Every card in JMC 2019

The difference

HintNot the ratio — that shrinks every birthday.

WhyIf one person is twice another's age now, they will not be twice it later — but the gap in years is fixed for life. Age puzzles are almost always solved by finding that gap first.

180°

HintThey lie between the parallels on the same side, like the two angles at the ends of a trapezium's slanted side.

WhyIn a parallelogram or rhombus, the two angles at the ends of any side add to 180° because the other two sides are parallel. In a trapezium it works only along the two non-parallel sides: each of those crosses the parallel pair, so the angles at its ends are co-interior. The angles at the ends of a parallel side need not add to 180° — a right-angled trapezium can have 90°, 90°, 120° and 60°. That one fact usually unlocks the angle beside a given 120° or 60°.

right; adjacent; parallel

HintDiamond, toy, table-top shape.

WhyA rhombus is a parallelogram with all four sides equal, so its diagonals bisect each other at right angles; a kite's diagonals also cross at right angles. Diagram questions use these properties to hand you parallel sides (for angle facts) and equal sides (for isosceles triangles).

They are equal

HintWrite both as fractions of 100 and look at the multiplications.

Whya% of b and b% of a are both a × b ÷ 100, so the two are always the same. This turns an ugly-looking calculation — 12% of 25 plus 25% of 12 — into 3 + 3 = 6 in your head.

9

HintThe fewer digits a number has, the smaller it is — so make each digit carry as much as possible.

WhyA number with fewer digits is always smaller than one with more, so pack the digit sum into the largest digit. For a digit sum of 30 that is 3999 — the remainder 3 goes first, the 9s after. Its last digit is 9 whenever the total is 9 or more.

A sum of the form 1 + 2 + 3 + … + n

HintBowling-pin or snooker-rack numbers.

WhyThe triangular numbers run 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91 … Learn the two-digit ones; they get asked for alongside squares and primes. 36 is the only two-digit number that is both square and triangular — 1 is square and triangular too, and the next one after 36 is 1225.

8000

HintCube the 2 and the 10 separately.

WhySo there are exactly ten cubes up to 10³ and twenty up to 20³. Knowing 10³ = 1000 and 20³ = 8000 lets you count cubes or squares in a range without listing them.

Exactly two factors needed

HintHow many different numbers divide into 1?

WhyA prime has exactly two factors, itself and 1; the number 1 has only one. So the smallest prime is 2, the smallest two-digit prime is 11, and 'prime digits' means 2, 3, 5 and 7 — never 1.

128

HintTake the two known angles away from one complete turn.

WhyWhen a square, an equilateral triangle and a regular hexagon share a vertex, their corners use 90° + 60° + 120° = 270°, leaving 90° — a right angle hiding in plain sight. Angles-at-a-point is the fastest route to a missing angle in a tiled diagram.

n(n + 1) ÷ 2

HintPair the first with the last, the second with the second-last…

Why1 + 2 + … + 12 = 12 × 13 ÷ 2 = 78, and 1 + … + 100 = 5050. Number-triangle and arrangement questions lean on this total: knowing the sum of all the pieces lets you say what the edges or corners must add to.

Subtract, working backwards

HintUndo the rule one step at a time from the far end.

WhyIf each term is the sum of the ones before it, then each earlier term is the later one minus the others. Start at the last known term and peel off one unknown per step — no algebra needed.

10000

HintBoth sides of the square convert, so the length factor is applied twice.

Why1 m = 100 cm, so 1 m² = 100 × 100 = 10 000 cm², not 100 cm². Area units square the length factor and volume units cube it: 1 m³ = 1 000 000 cm³. Convert before comparing, or the ratio is a hundred times out.

Add the minutes either side of midnight

HintWork out each leg separately, then combine them.

WhySubtracting the two clock times directly across midnight gives a negative or meaningless result. Instead find the minutes from the first time up to midnight, then the minutes from midnight to the second time, and add the two — it also works for any awkward boundary, such as crossing into a new hour or a new day.

bracketed group

HintOrder of operations always deals with what's grouped together before anything separate.

WhyThis is the first rule of BIDMAS/BODMAS: brackets before division, multiplication, addition or subtraction. It matters even when the bracket holds several additions and subtractions in a row — do all of them, get one number, then apply what's outside. Skipping this and dividing only part of the bracket on its own gives a different, wrong, answer.

Use 1 minus the fraction, then compare numerators

HintFind what's left over first — both amounts already share a denominator.

