📅 Last Reviewed: 19 August 2026 · ✍️ By: PSLE Hero Editorial Team
The NMOS 2026 competition was held on 14 July 2026. Under official NUS High rules, results are released within 10 working days directly to participating primary schools (dispatched by late July / early August 2026).
No individual results are published publicly online, and NUS High does not entertain phone enquiries. Check directly with your child's primary school Mathematics department, or visit the NMOS 2026 Information Hub for full results details, award tiers, official past papers, and format breakdown.
The National Mathematical Olympiad of Singapore (NMOS) is a primary-school Mathematics competition run by NUS High School in partnership with the Singapore Mathematical Society. This evergreen guide focuses on the problem-solving methods students can practise through five original worked examples. For current-year date, eligibility, format and results information, use the NMOS 2026 information hub. These examples are not the official 2026 question paper, answer key or past-year paper.
Check the official NMOS introduction, 2026 rules, awards page and official FAQ for updates.
There is no verified official NMOS 2026 answer key on this page. Students can use the five original questions below to review transferable problem-solving methods, but should not treat them as a reconstruction of the 2026 paper. Parents should check with their school's Mathematics teacher for the student's result once schools receive the organiser's release.
NMOS may help a student demonstrate mathematical aptitude, but the official FAQ does not promise a DSA place. DSA decisions depend on each school's current criteria and selection process, so an NMOS award should be treated as one possible part of a broader profile, not as a guaranteed advantage.
For parents: NMOS is enrichment, not a replacement for school Mathematics or PSLE preparation. Ask the school Mathematics department about selection and registration, confirm the current rules before paying any fee, and choose practice that builds curiosity rather than pressure.
For students: Try each question without looking at the solution. Write down what is known, identify a useful representation or pattern, and check that the final answer satisfies every condition. Then compare your method with the worked solution and note one improvement.
They are different assessments. NMOS is a competition designed to stretch mathematical thinking, while the PSLE Mathematics examination assesses the primary-school syllabus. In the 2026 SEAB format, PSLE Mathematics has two written papers: Paper 1 does not allow calculators, while calculators are allowed in Paper 2. Olympiad-style practice may help students practise reasoning, pattern spotting and clear working, but it does not guarantee a PSLE result and should not replace syllabus-based revision.
See the SEAB 2026 PSLE Mathematics syllabus and examination format.
These are 5 original Olympiad-style sample questions inspired by common primary-school problem-solving ideas. They are educational examples created for this page, not official NMOS questions, past papers or an answer key. For organiser-published sample material, use the official NMOS pages linked in the facts section above. Each worked solution focuses on a method students can explain and check.
Working backwards is a useful strategy when a problem describes a series of operations but only gives the final, end state. It can help with unfamiliar "Before-After" and fraction-of-remainder problems, although students should choose the method that best fits the question.
Alan, Ben, and Carl have a collection of marbles. First, Alan gives half of his marbles to Ben. Next, Ben gives one-third of the marbles he now has to Carl. Finally, Carl gives one-quarter of the marbles he now has to Alan. In the end, each boy has exactly 24 marbles. How many marbles did Alan have at the start?
Step 1: Start from the final state.
Alan = 24, Ben = 24, Carl = 24. (Total marbles = 72)
Step 2: Undo the final transaction (Carl gives 1/4 of his marbles to Alan).
Before giving 1/4 to Alan, Carl kept 3/4 of his marbles. Thus, Carl's ending count of 24 marbles represents 3/4 of what he had before this step.
Carl's marbles before transaction = 24 ÷ 3/4 = 32 marbles(Carl had 32)
Marbles Carl gave to Alan = 32 - 24 = 8 marbles(Alan received 8)
Alan's marbles before transaction = 24 - 8 = 16 marbles(Alan had 16)
State 3: Alan = 16, Ben = 24, Carl = 32
Step 3: Undo the second transaction (Ben gives 1/3 of his marbles to Carl).
Before giving 1/3 to Carl, Ben kept 2/3 of his marbles. Thus, Ben's count of 24 marbles in State 3 represents 2/3 of what he had before this step.
Ben's marbles before transaction = 24 ÷ 2/3 = 36 marbles(Ben had 36)
Marbles Ben gave to Carl = 36 - 24 = 12 marbles(Carl received 12)
Carl's marbles before transaction = 32 - 12 = 20 marbles(Carl had 20)
State 2: Alan = 16, Ben = 36, Carl = 20
Step 4: Undo the first transaction (Alan gives 1/2 of his marbles to Ben).
Before giving half of his marbles to Ben, Alan kept half. Thus, Alan's count of 16 marbles in State 2 represents half of his starting marbles.
