JMC 2021

Its digit sum is a multiple of 9

提示Add the figures together and check the total against the nine times table.

为什么Dividing 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

提示The digit-sum idea, with a twist.

为什么Starting 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

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

为什么An 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

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

为什么The 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

提示The word "positive" has been left out on purpose.

为什么The 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

提示Some numbers cannot be done with three.

为什么Zero 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

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

为什么Squaring 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

提示Check it on a cube before you trust it.

为什么A 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

提示Top, bottom, and the uprights joining them.

为什么n 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

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

为什么Ten 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

提示Squares, not roots — write them out.

为什么They 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

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

为什么Edge 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

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

为什么Addition 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

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

为什么Convert 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

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

为什么BIDMAS 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

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

为什么This 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

提示Multiply n by itself, then take away one.

为什么This 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

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

为什么Work 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

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

为什么Adding 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

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

为什么Find 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

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

为什么Picture 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

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

为什么Because 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

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

为什么If 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

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

为什么Substituting 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

提示Line 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.

为什么The 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.