Every card in JMC 2021

Its digit sum is a multiple of 9

HintAdd the figures together and check the total against the nine times table.

WhyDividing exactly by 9 means the number sits in the 9 times table, and the digit-sum test spots it at a glance: 738 has digit sum 18, so 738 = 9 × 82. Pair it with the boundaries — the three-digit multiples of 9 run from 108 upwards in steps of 9. A clue saying "the answer is also a whole number" is usually this fact in disguise.

The alternating sum of the digits is a multiple of 11

HintThe digit-sum idea, with a twist.

WhyStarting from the right, add and subtract the figures in turn. If the result is 0 or another multiple of 11, the number is a multiple of 11. For 2475: 5 − 7 + 4 − 2 = 0, and indeed 2475 = 11 × 225. The test lets you check five candidates in the time it takes to do one long division.

Two equal base angles

HintFold the triangle along its line of symmetry and see which corners land on each other.

WhyAn isosceles triangle has a mirror line through its apex, so the two angles opposite the equal sides match. Nearly every JMC angle-chasing question is a chain of isosceles triangles: mark each pair of equal sides, write the matching angles as the same letter, and the unknown falls out of the 180° in a triangle.

Length plus width

HintRun your eye along one outer edge of the picture: it is made of one long side and one short side.

WhyThe classic windmill arrangement. Each outer edge is a long side followed by a short side, so the outer square has side length + width, while the inner square has side length − width. Those two facts turn an area question into a two-line calculation.

Negatives and zero

HintThe word "positive" has been left out on purpose.

WhyThe integers are the whole numbers in both directions: …, −3, −2, −1, 0, 1, 2, 3, …. When several different integers multiply to a positive result, an even number of them must be negative — and small negatives are often what make it possible: three different integers with product 2 must be −2, −1 and 1, because three different positive integers multiply to at least 1 × 2 × 3 = 6. Setters use this exact wording to reward careful readers.

four

HintSome numbers cannot be done with three.

WhyZero counts as a square, so 12 = 4 + 4 + 4 and 30 = 25 + 4 + 1 use fewer. Some numbers genuinely need the full set — 23 = 9 + 9 + 4 + 1 cannot be made with three squares. Lagrange proved the theorem in 1770; JMC setters like to ask which small numbers manage with fewer.

8

HintWork out the last digits of 1², 2², 3² … 9² and see which figures never show up.

WhySquaring the digits 0 to 9 gives last digits 0, 1, 4, 9, 6, 5, 6, 9, 4, 1 — so a square ends in 0, 1, 4, 5, 6 or 9 and never in 2, 3, 7 or 8. In a crossnumber whose clue says "a square", this rules out most cells before you write anything.

V + F = E + 2

HintCheck it on a cube before you trust it.

WhyA cube has 8 vertices and 6 faces, 14 altogether, which is its 12 edges plus 2. The formula holds for every prism and pyramid, so if a question gives you two of the counts you can find the third without drawing. Named after Leonhard Euler, the Swiss mathematician who found it in the 1700s.

3n

HintTop, bottom, and the uprights joining them.

Whyn edges round the top, n round the bottom and n joining the two ends. A prism also has n + 2 faces and 2n vertices: a pentagonal prism has 15 edges, 7 faces and 10 vertices, and Euler agrees — 10 + 7 = 15 + 2. So "a prism with seven faces" is telling you the base is a pentagon.

210

HintPair the smallest with the largest, then the next pair inwards, and count how many pairs there are.

WhyTen pairs of 21. The same pairing trick gives 1 + 2 + … + 9 = 45 (four pairs of ten plus the 5 in the middle) — the total behind every puzzle that places the digits 1 to 9 once each: three rows of equal sum must each come to 15, the classic magic-square line.

6

HintSquares, not roots — write them out.

WhyThey are 16, 25, 36, 49, 64 and 81. Knowing the list by heart, along with which of them are odd, makes crossnumber clues such as "a square" or "an odd square" instant.

12; 9; 8

HintCount the lines where two faces meet — top, bottom and the uprights joining them.

WhyEdge counts start most straw-model and sliced-corner questions. A prism on an n-sided base has 3n edges, n + 2 faces and 2n vertices; a pyramid on an n-sided base has 2n edges, n + 1 faces and n + 1 vertices. The cube’s 6 faces and 8 vertices then check with Euler: 8 + 6 = 12 + 2.

Reorder the terms, carrying each sign with it

HintEvery number is glued to the mark in front of it — move the pair together.

WhyAddition is commutative, so 50 minus 80 plus 90 can be regrouped as 50 plus 90 minus 80 = 60, as long as each number keeps the sign written just before it — the minus stays glued to the 80, not to the 50. This turns a calculation with an early negative into two easy additions followed by one clean subtraction, which is how JMC setters expect you to work fast without a calculator.

Divide the total by the value of one coin

HintSame move as finding how many equal boxes fit inside a given weight or height.