Alan's marbles at first = 16 × 2 = 32 marbles(Alan had 32)
Marbles Alan gave to Ben = 32 - 16 = 16 marbles(Ben received 16)
Ben's marbles at first = 36 - 16 = 20 marbles(Ben had 20)
Carl's marbles at first = 20 (remained unchanged during this step)(Carl had 20)
State 1 (At Start): Alan = 32, Ben = 20, Carl = 20
Why this works: The total number of marbles (72) remains constant because they are only exchanged between the boys. Reversing operations is simple: adding becomes subtracting, and division becomes multiplication.
PSLE connection: Working backwards can help with unfamiliar Paper 2 structured or long-answer problems. The method should be chosen because it makes the quantities easier to track, not because every question requires the same heuristic.
The Assumption Method is a core Singapore Maths heuristic. While standard school problems usually feature two variables (e.g., chickens and rabbits), Olympiad-style questions often introduce a third variable (like unanswered questions or neutral scoring) to test a student's cognitive flexibility.
A mathematics quiz consists of 30 questions. For every correct answer, 5 marks are awarded. For every incorrect answer, 3 marks are deducted. For any unanswered question, 1 mark is deducted. Jerry attempted 28 questions and left the remaining questions blank. If his total score was 90 marks, how many questions did he answer correctly?
Step 1: Determine the number of unanswered questions.
The quiz has 30 questions. Jerry attempted 28, leaving the rest unanswered.
Unanswered questions = 30 - 28 = 2 questions
Marks lost from unanswered questions = 2 × (-1) = -2 marks
Step 2: Adjust Jerry's target score for the attempted questions.
Since the 2 unanswered questions deducted 2 marks, Jerry must have scored the remaining marks from his 28 attempts.
Score from attempted questions = 90 - (-2) = 92 marks
Step 3: Make an assumption. Assume Jerry got all 28 attempted questions correct.
Assumed score = 28 × 5 = 140 marks
Step 4: Find the difference between the assumed maximum score and Jerry's actual score.
Total score difference = 140 - 92 = 48 marks
Step 5: Calculate the score drop when one correct answer is replaced by an incorrect answer.
When you replace a correct answer (+5 marks) with an incorrect answer (-3 marks), you do not get the 5 marks and you lose 3 more marks.
Difference per replacement = 5 - (-3) = 8 marks
Step 6: Find the number of incorrect answers.
Number of incorrect answers = 48 ÷ 8 = 6 questions
Step 7: Find the number of correct answers.
Number of correct answers = 28 - 6 = 22 questions
Why this works: By immediately calculating the exact penalty of the third variable (unanswered questions), we reduce the problem to a standard two-variable scenario. We can then apply the 4-step Assumption Method: Assume all correct → Find total difference → Find individual difference → Divide.
PSLE Connection: A classic mistake in PSLE is using "Guess and Check", which consumes valuable time and increases calculation errors. Master the Assumption Method, and students can solve two-variable (and three-variable) word problems in under 90 seconds.
Cryptarithms (or alphametics) are puzzles where letters stand for digits. They develop deep number sense, place-value comprehension, and logical constraint checking.
In the addition sum below, different letters represent different non-zero digits (1 to 9):
A B A
+ B A B
-------
C D D C
Find the digits represented by the letters C and D.
Step 1: Analyze the maximum possible sum to find the thousands digit (C).
ABA and BAB are 3-digit numbers. Even if we use the largest possible digits (999 + 999), the sum is 1998.
Since the sum is a 4-digit number (CDDC), the thousands digit C must be 1.
C = 1
Step 2: Replace C with 1 and examine the units column.
Our addition looks like this:
A B A
+ B A B
---------
1 D D 1
Looking at the ones/units column (A + B), the sum must result in a units digit of 1. Since A and B are positive digits, A + B cannot equal 1. Thus, A + B must equal 11, carrying over 1 to the tens column.
A + B = 11 (with a carryover of 1)
Step 3: Analyze the tens column to find D.
In the tens column, we add B + A + 1 (the carryover from the units column).
Tens Column Sum = B + A + 1
Since we established that A + B = 11, the sum is:
Tens Column Sum = 11 + 1 = 12
This means the tens digit D must be 2, carrying over 1 to the hundreds column.
D = 2
Step 4: Verify the hundreds column.
In the hundreds column, we have A + B + 1 (the carryover from the tens column).
Hundreds Column Sum = A + B + 1 = 11 + 1 = 12
This yields a hundreds digit of 2 (which is indeed D) and carries over 1 to the thousands column (which is indeed C = 1). The math is fully consistent!
Final Sum = 1221 (e.g., 747 + 474 = 1221 or 838 + 383 = 1221)
Why this works: Instead of guessing digits randomly, we use structural mathematical boundaries (like the maximum limit of adding 3-digit numbers) to establish fixed parameters (C = 1), then cascade the logic through place-value constraints.
PSLE connection: Constraint checking is useful when a Paper 2 problem involves number properties, factors, multiples or place value. NMOS-style puzzles may go beyond the PSLE syllabus, so students should treat this as enrichment.
