Every card in IMC 2019

By two faces

HintNothing is added; something is hidden — on both sides.

WhyEvery join hides one face from each cube, so a shape built from k cubes with j face-to-face joins has surface area 6k − 2j. The shape with the most joins has the smallest surface area — count joins, not exposed faces.

The units digit of n

HintCube 12 and 22 and compare the endings.

WhyThe same is true of every power, so a cube ends in 8 only when the number cubed ends in 2, and a square ends in 9 only when its root ends in 3 or 7. Unlike squares, cubes can end in any digit — each of 0 to 9 appears exactly once among the cubes of 0 to 9.

The triangular numbers

Hint1, then 1 + 2, then 1 + 2 + 3 …

WhyThe nth triangular number is n(n + 1)/2 — the sum of the whole numbers from 1 to n. Their parities repeat in blocks of four — odd, odd, even, even — so the parity of any term can be read off from n alone without adding anything up.

positive; negative

HintSame signs, different signs, and the one number that swallows everything.

WhyThe sign rules apply to × and ÷ only — adding two negatives gives a negative, and subtracting a negative is adding. Order-of-operations questions packed with brackets and minus signs are testing whether you apply each rule to the right operation.

9

HintA thousand millions: add the zeros.

WhyA billion has nine zeros. Dividing it by a smaller round number is best done by cancelling powers of ten first — 10⁹ ÷ (4 × 10⁴) = 10⁵ ÷ 4 = 25 000 — so the zeros are never miscounted.

multiplies

HintWrite (a²)³ out in full and look at how many a's you get.

WhyCompare 8⁵ with 4⁷ by writing both as powers of 2: 2¹⁵ against 2¹⁴. Keeping everything as a power of the smallest base is the whole technique for index questions with no calculator.

Watch the square's colour

HintEvery single move flips something that only has two states.

WhyColour the board like a chessboard: every move — a step to a neighbour, or a reflection in a line running between rows or columns — swaps black for white. After an odd number of moves the token is on the opposite colour to where it started; after an even number, the same colour. Parity halves the search before you list anything.

Factorise x out

HintBoth terms contain x — what does that let you write?

WhyFrom ab = 2a + b: a(b − 2) = b, so a = b ÷ (b − 2); check that b = 2 is impossible in the original before dividing by the bracket.

7/9

HintNot tenths — the denominator that makes one digit repeat forever.

WhyA single recurring digit d is d/9; a recurring pair ab is ab/99. Add recurring decimals as fractions — the digit-by-digit method has no right-hand end to start from.

Take the square root

HintAreas came from lengths by one operation; go back the other way.

WhyAreas of similar shapes are in the ratio of the squares of their sides, so going from areas back to sides means taking a square root: areas 1 : 7 means sides 1 : √7 — not 1 : 3.5.

13

HintThree consecutive odd primes.

WhySo 123 123, 987 987 — any block-repeated six-digit number — is divisible by 7, 11 and 13, because it equals the block × 1001.

Longest side < sum of the other two

HintTry 2, 3 and 10 with real sticks.

WhyThe triangle inequality — and every length must be positive. When side lengths are given as expressions in n, solving "two sides equal" is only half the job: each solution must then be checked against both conditions, and any that fails is discarded.

n ÷ 10

HintOne tenth is one part in ten — so dividing does the job.

WhyThe same idea extends to any named fraction of a whole: 'n hundredths' is n ÷ 100, 'n thousandths' is n ÷ 1000 — each extra zero in the name adds one more division by ten, which is why place-value wording like this is really just a disguised calculation.

In the same ratio as the percentages

HintYou don't need the total quantity at all to compare them.

WhyTwo parts sized 60% and 40% of a whole are in ratio 60:40 = 3:2, whatever the whole actually is — because both percentages are measured in the same 1% unit, so dividing one by the other cancels the whole out.

List every case in one fixed systematic order

HintWork smallest to largest (or first position to last) and never skip ahead.

WhyFaster still: split by which value can legally sit in the most restricted position, count the free choices left for the rest, then multiply — this turns a long list into a short multiplication.

left to right

Hint× and ÷ sit on the same rung of the order-of-operations ladder.

WhyThe same left-to-right rule governs a chain of only + and −. It's why 20 ÷ 4 × 5 must be read as (20 ÷ 4) then × 5, giving 25 — whereas grabbing the multiplication first, as 20 ÷ (4 × 5), would give 1. Division never waits its turn behind multiplication: they share a rung, so position on the line decides.

The largest number of regions meeting at one point

HintFind the busiest shared corner — every region touching it needs its own colour.

WhyIf k regions all meet at a single point, no two of them can share a colour, so at least k colours are required. Showing k is also enough still needs a working colouring with exactly k colours to be exhibited.

Turn each mean back into a total by multiplying by the count

HintUndo the averaging before you subtract anything.

WhyOnce every running mean becomes a running total (mean × count), a newly added number is just this total minus the previous total — because 'total = mean × count' always reverses 'mean = total ÷ count'.

Test each possible total for self-consistency

HintGuess a count, then check whether that guess makes exactly that many claims true.

WhyA guessed total only works if the number of true statements it produces equals the guess itself — check every candidate this way instead of reasoning about individual speakers one at a time.

s√3 / 2

HintHalve the base, then use the 30-60-90 ratio or Pythagoras on the two new right triangles.

WhyThe altitude bisects the base and the apex angle, giving a right triangle with base s/2 and hypotenuse s; from sin 60° = height ÷ s, height = s × sin60° = (s√3)/2 — a ratio worth knowing cold for no-calculator problems.

Cross-multiply the ratio into bx = ay

HintClear the fraction before you combine it with the other equation.

WhyOnce the ratio is rewritten as a fraction-free linear equation, solve the pair exactly as any simultaneous equations — express one variable in terms of the other and substitute into the sum equation.

Replace every term with its largest or its smallest

HintYou only need to know which sum is bigger, not the exact value of either.

WhyReplacing each term by the biggest gives an upper bound (count × biggest term); by the smallest gives a lower bound. Comparing bounds often settles which sum wins without a single fraction being added — vital with no calculator allowed.

Draw legs a and b, then use Pythagoras for the hypotenuse

HintTan already hands you two sides; the third comes from the theorem you already know.

WhyWith opposite = a and adjacent = b, the hypotenuse is √(a² + b²), so sin θ = a ÷ √(a²+b²) and cos θ = b ÷ √(a²+b²) follow at once — and scaling a and b by the same factor never changes the resulting sine or cosine.

Take each remainder off its own number, then factor the difference of those two multiples of N.

HintEliminate the unknown quotient by subtracting one division statement from the other.

WhyIf a = Nj + r1 and b = Nk + r2, then b − a = N(k − j) + (r2 − r1), so N divides that quantity once the remainder gap is removed. List its factors — N must be one of them, and also bigger than every remainder given.

90

HintThe radius meets the tangent exactly where they touch.

WhyThis right angle is why dropping a perpendicular from a circle's centre is the standard way to test whether a line is tangent, and why right-angled triangles appear naturally wherever a radius meets a tangent line.