20 original questions · foundation and bridge practice
IOQM Topic-wise Practice Questions: Number Theory, Algebra, Geometry and Combinatorics
Use these original, independently written practice problems to check method selection across the four major Olympiad domains. Attempt each question before opening the worked solution, and record every answer in two-digit form.
Allow a serious attempt before viewing the solution. Write the main idea, calculate carefully, record the two-digit answer and then compare the reasoning—not only the number. These are original practice questions, not official IOQM questions or a prediction of a future paper.
Questions 1–3
Number theory practice
Q1 · Congruence
Find the smallest positive integer n such that 7n leaves remainder 1 when divided by 20.
Show solution
Test the inverse of 7 modulo 20. Since 7×3=21≡1 (mod 20), the smallest positive n is 3. Answer: 03.
Q2 · GCD structure
Find gcd(2¹⁵−1, 2⁹−1).
Show solution
Use the Euclidean algorithm to see why the gcd is preserved: 2¹⁵−1 = 32767 and 2⁹−1 = 511. Now 32767 = 64 × 511 + 63, 511 = 8 × 63 + 7, and 63 = 9 × 7. The last nonzero remainder is 7, so the gcd is 7. Answer: 07.
Q3 · Divisors
How many positive divisors does 360 have?
Show solution
360=2³·3²·5. A divisor chooses exponents 0–3, 0–2 and 0–1, giving (3+1)(2+1)(1+1)=24. Answer: 24.
Questions 4–6
Algebra practice
Q4 · Symmetric expression
A nonzero real number x satisfies x+1/x=5. Find x²+1/x².
Show solution
Square the given relation: x²+2+1/x²=25. Therefore x²+1/x²=23. Answer: 23.
Q5 · Arithmetic progression
The first term of an arithmetic progression is 3 and the common difference is 2. Find the sum of its first 6 terms.
Show solution
The terms are 3,5,7,9,11,13. Pairing ends gives three pairs of 16, so the sum is 48. Answer: 48.
Q6 · Roots without solving
If r and s are the roots of t²−7t+10=0, find r²+s².
Show solution
By Vieta, r+s=7 and rs=10. Hence r²+s²=(r+s)²−2rs=49−20=29. Answer: 29.
Questions 7–9
Geometry practice
Q7 · Right-triangle area
A right triangle has perpendicular sides 6 and 8. Find its area.
Show solution
The perpendicular sides are base and height. Area=½×6×8=24. Answer: 24.
Q8 · Cyclic angles
ABCD is a cyclic quadrilateral and angle A is 112°. Find angle C in degrees.
Show solution
Opposite angles of a cyclic quadrilateral are supplementary, so C=180°−112°=68°. Answer: 68.
Q9 · Similarity and area
Two similar triangles have corresponding side ratio 3:5. If the smaller area is 27, find the larger area.
Show solution
Areas are in the square of the side ratio, 9:25. The larger area is 27×25/9=75. Answer: 75.
Questions 10–12
Combinatorics practice
Q10 · Handshakes
Ten people each shake hands once with every other person. How many handshakes occur?
Show solution
Each handshake is an unordered pair, so the count is C(10,2)=10×9/2=45. Answer: 45.
Q11 · Unordered selection
How many two-person teams can be selected from 8 students?
Show solution
Order does not matter, so choose 2 from 8: C(8,2)=8×7/2=28. Answer: 28.
Q12 · Inclusion–exclusion
How many positive integers not exceeding 100 are divisible by 4 or by 6?
Show solution
There are 25 multiples of 4 and 16 multiples of 6. Multiples of lcm(4,6)=12 were counted twice; there are 8. Thus 25+16−8=33. Answer: 33.
Questions 13–20 · bridge to mixed reasoning
Combine ideas in a longer solution
The first twelve questions are foundation checks. These eight bridge questions ask for more than a direct formula. They are original exercises, not official past-paper questions or a prediction of IOQM difficulty. Try a hint before opening the full solution.
Q13 · Remainder cycle
Find the remainder when 72026 is divided by 100.
Show a hint
Calculate 7² and 7⁴ modulo 100 before considering the large exponent.
Read the step-by-step solution
Idea: a short power cycle replaces a huge calculation.
- 7² = 49 and 7⁴ = 2401, so 7⁴ leaves remainder 1 on division by 100.