Grid path puzzles are a staple of Olympiad combinatorics. While they look intimidating, they can be solved easily using the Addition Principle (or Pascal's grid labeling), showing students how complex combinations can be broken down into simple addition steps.
A student wants to walk from junction A to junction B along the grid of streets shown below. The student can only walk East (Right) or North (Up). However, junction X is closed due to road construction. How many different paths can the student take from A to B without passing through X?
Step 1: Understand the Addition Principle.
To reach any junction, a student must come from either the junction directly to its West (Left) or the junction directly to its South (Down). Thus, the number of ways to reach a junction is the sum of the ways to reach its left neighbor and its bottom neighbor.
Step 2: Label the starting boundaries.
Junction A is the start, so it is labeled 1. The bottom edge can only be reached by walking straight East, so all junctions along the bottom are labeled 1. Similarly, the leftmost edge can only be reached by walking straight North, so all junctions along the left are labeled 1.
Step 3: Propagate the addition through the grid.
Total routes avoiding X = 11 routes
Why this works: Instead of memorising complex permutations and combination formulas (which fail when road blocks are introduced), the Addition Principle allows students to solve highly customized combinatorics questions using simple, error-free addition.
PSLE Connection: Non-routine pattern drawing or pathway problems appear in both Paper 1 (Section B) and Paper 2 of the PSLE. Students who can list and count systematically using node addition are far more likely to get these correct than those who try to list paths manually in writing.
Geometry questions in primary school exams do not require trigonometry. Instead, they require spatial reasoning, specifically the ability to break complex shapes down into simpler building blocks and analyze overlapping areas.
The figure below shows a square ABCD with a side length of 14 cm. Two quarter circles are drawn inside the square. One is centered at D (covering corner A to C), and the other is centered at B (covering corner C to A). Find the area of the shaded leaf-shaped region in the middle. (Take π = 22/7)
Step 1: Calculate the area of the entire square.
Area of square = 14 cm × 14 cm = 196 cm²
Step 2: Calculate the area of one quarter circle.
The radius of the quarter circle is equal to the side length of the square (14 cm).
Area of 1 Quarter Circle = 1/4 × π × r²
Area = 1/4 × 22/7 × 14 × 14 = 154 cm²
Step 3: Look at the overlapping region.
If we add the areas of the two quarter circles together, the leaf-like overlapping region is counted twice, while the unshaded corners are counted once.
Therefore, the sum of the two quarter circles is equal to the area of the square plus the area of the overlap:
Sum of Quarter Circles = Area of Square + Shaded Overlap
154 cm² + 154 cm² = 308 cm²
Step 4: Subtract the square's area to find the overlap.
Shaded Overlap Area = 308 cm² - 196 cm² = 112 cm²
Why this works: Conceptualizing shapes as overlaps (`A + B - Overlap = Total`) is much faster and mathematically cleaner than attempting to calculate the area of the leaf shape directly using circular segments.
PSLE connection: Area questions in Paper 2 may reward students who split a complicated figure into familiar shapes and show clear working. This original example uses circles and is enrichment; it should not be treated as a prediction of a future PSLE question.
Q: What is the National Mathematical Olympiad of Singapore (NMOS)?
NMOS is a NUS High School initiative in partnership with the Singapore Mathematical Society. The organiser's current rules say it is open to Primary 5 students, with registration through participating primary schools. For the latest eligibility and dates, use the official NMOS pages linked above.
Q: How does Olympiad Maths help with PSLE Mathematics?
They are different assessments. NMOS practice can develop reasoning, pattern spotting and clear working, but it does not replace learning the MOE Primary Mathematics syllabus or practising the SEAB PSLE format.
Q: How can a student prepare for NMOS?
Practise without a calculator, start with official sample material or reputable competition-style problems, explain each step, and review errors. Keep school Mathematics fundamentals strong, and confirm the current rules and registration process with the school.
Q: When were the NMOS 2026 results released?
The NMOS 2026 competition was held on 14 July 2026. Under official rules, results are released within 10 working days directly to participating primary schools (dispatched by approximately late July to early August 2026). Individual results are not posted online publicly — contact your child's primary school Mathematics department to receive official results and certificates.
Q: How are NMOS Gold, Silver and Bronze awards allocated?
The organiser's awards page lists Gold, Silver, Bronze, Honourable Mention and Participation certificates, but does not publish a fixed percentage allocation there. Detailed results are made known through the respective schools.
Q: Does an NMOS award guarantee DSA admission?
No. The official FAQ says NMOS gives students a platform to showcase mathematical aptitude and that outstanding performance will be noted. It does not guarantee a DSA offer. Families should check the target secondary school's current DSA criteria and application process.
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Many mathematical thinking skills developed through Olympiad-style problems are also valuable for solving challenging PSLE Maths questions. Continue practising similar problem-solving skills in PSLE Hero.
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