $12 + a$ is a multiple of 9, giving $a = 6$; this satisfies $a > b$ ($6 > 0$) and yields $a + b = 6 + 0 = 6$. If $b = 4$, then $16 + a$ requires $a = 2$, but this violates $a > b$. If $b = 8$, then $20 + a$ requires $a = 7$, which also violates $a > b$. Thus, $a = 6, b = 0$, and $a + b = 6$.
2What is the remainder when the sum $S = 1 + 2 + 3 + \dots + 2023$ is divided by $7$?
A.0
B.1
C.3
D.5
Explanation: The sum of the first $n$ positive integers is given by $S = \frac{n(n+1)}{2}$. For $n = 2023$, $S = \frac{2023 \times 2024}{2} = 2023 \times 1012$. Notice that $2023 = 7 \times 289$, meaning 2023 is an exact multiple of 7. Therefore, $S = 7 \times 289 \times 1012$ is an exact multiple of 7, leaving a remainder of 0.
3How many positive integer divisors of $N = 2^4 \times 3^3 \times 5^2$ are multiples of $18$?
A.12
B.16
C.20
D.24
Explanation: A divisor $d$ of $N$ is a multiple of $18 = 2^1 \times 3^2$ if and only if $d = 18k$, where $k$ divides $\frac{N}{18} = \frac{2^4 \times 3^3 \times 5^2}{2^1 \times 3^2} = 2^3 \times 3^1 \times 5^2$. The number of positive divisors of $2^3 \times 3^1 \times 5^2$ is given by $(3+1)(1+1)(2+1) = 4 \times 2 \times 3 = 24$.
4Two positive integers $a$ and $b$ have $\gcd(a, b) = 15$ and $\text{lcm}(a, b) = 450$. If $a = 75$, what is the value of $b$?
A.60
B.90
C.120
D.150
Explanation: For any two positive integers, the product of their greatest common divisor and least common multiple is equal to the product of the two numbers: $\gcd(a, b) \times \text{lcm}(a, b) = a \times b$. Substituting the given values gives $15 \times 450 = 75 \times b$, which simplifies to $6750 = 75b$. Dividing both sides by 75 gives $b = \frac{6750}{75} = 90$.
5What are the last two digits of $7^{2026}$ in its standard decimal representation?
A.49
B.43
C.07
D.01
Explanation: The last two digits of a number correspond to its value modulo 100. We examine powers of 7 modulo 100: $7^1 = 7$, $7^2 = 49$, $7^3 = 343 \equiv 43$, and $7^4 = 2401 \equiv 1 \pmod{100}$. Since $7^4 \equiv 1 \pmod{100}$, the powers repeat in a cycle of length 4. Dividing the exponent 2026 by 4 yields $2026 = 4 \times 506 + 2$. Therefore, $7^{2026} = (7^4)^{506} \times 7^2 \equiv 1^{506} \times 49 \equiv 49 \pmod{100}$.
6When the repeating decimal $0.2\overline{36} = 0.2363636\dots$ is written as an irreducible fraction $\frac{p}{q}$, what is the value of $p + q$?
A.58
B.64
C.68
D.72
Explanation: Let $x = 0.2363636\dots$. Multiplying by 1000 gives $1000x = 236.3636\dots$, and multiplying by 10 gives $10x = 2.3636\dots$. Subtracting these equations gives $990x = 234$, which means $x = \frac{234}{990}$. Dividing both the numerator and the denominator by their greatest common divisor $\gcd(234, 990) = 18$ gives $\frac{234 / 18}{990 / 18} = \frac{13}{55}$. Since $\gcd(13, 55) = 1$, the fraction is irreducible, and $p + q = 13 + 55 = 68$.
7For any three-digit positive integer $N = \overline{abc}$, the six-digit integer $\overline{abcabc}$ is formed by repeating the digits. This six-digit integer is always divisible by three distinct prime numbers greater than $5$. What is the sum of these three prime numbers?
A.23
B.27
C.29
D.31
Explanation: Any six-digit number of the form $\overline{abcabc}$ can be expanded as $\overline{abc} \times 1000 + \overline{abc} = \overline{abc} \times (1000 + 1) = \overline{abc} \times 1001$. The prime factorization of 1001 is $1001 = 7 \times 11 \times 13$. The three prime factors are 7, 11, and 13 (all greater than 5), and their sum is $7 + 11 + 13 = 31$.
8What is the smallest positive integer $k$ such that the product $1080 \times k$ is a perfect cube?
A.15
B.25
C.50
D.75
Explanation: The prime factorization of 1080 is $1080 = 2^3 \times 3^3 \times 5^1$. For $1080 \times k$ to be a perfect cube, the exponents of all prime factors in its prime factorization must be multiples of 3. The primes 2 and 3 already have exponent 3. The prime 5 has exponent 1, so it needs an additional exponent of $3 - 1 = 2$. Therefore, the minimal positive integer is $k = 5^2 = 25$.
9When the decimal number $2024$ is converted into base $5$, what is the sum of its digits?
A.12
B.14
C.15
D.16
Explanation: We perform successive divisions by 5: $2024 = 404 \times 5 + 4$, $404 = 80 \times 5 + 4$, $80 = 16 \times 5 + 0$, $16 = 3 \times 5 + 1$, and $3 = 0 \times 5 + 3$. Reading the remainders from last to first gives $2024_{10} = 31044_5$. The sum of the digits in base 5 is $3 + 1 + 0 + 4 + 4 = 12$.
10A farmer counts eggs in a basket. When grouped in sets of $3$, $2$ eggs remain; when grouped in sets of $5$, $3$ eggs remain; and when grouped in sets of $7$, $2$ eggs remain. What is the smallest possible positive number of eggs in the basket?
A.17
B.18
C.21
D.23
Explanation: We set up a system of linear congruences: $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 5$, and $x \equiv 2 \pmod 7$. Combining $x \equiv 2 \pmod 3$ and $x \equiv 2 \pmod 7$ gives $x \equiv 2 \pmod{21}$ since $\gcd(3,7)=1$. The positive integers congruent to $2 \pmod{21}$ are $2, 23, 44, 65, \dots$. Testing these modulo 5: $2 \equiv 2 \pmod 5$, but $23 = 4 \times 5 + 3 \equiv 3 \pmod 5$. Thus, 23 satisfies all three conditions simultaneously.

