Every card in IMC 2016

Its digit sum is 45, a multiple of 9

HintAdd up 1 + 2 + … + 9 and apply the test for three and nine.

WhyThe digits 1 to 9 add to 45, so every such number is a multiple of 9 — and 9 is a proper factor, so none of them is prime. This settles a whole family of competition questions: when the digits are a fixed set, the digit sum is fixed too, so divisibility by 3 and by 9 is decided before a single digit is placed. It also explains why rearranging the digits of a number never changes its remainder on division by 9.

Lowest common multiple

HintTwo events repeat on different cycles; when do they first happen together?

WhySomething that happens every 4 days and a weekday that comes round every 7 days next coincide after the lowest common multiple of 4 and 7, which is 28 days. When the two numbers share no factor, the LCM is simply their product; when they do, it is smaller.

36

HintMetres per second to metres per hour first, then to kilometres.

WhyThere are 3600 seconds in an hour and 1000 metres in a kilometre, so a speed in m/s becomes metres per hour when multiplied by 3600 and kilometres per hour when that is divided by 1000 — the whole conversion is × 3.6. Going the other way, divide by 3.6: 90 km/h is 25 m/s. For an approximate answer, round the awkward numbers first.

720

HintCut it into triangles from one corner and count them.

WhyFour triangles from one vertex, each 180°, gives 720° — the formula 180(n − 2) with n = 6. When an angle is marked round the outside of a vertex, swap it for 360° minus it (an exterior angle proper is 180° minus the interior), then use this total.

They are parallel

HintCo-interior angles behave this way for exactly one kind of pair of lines.

WhyAngles that add to 180° between two lines and a transversal mean the lines are parallel — and if the other pair does not add to 180°, those sides are not. One pair parallel and the other not is the definition of a trapezium, which is how a quadrilateral can be classified from its angles alone.

Contains 2³ × 3

HintFactorise the divisor first, then look for every one of its prime powers.

WhyA number is a multiple of d exactly when its prime factorisation contains every prime power in d's. For 24 that means at least three 2s and at least one 3; 2² × 3² × 5 fails, however large the other powers are. The same check works for any divisor you can factorise.

Big circle minus small circle

HintTwo discs, one take-away.

WhyRing areas are a subtraction of two circle areas: π(R² − r²), which factorises as π(R + r)(R − r). Keep the π as a common factor to the end — it usually cancels or stays in the answer.

mode; median; mean

HintMost common, middle, and fair share — three different words.

WhyQuestions that give a mode, a median and a mean and ask for the smallest or largest possible value are logic puzzles built on these definitions: a mode must appear at least twice, a median splits the ordered list in half, and a mean fixes the total. Each fact forces something about the list.

Perpendicular bisectors of each other

HintTwo lines through the centre; think about the symmetry of the shape.

WhyThe diagonals of a rhombus cross at right angles and each cuts the other in half, so they carve the shape into four congruent right-angled triangles. Two circles of equal radius that cross: their centres and crossing points form a rhombus, so the line of centres and the common chord bisect each other at right angles.

Their own speed plus the walkway's speed

HintBoth motions carry the person forward at once.

WhySpeeds add, times do not: convert both to speeds (distance ÷ time), add, convert back. Walk against the walkway instead and the speeds subtract: walking speed minus walkway speed, which comes out negative — carrying you backwards — if the walkway is faster.

k²

HintLength counts once, area twice.

WhyDoubling every length quadruples the area — for a hexagon, a triangle, or any shape at all.

Take out the common factor

HintShift a power of ten so both products share a piece, then factorise.

WhyNon-calculator papers plant these on purpose: multiplying out is slow and error-prone, but rewriting one number as the other times a power of ten turns the whole thing into (common factor) × (a simple subtraction). If a calculation looks brutal, look for what the two halves share.

Its distance from the target number

HintDon't compare the fractions to each other — compare each one to something else first.

WhyComparing 5/6 and 8/9 directly is slow. Subtract each from the target instead: 1 − 5/6 = 1/6 and 1 − 8/9 = 1/9. The smaller gap wins — and when every gap has a numerator of 1, a bigger denominator means a smaller gap, so you can rank them without doing the subtraction at all.

One number is even and the other odd, so their squares are even and odd, and even + odd = odd

HintSquaring preserves evenness or oddness — think about what that means for two neighbouring integers.