WhyConvert both amounts to the same unit first — pence, not a mix of pounds and pence — then divide: a total of 840 pence in 2p coins is 840 divided by 2, giving 420 coins. Do this separately for each coin type before adding the counts together; skipping the unit conversion is the single most common slip in these problems.

BIDMAS

HintSix letters, one for each stage, starting with the shapes that come first.

WhyBIDMAS stands for Brackets, Indices, Division, Multiplication, Addition, Subtraction, and fixes the order everyone uses so an expression like 3 + 4 x 2 is read the same way by every person: work out the x first, then the +, giving 11 rather than 14. Division and multiplication rank equally and go left to right, and so do addition and subtraction.

Find what 1% of it is worth

HintDivide the known amount by its percentage, then multiply up to 100.

WhyThis is the unitary method: if 30% of a length is known to be 90 m, then 1% is 90 divided by 30, that is 3 m, and the whole length is 100 times 3, giving 300 m. It works for any "this percentage is worth this amount" set-up, and is faster than writing and solving a fraction equation.

n squared minus 1

HintMultiply n by itself, then take away one.

WhyThis identity turns any "integer minus its own reciprocal" into one fraction at a stroke: 5 minus 1/5 becomes (25 minus 1) over 5, giving 24/5. Chaining several such fractions together and multiplying is a favourite JMC trick, because most of the top and bottom numbers then cancel against neighbouring terms.

Adding the positive of that number

HintTwo minus signs sitting next to each other flip into a plus.

WhyWork from the innermost bracket outward. If an inner result comes out negative, subtracting it flips to addition: 3 minus (−1) becomes 3 + 1 = 4. Miss this and a long chain of nested brackets is almost guaranteed to end on the wrong sign, since every layer either keeps or flips the sign of everything inside it.

Draw extra lines so the shape splits into equal congruent parts

HintThen the shaded fraction is just how many of those pieces are shaded, out of the total.

WhyAdding lines along a shape's natural lines of symmetry — diagonals, medians, lines joining midpoints — often divides it into a set of identical smaller triangles. Counting how many of those are shaded, out of the total, turns a fiddly area calculation into simple counting: three shaded out of eight identical triangles is just three eighths.

Convert both rates to the same unit of time

HintWork out how much each does per single minute or second, then compare like for like.

WhyFind a common unit — say, questions per minute, or seconds per question — for both rates before doing anything else. Once the units match, you can find how long each person needs for the same-sized job and subtract the two times directly, rather than comparing raw counts like '3 questions every 20 seconds' against '5 questions every 45 seconds' as they stand.

It splits evenly into two 180-degree shares

HintVertically opposite angles are always identical, so matched groups must total the same.

WhyPicture the point where the segments cross: going around it, the angles alternate between the ones sitting inside a triangle and the ones sitting in the gaps between triangles. Each inside angle has an exactly equal gap angle vertically opposite it, so the two families add up to the same amount, 180 degrees each out of the full 360. That 180 is then subtracted from the triangles' own angle total to leave the marked angles.

The simple mean of the two coin values, ignoring how many coins there are

HintAdd the two amounts and divide by two — the count cancels out of the working.

WhyBecause the two coin types appear in equal numbers, the weighted average collapses to the ordinary mean: with equal counts of 20p and 50p coins, the average per coin is (20 + 50) divided by 2, giving 35p, whatever the total number of coins turns out to be. Multiply that mean by the coin count to reach the total value — a shortcut that skips setting up a pair of simultaneous equations.

Let the quantity equal 3k instead of a plain letter

HintChoose a variable that is already a multiple of the denominator in the fraction.

WhyIf a problem says 'one third of the girls do something', let the number of girls be 3k rather than g — then a third of them is exactly k, a whole number, with no fraction in sight. Every other quantity in the problem can then be built from k too, and the final total often reveals itself as a multiple of some number, exactly the clue a multiple-choice question rewards.

whole number

HintCombine the clue with the equation, then check which integers in the resulting range actually fit every condition.

WhySubstituting the equation’s relationship into the inequality usually produces a narrow range for one unknown; because the answer must be a whole number, only one or two integers in that range survive, and a second condition often eliminates all but one. This two-step combination, algebra first then integer-checking, is common wherever a problem describes real, countable things.

a + d and b + c

HintLine the six totals up using only the sorting. Four of them slot into place unambiguously; the two that both land in the middle are the ones you cannot separate.

WhyThe sorting forces a + b < a + c < a + d < b + d < c + d and also a + b < a + c < b + c < b + d < c + d. So a + d and b + c both sit between a + c and b + d, and the ordering never says which is bigger: 1, 2, 3, 100 gives a + d = 101 but b + c = 5, while 1, 2, 3, 4 gives a + d = b + c = 5. That's why a problem giving six pairwise sums often states that only five different values appear, so one must repeat. The only pair that can ever repeat is a + d and b + c, so that repeat is what pins the equality down.