WhyIf two-fifths of an amount is used, three-fifths remains: the ratio remaining : used is 3 : 2, straight from the numerators once both fractions are over the same denominator. This turns any "used this much, what's left" problem into a fraction subtraction followed by a same-denominator ratio read-off — no cross-multiplying needed.

The line crosses its outline more than twice

HintThink about a zig-zag or notched edge, not a simple oval.

WhyA convex shape's outline can only be crossed twice by a straight line, giving exactly two pieces. A shape with an inward dent or a notch lets a single straight line cross the outline four, six or more times — each extra crossing pair adds one more piece. So a five-pointed star or a zig-zag outline can be sliced into several pieces with one straight cut, while a circle or square never can.

How far it is from the target

HintIgnore whether a number is above or below — only the size of the gap matters.

WhySubtract the target from each option (or the option from the target) and ignore any minus sign — that gap size is what decides 'closest'. A number just above the target can beat one just below it, or vice versa; only the smallest gap wins, regardless of direction.

Rule out every option that breaks a constraint

HintCheck each candidate spot against all the rules before accepting it.

WhySudoku-style grids and 'forced move' puzzles are both solved by process of elimination: for a given value, check each rule (row, column, box, or an adjacency restriction) in turn, and cross off any position that would break one. Often only a single position survives every rule at once, so it must be correct even before you've reasoned out why directly. The same logic finds a forced losing move in a two-player placement game — a square already ruled out by every neighbour has become impossible to fill.

Multiply within each group, then add the groups

HintWork out a route count per starting point, and only afterwards combine the totals.

WhyThis is the basic counting principle: if a task has several stages, multiply the number of choices at each stage to get the routes from one starting point. When there are different types of starting point offering different route counts — say, one group of squares giving six routes each and another group giving two routes each — work out each group's total separately, then add the groups together, rather than multiplying everything into one giant product.

The whole route has just one beginning and one end

HintA branch you enter and leave again eats up your single visit to it — you only pass through once, in one direction.

WhyAny dead-end branch or dense cluster of edges can only be fully explored if the route begins or ends right there — passing through and coming back out uses up the one beginning or one end the whole journey is allowed. To find the true longest route, add up the maximum edges each separate part could contribute assuming the route begins or ends there, then check that a single route achieving that total actually exists.

Rotate or reflect the whole layout

HintTurn the entire picture, don't hunt for a brand-new one from scratch.

WhyIf a square (or any symmetric grid) admits one valid tiling, then rotating that whole tiling by 90°, 180° or 270° — or reflecting it — gives more valid tilings for free, because the grid itself looks the same after those moves. This turns 'find every possible arrangement' into 'find one arrangement, then apply the grid's symmetries', which is far faster than searching from nothing each time.

The square root of that ratio

HintUndo a squaring: if one shape is four times the area of the other, its sides are twice as long.

WhyArea scales with the SQUARE of the linear scale factor, so to go the other way you take a square root: if two similar shapes have areas in the ratio 4:1, their side lengths — and therefore their perimeters — are in the ratio 2:1, not 4:1. Mixing up 'area ratio' and 'length ratio' is one of the most common errors with similar shapes.

1

HintAny first digit of 2 or more would push the product past 9999.

WhyMultiplying a four-digit number by 9 nearly multiplies it by 10, so the product is close to ten times the original. For the product to still fit in four digits, the original number can be at most 1111 (since 1111 × 9 = 9999) — so its first digit has to be 1, whatever the other three digits are. This kind of bound — working out the largest value that still fits — is the fastest way into a 'find the missing digit' cryptarithm.

enclosing rectangle; subtract

HintDraw a simple shape around the tricky region, work out its area, then take away what spills over.

WhyWhen a shaded area is bounded by parts of several different shapes, extend some of its edges until they meet and form a rectangle (or other simple shape) around it. Find that simple shape's area, then remove any extra regions you've included that aren't part of the shaded area. This turns an awkward, irregular outline into ordinary rectangle and square arithmetic.

360°

HintImagine walking all the way round the outside, turning at every corner.

WhyThis is true for a triangle, a square, a regular heptagon, or any convex polygon at all — walk around the outside and you turn through a full 360° by the time you're back where you started, no matter how many sides there are. It is the fact behind why a 'star' of angles cut off by diagonals inside a polygon reduces to a simple formula: the answer only ever depends on the number of sides through this same 360° total, never on the specific shape drawn.