About the OBMEP Exam

The Olimpíada Brasileira de Matemática das Escolas Públicas (OBMEP) is the largest academic competition in Brazil and one of the largest mathematical olympiads in the world, engaging over 18 million students from more than 55,000 public and private schools across 99.8% of Brazilian municipalities each year. Created in 2005 by the Instituto de Matemática Pura e Aplicada (IMPA) in partnership with the Sociedade Brasileira de Matemática (SBM) and supported by the Ministry of Science, Technology and Innovation (MCTI) and the Ministry of Education (MEC), OBMEP aims to stimulate mathematical interest, identify young talents, foster educational equality, and enhance the quality of basic education across Brazil. The competition is structured into three competition tiers: Nível 1 (6th and 7th grades of Ensino Fundamental), Nível 2 (8th and 9th grades of Ensino Fundamental), and Nível 3 (all grades of Ensino Médio). Testing takes place across two distinct stages: an objective school-level Phase 1 (20 five-option multiple-choice questions, 2h30) where approximately the top 5% per school qualify for Phase 2, followed by a rigorous 6-problem discursive Phase 2 (3h00) evaluated by expert regional grading panels. Top national performers earn Gold, Silver, and Bronze medals, honorable mentions, and enrollment in the prestigious Programa de Iniciação Científica Jr. (PIC) with a monthly CNPq scholarship stipend, as well as qualification for direct university entrance via 'Vagas Olímpicas' at premier institutions such as USP, UNICAMP, and UNESP.

Assessment

Question count varies by exam level

Time Limit

Phase 1: 2h30 (20 MCQs); Phase 2: 3h00 (6 Discursive Problems)

Passing Score

Quota-based school advancement (Phase 1) and national merit ranking (Phase 2)

Exam Fee

Free (Gratuito) (Instituto de Matemática Pura e Aplicada (IMPA) & Sociedade Brasileira de Matemática (SBM))

OBMEP Exam Content Outline

Not published

Aritmética e Teoria dos Números

Number theory fundamentals and computational arithmetic: divisibility criteria, modular arithmetic, Euclidean algorithm, prime factorization, GCD and LCM properties, repeating decimals, Diophantine equations, digit sums, and base positional representations.

Not published

Geometria Plana e Espacial

Plane Euclidean geometry, spatial mensuration, and geometric transformations: angle properties in intersecting lines and polygons, triangle similarity and congruence, Pythagorean theorem, area partitioning and dissections, circle theorems, inscribed/circumscribed polygons, and surface areas and volumes of prisms, pyramids, cylinders, cones, and spheres.