- Write 2026 = 4 × 506 + 2. Thus 72026 = (7⁴)506 × 7².
- The remainder is 1506 × 49 = 49.
Answer: 49. Check the modulus: a last-digit cycle alone would only give a remainder modulo 10.
Q14 · Divisibility with consecutive integers
How many integers n with 1 ≤ n ≤ 60 satisfy 12 | n(n + 1)?
Show a hint
Check divisibility by 3 and by 4 separately. The condition repeats every 12 integers.
Read the step-by-step solution
Idea: list the allowed remainders, then count complete blocks.
- For divisibility by 3, n must leave remainder 0 or 2 modulo 3. If n leaves remainder 1, neither n nor n + 1 is divisible by 3.
- For divisibility by 4, n must leave remainder 0 or 3 modulo 4. In the other cases the product has only one factor of 2.
- Checking remainders 0–11 gives the four possibilities 0, 3, 8 and 11 modulo 12.
- Each block of 12 consecutive integers contains four valid values. There are five blocks from 1 to 60, giving 5 × 4 = 20.
Answer: 20. Because 3 and 4 are coprime, satisfying both conditions is sufficient for divisibility by 12.
Q15 · Turn fractions into factors
How many ordered pairs of positive integers (x, y) satisfy 1/x + 1/y = 1/6?
Show a hint
Multiply by 6xy, then add the same constant to both sides to form a product.
Read the step-by-step solution
Idea: convert the equation into a divisor-counting problem.
- Multiplying gives 6x + 6y = xy, hence (x − 6)(y − 6) = 36.
- Since 1/x < 1/6 and 1/y < 1/6 in the original equation, both x and y exceed 6.
- Every positive divisor d of 36 gives exactly one ordered pair x = d + 6, y = 36/d + 6.
- Since 36 = 2² × 3², it has (2 + 1)(2 + 1) = 9 positive divisors.
Answer: 09. The question asks for ordered pairs, so swapping unequal x and y counts again. The pair (12, 12) is counted once.
Q16 · Understand a recurrence
A sequence satisfies a₀ = 2, a₁ = 3 and an+2 = 3an+1 − 2an for n ≥ 0. Find the remainder when a₁₀ is divided by 100.
Show a hint
Subtract an+1 from both sides and examine consecutive differences.
Read the step-by-step solution
Idea: the differences follow a simpler rule than the terms.
- Rearranging gives an+2 − an+1 = 2(an+1 − an).
- The first difference is a₁ − a₀ = 1. The next differences are 2, 4, 8, and so on.
- Therefore a₁₀ = 2 + (1 + 2 + 4 + ··· + 2⁹) = 2 + (2¹⁰ − 1) = 1025.
- The remainder on division by 100 is 25.
Answer: 25. A few initial terms suggest the pattern; the difference equation explains why it continues.
Q17 · Find a hidden height
Triangle ABC has AB = 13, AC = 15 and BC = 14. Find its area.
Show a hint
Drop perpendicular AD to BC and write Pythagoras in the two right triangles.
Read the step-by-step solution
Idea: use the same height in two equations.
- Put B = (0, 0), C = (14, 0) and A = (x, h), with h > 0. Then x² + h² = 169 and (14 − x)² + h² = 225.
- Subtracting eliminates h²: 196 − 28x = 56, so x = 5.
- Now h² = 169 − 25 = 144, giving h = 12. Since 0 < 5 < 14, the perpendicular foot lies on BC as drawn.
- The area is ½ × 14 × 12 = 84.
Answer: 84. The height is perpendicular to the chosen base; the sloping side 13 is not the height.
Q18 · A bisector and two areas
In triangle ABC, AB = 10 and AC = 15. The internal bisector of angle BAC meets BC at D. If the area of ABC is 60, find the area of ABD.
Show a hint
First find BD : DC. The two small triangles have the same perpendicular height from A.
Read the step-by-step solution
Idea: a side ratio becomes an area ratio when the heights agree.
- By the internal angle-bisector theorem, BD : DC = AB : AC = 10 : 15 = 2 : 3.
- Triangles ABD and ACD share the altitude from A to line BC. Their areas are therefore in the ratio 2 : 3.
- ABD occupies 2/(2 + 3) = 2/5 of the total area.