WhySquaring never changes parity: an even number squared is even, an odd number squared is odd. Any two consecutive whole numbers are one even and one odd, so their squares are one even and one odd too — and even + odd always gives an odd total. That doesn't make the sum prime, but it does rule out 2 as a factor before you check anything else.

Double it

HintThe midpoint splits the trip into two matching halves.

WhyBy definition, a midpoint is equally far from both ends, so the distance from it to either end is exactly half the total. Once you know that one distance, multiply by two to get the whole journey — you never need to know where the two ends actually are. This shortcut also works for the midpoint of any line segment, not just a journey.

The whole is three times that difference

HintThe smaller share equals exactly one 'part' of the ratio — what does one part equal?

WhySplitting something in the ratio 1 : 2 means the smaller share is exactly one ratio-unit and the larger is two, so the gap between them equals one unit. If you're told that gap, you instantly know the size of one unit — and the whole is three units, so multiply the gap by three. The same idea rescues any ratio problem that gives you a difference instead of a total.

Count exactly halfway around the ring

HintPicture walking round the band of faces until you reach the far side.

WhyWhen a net folds up, a strip of faces that closes into a loop wraps around the solid like a belt. The face directly opposite any face in that belt is the one reached by travelling exactly half the belt's length in either direction — you don't need to build the whole solid to work it out, just count round the flat net. The same trick identifies opposite faces on a cube's net when the net is drawn as a strip of four squares.

Add them to get twice the larger; subtract to get twice the smaller

HintTreat the sum and the difference as two equations you can combine.

WhyAdding 'sum' and 'difference' cancels out the smaller number, leaving twice the larger one; subtracting them cancels the larger, leaving twice the smaller. Halve each result and you have both numbers — without ever setting up x and y from scratch. This is the elimination method for simultaneous equations, applied to the simplest possible pair.

Sum every number once for each line it lies on, then divide by the number of lines

HintWhatever sits on two of these spots gets added in twice before you share the grand total out.

WhyThis kind of diagram repeats some values across more than one line, so counting every entry once per line it touches, then adding it all up, double-counts on purpose. Divide that grand sum by how many lines there are and you get the shared line-total — without ever working out where a single missing number goes. It's the same 'total then share equally' logic behind finding a mean.

1.5

HintTurn the percentage words into a decimal multiplier first.

Why'50% more' is a 50% increase, and any percentage increase becomes a single multiplier: 100% (the original) + 50% (the increase) = 150%, which is ×1.5. The same idea gives '20% more' as ×1.2 and '50% less' as ×0.5 — always start from 100% of the original.

No — an area is already built from the squared value

HintAsk what the final quantity you want is actually made of.

WhyPythagoras' theorem naturally produces a squared length (like x²) partway through working. If the thing you actually need — an area, or another squared quantity — is built from that same square, square-rooting to find x and then squaring it again is wasted work. Stop as soon as you have the value you need.

2

HintBoth legs are equal, so Pythagoras adds the same squared length to itself.

WhyPythagoras says hypotenuse² = leg² + leg² for a right-angled isosceles triangle, since both shorter sides are equal. That's leg² + leg² = 2 × leg², so the square on the hypotenuse is always exactly double the square on either leg — a handy shortcut whenever a right-angled isosceles triangle turns up.

Express both areas using one shared unknown length, then simplify the ratio

HintGive the smallest measurement in the diagram a single letter and build everything from it.

WhyPick the shortest length in the figure and call it a single letter, then write every other length in the diagram as a multiple of it — often using Pythagoras or similar-triangle relationships to connect them. Once both areas are algebraic expressions in that one letter, the letter itself cancels out of the ratio, so you never need to know its actual value.

The block length must divide evenly into the number of people, or the pattern doesn't join up

HintPicture the repeating sequence wrapping around and meeting its own start.

WhyIf a claim forces a repeating cycle of, say, three answers before it repeats, that block of three has to fit a whole number of times around the circle — otherwise the pattern breaks where it should join back to the start. So only cycle lengths that are factors of the total number of people are possible, and checking each valid cycle length is how you find every consistent arrangement, including the ones with the most and fewest truth-tellers.

360° ÷ n

HintDivide a full turn equally among every shape in the set.

WhyCongruent shapes arranged with full rotational symmetry around a shared centre are evenly spaced in angle as well as in position, so the angle from the centre to one shape's centre, then round to the next, is always a full turn divided by how many shapes there are. This locks in extra angle facts for free wherever that kind of symmetry appears in a diagram.