IMC 2016：全部卡片

Its digit sum is 45, a multiple of 9

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

为什么The 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

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

为什么Something 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

提示Metres per second to metres per hour first, then to kilometres.

为什么There 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

提示Cut it into triangles from one corner and count them.

为什么Four 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

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

为什么Angles 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

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

为什么A 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

提示Two discs, one take-away.

为什么Ring 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

提示Most common, middle, and fair share — three different words.

为什么Questions 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

提示Two lines through the centre; think about the symmetry of the shape.

为什么The 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

提示Both motions carry the person forward at once.

为什么Speeds 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

提示Length counts once, area twice.

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

Take out the common factor

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

为什么Non-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

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

为什么Comparing 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

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

为什么Squaring 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

提示The midpoint splits the trip into two matching halves.

为什么By 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

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

为什么Splitting 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

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

为什么When 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

提示Treat the sum and the difference as two equations you can combine.

为什么Adding '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

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

为什么This 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

提示Turn the percentage words into a decimal multiplier first.

为什么'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

提示Ask what the final quantity you want is actually made of.

为什么Pythagoras' 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

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

为什么Pythagoras 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

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

为什么Pick 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

提示Picture the repeating sequence wrapping around and meeting its own start.

为什么If 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

提示Divide a full turn equally among every shape in the set.

为什么Congruent 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.