IMC 2017

2, 3, 7 or 8

HintFour of the ten digits never appear at the end of a square.

The whyThe last digit of n² depends only on the last digit of n, so ten small squares settle it: a square always ends in 0, 1, 4, 5, 6 or 9. Questions asking how many squares end in a certain digit, or whether a total could be a square, are decided by this list alone.

2

HintOdd plus odd is even, so one of the two primes must be the single even one.

The whyEvery prime except 2 is odd, and odd + odd = even, so an odd sum of two primes has to use the even prime. That turns the whole question into one primality test: 13 works because 13 − 2 = 11 is prime, while 51 fails because 51 − 2 = 49 = 7 × 7. The same parity argument shows 2 and 3 are the only two primes that differ by 1.

Multiplied by four

HintThe area formula uses the radius twice.

The whyArea = πr², so doubling r multiplies the area by 2² = 4 — and the region added by that doubling is three quarters of the bigger circle, a fact many shaded-fraction questions rest on.

The range

HintHow spread out the data is, in one number.

The whyThe range is a measure of spread, not an average. For a set of different whole numbers the smallest possible range is one less than the count, and only when the numbers are consecutive.

9

HintEach digit is worth ten times as much in one number as in the other.

The whyWith tens digit a and units digit b the number is 10a + b and its reversal is 10b + a; subtracting gives 9(a − b), so only the gap between the two digits matters. Writing digits as an expression is what unlocks every reversal puzzle.

3

HintBoth are small powers of the same number; find it before you touch the exponents.

The whyOnce both sides are powers of one base, the bases can be dropped and the exponents equated, using (aᵐ)ⁿ = aᵐⁿ on each side. The answer can be a fraction — a non-integer exponent is not a sign of a mistake.

540

HintSplit it from one corner and count the triangles.

The why180(n − 2) with n = 5. Any five-sided shape — irregular, even with a reflex corner — has this total. A regular pentagon therefore has interior angles of 108° and exterior angles of 72°.

Subtract the fractions

HintBoth levels are measured against the same container.

The whyThe amount used equals the difference of the two fractions of the full battery, so a common denominator gives it as one fraction — and the full capacity is the amount used divided by that fraction. Reverse-fraction questions are the unitary method in disguise.

5 and 6

HintWhich two square numbers sit either side of it?

The whyTrap a root between neighbouring squares: 25 < 30 < 36. To decide which side it is closer to, try the halfway value — 5.5² = 30.25, so √30 is just under 5.5. "Which is closest to" questions with a root in the answer are settled this way.

144

HintCut the shape into two triangles for the total, then share it among ten equal parts.

The whyWhen a quadrilateral's angles are given as a ratio, share 360° in that ratio: each part is 360 divided by the total of the ratio's parts. The same works for any polygon once you know its angle sum.

60; 120; 135

HintHow the tiles meet: triangles six to a point, hexagons three, octagons with squares.

The whyHalving an equilateral triangle of side 2 gives a right-angled triangle with sides 1, √3 and 2 — so the height of an equilateral triangle of side s is (√3/2)s and its area is (√3/4)s².

Its angle's share of 360°

HintThe bigger the turn at the centre, the bigger the slice — in exact proportion.

The whyCorner sectors of one radius at every vertex of a polygon add up to (angle sum ÷ 360) circles — half a circle for a triangle, one whole circle for a quadrilateral.

2 and 5

HintThink about which primes are hiding inside every power of ten.

The whyThis is why fractions with denominators like 4, 20 or 25 always terminate — each is built only from 2s and 5s — while a denominator with any other prime factor (3, 7, 11…) forces a recurring decimal. Recognising terminating denominators lets you convert several such fractions to decimals and add them by lining up place value, rather than hunting for a common denominator.

Compare the first digit where they differ

HintLook past the matching part — the answer sits at the next new value along.

The whyRecurring-decimal options often share their first several digits, so scan along until you find the first place they disagree — the larger value there decides which decimal is bigger, no matter how the pattern continues afterwards. Writing out a few extra repeats of the recurring block on each option makes this place easy to spot.

