IMC 2026

2

HintTwo staircases make a rectangle.

The whyTriangular numbers: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91. The mean of 1 to n is (n + 1) ÷ 2, so those means grow by a half each step.

It factorises as (n − 1)(n + 1)

HintWhat is 1 the square of?

The why(n − 1)n(n + 1) is a product of three consecutive numbers, so it is always a multiple of 6; for n not divisible by 2 or 3, (n − 1)(n + 1) is a multiple of 8 and of 3, hence of 24 — which is why 17, 19 and 23 work while 21 and 27 do not.

Clear the fractions first

HintA ratio survives one particular operation on every part.

The why1 : ½ : ⅓ is 6 : 3 : 2 (multiply by 6), eleven parts in all. A ratio is unchanged when every part is multiplied by the same number — so fractions and decimals never need to stay in it.

Equal to the rectangle's

HintSlide every inward edge outwards.

The whyEach horizontal edge projects onto the top or bottom of the enclosing rectangle, each vertical edge onto a side, with nothing double-counted — so a staircase shape has perimeter 2(width + height). Add any sloping edges separately.

a√2

HintHalf a square, cut corner to corner.

The whySo on a figure whose angles are all multiples of 45°, a sloping edge is √2 times its horizontal (= vertical) span, and two sloping edges that together rise 3 have total length 3√2. Split the perimeter into level, upright and sloping parts.

(√3 ÷ 4) s²

HintHalf the base times a height found by Pythagoras.

The whyHeight = (√3 ÷ 2) s, so area = ½ × s × (√3 ÷ 2) s. A parallelogram's area is base × perpendicular height — never base × slant side. Ratios of areas built from the same √3 cancel it, leaving a whole-number ratio.

Two equal halves

HintWhat does the line change about the two pieces, and what does it not?

The whyEqual bases plus a shared perpendicular height means equal areas. Two such lines from two vertices split the triangle into four pieces you can pair up.

3xy

HintExpand the cube and collect the mixed terms.

The whyWith x + y = 5 and x³ + y³ = 95: 125 = 95 + 15xy, so xy = 2 — no need to find x or y. Its square cousin: (x + y)² = x² + y² + 2xy.

100

HintWrite the percentage as a fraction in lowest terms and look at the bottom.

The why97/100 × N is a whole number only when 100 divides N, because 97 and 100 are coprime. 90% is different: 9/10 × N only needs N to be a multiple of 10. Reduce the percentage to lowest terms and read the denominator.

Division by the number of orders

HintThree items can be lined up in how many ways?

The whyPicking three things in order gives n × (n − 1) × (n − 2) sequences, and each set of three appears in 3 × 2 × 1 = 6 of them, so divide by 6. Then count the favourable sets systematically and divide.

1; even

HintLargest digit plus largest digit plus a carry — how big can that be?

The why9 + 9 + 1 = 19, so a carry is never more than 1; a column total of 2A or 2A + 1 tells you at once whether a carry came in; a leading digit is never 0. Three facts crack most letter-sum puzzles.

Peel from the outside inwards

HintWhich layer was built last?

The whyMove the outer constant, take reciprocals, and the next layer appears; three or four repeats reach x. Simplifying (rather than solving) goes the other way, from the innermost fraction out.

1440

HintMultiply the minutes in one hour by the hours in one day.

The whyA clock's minute hand completes 60 minutes per hour, and a day holds 24 hours, so 60 × 24 = 1440 minutes make a full day. Reducing any 'minutes later' problem by that number first (taking the remainder) shows how far into the next day's cycle you land — far quicker than counting forward minute by minute.

Its cube root

HintWhat is the inverse of cubing a number?

The whyFor (n − 1) × n × (n + 1) = V, this is close to n³ = V since n dominates the product, so ∛V pinpoints where to start testing consecutive integers — for example ∛990 is close to 10, so trying 9, 10 and 11 finds the answer fast rather than guessing from 1 upward.

1/(n + 1)

HintRewrite the fraction as 1 minus something small.