Not published

Álgebra e Funções

Algebraic manipulation, functional equations, and sequences: notable algebraic identities, linear and quadratic equations and inequalities, Vieta's formulas, arithmetic and geometric progressions, function optimization, logarithmic properties, and polynomial remainder theorems.

Not published

Combinatória e Probabilidade

Discrete enumeration and probability theory: fundamental counting principle, permutations and combinations, grid paths, circular arrangements, stars and bars, inclusion-exclusion principle, classical and conditional probability, and the Pigeonhole Principle (Princípio da Casa dos Pombos).

Not published

Raciocínio Lógico e Jogos Matemáticos

Mathematical deduction and game theory: knights and knaves truth deduction, parity arguments, invariant and monovariant analysis, balance scale weighings, hat color deduction puzzles, and combinatorial subtraction and Nim game strategies.

How to Pass the OBMEP Exam

What You Need to Know

  • Passing score: Quota-based school advancement (Phase 1) and national merit ranking (Phase 2)
  • Assessment: Question count varies by exam level
  • Time limit: Phase 1: 2h30 (20 MCQs); Phase 2: 3h00 (6 Discursive Problems)
  • Exam fee: Free (Gratuito)

Keys to Passing

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

OBMEP Study Tips from Top Performers

1Focus on Mathematical Logic Over Rote Memorization: OBMEP problems emphasize creative deduction, pattern recognition, and invariant analysis rather than mechanical calculation. Always look for symmetries, parity arguments, or simple extreme cases.
2Master Area Partitioning and Geometric Dissections: In plane geometry, decompose irregular or shaded polygons into basic triangles and rectangles, or calculate the target area by subtracting unshaded regions from a bounding polygon.
3Apply Systematic Case-by-Case Counting: For combinatorics and probability questions, construct clear counting trees or categorize mutually exclusive cases to prevent omissions and duplicate counting.
4Practice with Past Exams and IMPA's Banco de Questões: Solve official past Phase 1 and Phase 2 exams under timed conditions (2h30 for Phase 1, 3h00 for Phase 2) to become fluent with olympiad phrasing and problem pacing.
5Write Complete, Step-by-Step Mathematical Proofs: When preparing for Phase 2 discursive questions, always write out the full reasoning, clearly defining variables, stating used theorems, and showing why no additional cases exist.

Frequently Asked Questions

What is OBMEP and who is eligible to participate?

The Olimpíada Brasileira de Matemática das Escolas Públicas (OBMEP) is Brazil's premier national mathematics competition, organized by IMPA and SBM with support from MCTI and MEC. Over 18 million students from 6th grade of elementary school through the 3rd/4th year of high school (Ensino Médio) participate annually across public and private schools nationwide.

How is the official OBMEP exam structured between Phase 1 and Phase 2?

OBMEP is held in two progressive phases. Phase 1 (1ª Fase) consists of 20 multiple-choice questions (5 options each) administered at each student's home school with a 2h30 time limit. Approximately the top 5% of students per school advance to Phase 2 (2ª Fase), which consists of 6 multi-part discursive problem-solving questions administered at regional centers with a 3h00 time limit.

What are the three official competition levels (Níveis) of OBMEP?

OBMEP categorizes participants into three academic tiers: Nível 1 for students enrolled in the 6th and 7th grades of Ensino Fundamental; Nível 2 for students in the 8th and 9th grades of Ensino Fundamental; and Nível 3 for students enrolled in any year of Ensino Médio.

How are winners determined and what awards are granted?

Advancement and awards are ranking- and quota-based rather than a universal pass score. Top Phase 2 performers are awarded national Gold (650), Silver (1,950), and Bronze (3,900) medals, as well as up to 46,000 Honorable Mention certificates and state-level medals. Medalists receive invitations to the prestigious Programa de Iniciação Científica Jr. (PIC), which includes advanced mathematics mentoring and a monthly CNPq scholarship stipend.

How does the 'Vagas Olímpicas' university admission system work with OBMEP medals?

Leading Brazilian public research universities, including USP, UNICAMP, UNESP, and UNIFEI, allocate specific undergraduate degree slots ('Vagas Olímpicas') each year for olympiad medalists. OBMEP medal winners can earn direct admission into competitive STEM, Engineering, and Computer Science degrees without sitting for traditional vestibular entrance exams.

Is this online practice bank an official test or a study adaptation?

This practice question bank is an English-language study adaptation comprising 100 high-yield multiple-choice questions (4 options each) modeled on official IMPA OBMEP curricula, past Phase 1 and Phase 2 problem types, and Banco de Questões materials to help students master core mathematical problem-solving skills.