- Its area is (2/5) × 60 = 24.
Answer: 24. An angle bisector does not generally divide the opposite side into equal halves.
Q19 · Count after fixing a condition
How many four-digit positive integers can be formed from the digits 1, 2, 3, 4, 5 without repetition, if the integer is divisible by 5?
Show a hint
Which digit must be last? Count the remaining positions only after fixing it.
Read the step-by-step solution
Idea: apply the restriction before multiplying choices.
- A number divisible by 5 ends in 0 or 5. Since 0 is unavailable, the last digit must be 5.
- The thousands digit has four choices from 1, 2, 3, 4.
- There are then three choices for the hundreds digit and two for the tens digit.
- The count is 4 × 3 × 2 = 24.
Answer: 24. No leading-zero adjustment is needed because zero is not in the digit set.
Q20 · Count paths by subtraction
A path starts at (0, 0) and ends at (3, 3). Each step is one unit right or one unit up. How many such paths avoid the point (1, 1)?
Show a hint
Count all paths, then subtract those that pass through the forbidden point.
Read the step-by-step solution
Idea: the unwanted paths split naturally at (1, 1).
- Every complete path has three right and three up steps, so there are 6!/(3!3!) = 20 paths in total.
- There are two paths from (0, 0) to (1, 1): right then up, or up then right.
- From (1, 1) to (3, 3), choose two right steps among four steps: 4!/(2!2!) = 6 ways.
- Exactly 2 × 6 = 12 paths pass through (1, 1). Subtracting gives 20 − 12 = 8.
Answer: 08. Monotone paths cannot revisit (1, 1), so each unwanted path is counted exactly once.
Second pass
Remove the topic labels and test recognition
On a later day, copy only the question statements into a mixed order. Without the section labels, write the first useful cue beside each:
- inverse modulo 20;
- exponent gcd identity or Euclidean algorithm;
- prime-exponent choices;
- square a symmetric relation;
- pair an arithmetic progression;
- Vieta before solving;
- perpendicular base and height;
- cyclic supplementary angles;
- area scales as side ratio squared;
- unordered pairs;
- unordered selection;
- subtract an overlap.
The ability to choose the method without a chapter heading is closer to the real examination demand.
Error log
Classify the earliest cause of every miss
| Error type | Next action |
|---|---|
| Concept not known | Study the smallest relevant lesson and solve a graded set |
| Method not recognised | Write the cue and compare a disguised variation |
| Execution | Slow the algebra or arithmetic and add a verification step |
| Condition missed | Rewrite the theorem with its domain or configuration |
| Format/transcription | Practise two-digit recording and the OMR check routine |
Progression
Move from original drills to official mixed papers
- Finish the twelve foundation checks, then attempt the eight bridge questions without open solutions.
- Reattempt every miss after two or three days.
- Use the detailed topic hubs for the weakest domain.
- Solve selected official previous-paper questions.
- Later take a timed official paper under the current pattern.
Original practice builds specific skills; official papers are essential for authentic style, balance and difficulty.
Questions answered
Frequently asked questions
Are these official IOQM questions?
No. They are original independent practice questions. Use the linked HBCSE archive for official previous papers.
Do the questions follow IOQM topics?
They practise number theory, algebra, geometry and combinatorics, the four major Mathematical Olympiad areas.
Should I look at the solution immediately?
No. Make a serious independent attempt, write the main idea and only then compare the worked reasoning.
Why are answers written as two digits?
It reinforces current IOQM OMR practice, where numerical answers are recorded in two-digit form.
Do all answers fit the two-digit practice format?
Yes. The results are between 00 and 99, matching the numerical-answer recording habit used in current IOQM instructions.
How should I review an incorrect answer?
Find the earliest cause—concept, recognition, condition, execution or transcription—and choose a specific reattempt or repair.
What should I solve after this set?
Use the four topic hubs for weak areas, then move into official previous-paper questions and later timed papers.
Can I use these questions for a mock test?
Use them as a short diagnostic or topic drill, not as a prediction or full replica of an IOQM paper.
Next step
Attempt first. Diagnose second. Reattempt later.
Use this set to identify the weakest reasoning step, then move into the relevant topic hub and official paper archive.
MENTOR PERSPECTIVE
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