The whyTo solve n/(n + 1) > k, rearrange to 1/(n + 1) < 1 − k, then n + 1 > 1/(1 − k) — for example, solving n/(n + 1) > 0.8 needs 1/(n + 1) < 0.2, so n + 1 > 5, meaning n ≥ 5 is the first term past 0.8.

A prime number's partner — it can only be 1

HintWhich of the numbers has just two possible factors?

The whyA prime number's only factors are 1 and itself, so in a puzzle demanding an exact whole-number pairing it must pair with 1 — fixing that link narrows every other choice. After the prime, look for a number with an unusually restrictive property (the only multiple of a particular value in the list) to fix the next link, and keep working outward from the most constrained numbers.

Equals the sum of the two remote interior angles

HintExtend one side past a vertex, then compare with the two non-adjacent corners.

The whyThis follows directly from the angle-sum rule: exterior + its adjacent interior angle = 180°, and the triangle's own angle sum is also 180°, so the exterior angle must equal the other two angles combined. It's the fastest way to angle-chase across two triangles that share a vertex.

The perpendicular from the apex to the midpoint of the base

HintIt cuts the shape into two congruent right-angled pieces.

The whyThat perpendicular bisects the base, creating two right-angled triangles that share the isosceles triangle's equal side as their hypotenuse and half the base as one leg — Pythagoras then finds the height. For sides 13, 13 and 10, the base splits into two lengths of 5, giving a 5-12-13 right triangle and a height of 12.

The formula only needs the radius squared, not the radius

HintPythagoras already produces exactly that value in one step, no root needed.

The whyIf the radius is the hypotenuse of a right-angled triangle with two known legs, Pythagoras gives radius² directly as leg₁² + leg₂² — exactly what the area formula needs, so there is no need to take an unnecessary square root. This is especially useful when subtracting two sectors that share a corner point: the squared-radius values often combine into a clean multiple of π.

It must match a value already present in the set

HintWhich numbers on the list are being repeated?

The whyThe mode is whichever value appears most often. A brand-new value starts at frequency 1, so it can only become the mode by matching a value already present — that's necessary, but not automatically enough. In the set 1, 1, 1, 2, 3, an extra 2 only reaches frequency 2, still short of 1's frequency of 3. When every value in the set currently appears equally often, though, one more copy of any of them takes sole lead, so the number of valid choices equals how many distinct values are present. Contrast the mean, which needs one exact algebraic value, and the median, which depends on where the new value slots into sorted order.

It shrinks at a steady, constant rate

HintEqual fixed speeds create a straight-line graph, not a curve.

The whyBecause the combined closing speed never changes, the gap between them is a straight-line (linear) function of time — so two known gap-and-time pairs are enough to work out that speed and then extrapolate to any other time, including right back to the very start.

AAS

HintTwo matching angles are already fixed before you even compare a length.

The whyWhen two lines cross, the angles on opposite sides of the crossing point are always equal, so a shared crossing point plus a right angle in each triangle gives two matching angles for free. Add one matching side length (not necessarily between those angles) and that is Angle-Angle-Side — enough to prove the triangles congruent and make their unknown sides equal too.

The two base angles, opposite the equal sides

HintLook across from the pair of matching lengths — what's there?

The whyBecause the triangle is symmetric about the line from its apex to the midpoint of the base, the two base angles mirror each other and so must be equal. Spotting this quickly is what lets you angle-chase through a figure built from several nested isosceles triangles, since marking one base angle often reveals several others at once.

Split the casework around one clearly defined bottleneck choice

HintPick a single busy junction and count arrivals from each neighbour separately.

The whyA route that must visit every point exactly once is easy to miscount by browsing at random. Anchoring the casework on one well-defined decision — for example, 'which neighbouring point does the route use immediately before reaching a particular busy junction' — guarantees every valid route falls into exactly one case, because that choice exists and is unique for every possible route.

horizontal; vertical

HintTrace the outline all the way round — you end up back where you began.

The whySplitting a closed polygon's perimeter into rightward/leftward and upward/downward pieces means each pair must cancel to zero — otherwise you would not return to your starting point. This turns unknown side lengths in a many-sided figure (even one with several 45°-sloped edges) into a solvable pair of equations.