360

HintA full turn all the way around a point.

The whyThis is different from angles on a straight line (180°) or in a triangle (180°) — it's the 'full turn' fact, and it often combines with parallel-line angle rules (alternate or corresponding angles) when a diagram is built from several shapes meeting at one vertex.

k²:1

HintSquaring the side ratio gives the area ratio.

The whyFor similar figures, every length scales by k, so area — built from two lengths — scales by k². Volume, built from three lengths, would scale by k³. It's the same idea as a circle's radius doubling and its area quadrupling, just for any similar shape, not only circles.

Add the equations covering every other unknown once, then subtract from the total.

HintFind the combination of clues that leaves only your target hidden.

The whyThis works because summing the right subset counts every unknown except the target exactly once; subtracting that subset's total from the grand total cancels everything else and leaves the target alone. It turns what looks like a large system into one line of arithmetic — spotting the right subset is the whole skill.

a² + b² = c², where c is the hypotenuse and a, b are the other two sides

HintIt relates the triangle's longest edge to the two shorter ones, squared and added.

The whyIt only holds for a right-angled triangle, with c always the side opposite the right angle (the hypotenuse) — the longest side. Once you have any two sides you can find the third; for the hypotenuse you add the squares, for a shorter side you subtract before rooting.

Write each relationship as its own equation

HintTurn every sentence of the problem into symbols before combining them.

The whyTwo independent facts about the same two unknowns give two equations — write both first, in terms of the same two letters, before trying to solve either. Once you have the pair, substitute one equation into the other to reduce it to a single-unknown equation, which may turn out to be a quadratic rather than linear.

zero

HintThis is exactly why factorising an equation into a product helps you solve it.

The whyThis is the zero-product rule, the reason factorising solves equations: rewrite an equation as (something) × (something else) = 0, and each factor gives its own mini-equation to solve. When the equation comes from a real situation — ages, counts, lengths — one of the resulting values is often impossible, so the context, not the algebra, decides which root is the real answer.

SAS (side, angle, side)

HintThe one equal turn must sit trapped between the two equal edges, not anywhere else.

The whySAS needs the angle to be the included angle — sitting exactly where the two known sides meet — otherwise two sides and a non-included angle can, in some cases, describe two different triangles (the ambiguous case). Once triangles are proved congruent, every corresponding side and angle can be assumed equal without further checking.

Their sum

HintYou don't need to isolate each letter separately to answer some questions.

The whySometimes a question only wants a combined total, not the individual values. If subtracting two equations happens to leave equal coefficients on both remaining letters, dividing the whole equation by that coefficient gives their sum straight away — a shortcut worth spotting before grinding through full elimination and substitution.

Every pair of positions they could occupy

HintWork through the numbered spots and cross off any two that sit side by side.

The whyFor a small number of positions, list every pair of slots the two restricted people could take, cross out any pair that's adjacent, and count what's left — that count becomes one factor in a bigger calculation, since the two people can usually swap between their chosen positions and everyone else can be arranged independently in the remaining spots.

Multiply the number of choices at each step

HintThink of the decisions as independent — the total ways is a product, not a sum, of how many options exist at every stage.

The whyThis is the fundamental counting principle: if the first placement can be made in a ways, the next in b ways (given the first is fixed), and so on, the total number of arrangements is a × b × c × … . It's the same reasoning that turns "three choices, then two, then one" into a product rather than a running total, and it applies to arranging the remaining items after any fixed or restricted items are already placed.

numerator; denominator

HintPicture the fractions in a chain — top of one term meets bottom of the next, and they wipe each other out.

The whyBecause each fraction is built as (something + 1) over that something, the value on top of one fraction reappears on the bottom of the next, so multiplying them together leaves it out entirely. Only the very first denominator and the very last numerator survive the chain — spotting this turns a long product into a single fraction, which is then easy to test against whole-number targets.