Every card in IMC toolkit
The whole deck, in order — so you can read it through before your child ever sees it.
- An IMC geometry question gives two angles and asks for a third, far across the diagram. What is the habit that gets you there?
Angle chasing — write every angle you can find on the diagram
HintFill in the ones you are not asked for as well; they are the stepping stones.
WhyMark each angle as you deduce it — angles on a straight line, in a triangle, on parallel lines — and the target usually falls out after three or four steps. A chain of small known facts is what competition geometry is.
- A shaded region is a square with a circle cut out of it. How do you find its area?
Subtract the circle's area from the square's
HintTwo shapes you can handle separately, then one take-away.
WhyArea by subtraction is the workhorse of competition geometry: find the big shape, find the hole, take one from the other. A circle of radius 3 inside a square of side 6 leaves 36 − 9π, and the answer choices are usually left in terms of π so you never have to compute it.
- A right-angled triangle has shorter sides 5 and 12. Which number should you recognise instantly?
13 — the 5, 12, 13 Pythagorean triple
HintThere is a famous trio that begins with these two numbers.
WhyCompetition papers reuse a handful of whole-number right-angled triangles: 3-4-5, 5-12-13, 8-15-17 and their multiples such as 6-8-10. Spotting one saves the square root entirely and often unlocks a length the question never stated.
- Two similar triangles have sides in the ratio 1 : 3. What is the ratio of their areas?
1 : 9
HintArea is two-dimensional, so the scale factor is used twice.
WhyLengths scale by k, areas by k squared, volumes by k cubed. An enlargement by 3 makes every area nine times bigger — a fact that appears in IMC questions about shadows, maps and nested shapes far more often than the word "similar" does.
- Written as a product of primes, 360 = 2³ × 3² × 5, so it has (3 + 1)(2 + 1)(1 + 1) = ____ factors.
24
HintAdd one to each power, then multiply the results together.
WhyCounting factors from a prime factorisation is a competition staple: each prime can appear in a factor anywhere from zero times up to its full power, so the choices multiply. It also tells you a number is a perfect square exactly when every power is even.
- How do you find the sum of the whole numbers from 1 to 100 in one line?
Pair them up: 100 × 101 ÷ 2 = 5050
HintMatch the first with the last, the second with the second-last, and notice what each couple adds to.
WhyFifty pairs each adding to 101 gives 5050 — the formula n(n + 1) ÷ 2 for the sum of 1 to n. These triangular numbers turn up disguised as handshakes, dots in a triangle and the number of matches in a league.
- The question asks for the 100th term of a sequence you do not recognise. What is the plan?
Work out the first few terms and spot the pattern
HintSmall versions of the problem reveal the rule; then jump to the big one.
WhyTry small cases is the most reusable competition strategy there is. Three or four terms are usually enough to see the rule — but check one more before you trust it, and watch the off-by-one trap between "term 0" and "term 1".
- The answer choices are 2⁸, 2⁹, 2¹⁰, 2¹¹ and 2¹². What tool decides between them fastest?
The laws of indices
HintKeep everything as powers of 2 and combine the exponents rather than multiplying out.
WhyWhen every option is a power of the same base, never evaluate: 4³ × 8² becomes 2⁶ × 2⁶ = 2¹², and the exponent alone gives the answer. Add to multiply, subtract to divide, multiply for a power of a power.
- An IMC question ends "Which of the following must be true?" Which strategy fits?
Test each option with an extreme or awkward case
HintTry to break every statement using a value at the edge — zero, one, a negative, a huge number.
Why"Must be true" questions are won by counterexamples: one edge value that breaks a statement rules it out. Zero and one behave unusually under multiplication, negatives flip inequalities, and very large numbers show which term dominates.
- What is the units digit of 3¹⁰⁰?
1
HintLook at how the last digits of 3, 9, 27, 81, 243 behave.
WhyPowers of 3 end 3, 9, 7, 1, 3, 9, 7, 1 … a cycle of four. 100 ÷ 4 leaves remainder 0, which is the fourth position in the cycle, so the units digit is 1 — the same as 3⁴ = 81. Remainder 0 means the end of the cycle, not the start: the classic off-by-one trap.
- In the Intermediate Maths Challenge, a wrong answer to one of questions 16 to 20 loses ____ mark.
1
HintThe lighter of the two penalties.
WhyMarks: 5 each for 1–15, 6 each for 16–25, blanks 0, with 1 lost for a wrong 16–20 and 2 lost for a wrong 21–25. A blind guess on 21–25 loses marks on average — leave it blank unless you can rule out at least two of the five options (three left is already worth a guess). On 16–20 the penalty is small enough that even a blind guess comes out slightly ahead on average.
- The interior angles of a polygon with n sides add up to ____ degrees.
180(n − 2)
HintSplit the shape into triangles from one corner and count them.
WhyEvery polygon splits into n − 2 triangles from one vertex, each worth 180°, so a hexagon's angles total 720° and each angle of a regular hexagon is 120°. The exterior angles of any polygon always add to 360° — often the faster route.
1★ KS3-MATH-IMC-0001
2★ KS3-MATH-IMC-0002
3★ KS3-MATH-IMC-0003
4★ KS3-MATH-IMC-0004
5★ KS3-MATH-IMC-0005
6★ KS3-MATH-IMC-0006
7★ KS3-MATH-IMC-0007
8★ KS3-MATH-IMC-0008
9★ KS3-MATH-IMC-0009
10★ KS3-MATH-IMC-0010
11★ KS3-MATH-IMC-0011
12★ KS3-MATH-IMC-0012
Keep what you learn
Here, nothing is saved. In your child’s own sky every card is scheduled — it comes back just before they’d forget it — and the professor who wrote it is one tap away.