Every card in JMC 2023

Exactly the same

HintSlide the two cut edges outwards until they sit on the old outline.

WhyThe two new edges of the notch are just the missing pieces of the old sides, moved inwards, so the total distance round is unchanged. That means a perimeter given for an L-shape tells you the full rectangle’s length plus width at once — and the area follows from subtracting the notch.

Bigger than 1

HintHow many times does the small one fit into the large one?

WhyDividing a larger number by a smaller always gives more than 1, whatever their size — 0.6 ÷ 0.3 = 2. With two numbers below 1, the product and the difference can never reach 1, so only the quotient and the sum are in the running: for 0.2 and 0.4 the sum is 0.6, the difference 0.2, the product 0.08, but 0.4 ÷ 0.2 = 2. The quotient always wins when the smaller number is at most a half; above that, check the sum as well — 0.6 + 0.7 = 1.3 beats 0.7 ÷ 0.6.

72°

HintEvery slice at the centre is the same size.

WhyThe centre of any regular polygon splits a full turn into equal slices: 72° for a pentagon, 60° for a hexagon, 45° for an octagon. The triangles from the centre are isosceles, so the interior angle of a regular pentagon is 2 × (180 − 72) ÷ 2 = 108°.

One is a multiple of 7

HintA prime cannot be split between two numbers.

WhyA prime factor of a product must come from one of the factors: 6 × 35 = 210 is a multiple of 7 only because 35 is. In a grid where the product of each row is given, the rows whose products are multiples of 5 or of 7 pin down where the 5 and the 7 must sit — that is usually the way in.

odd

HintThink about the parity of two neighbouring numbers.

WhyConsecutive whole numbers are one even and one odd; their squares are too, and even plus odd is odd. So such a sum is never divisible by 2, and it may or may not be prime — 49 + 64 = 113 is prime, 64 + 81 = 145 is not. Check the small primes by hand rather than guessing.

110

HintAdd one more eleven to the last of the two-digit ones.

WhyFrom 11 up to 99 the multiples of 11 are the repeated digits, so a two-digit multiple of 11 always has an even digit sum — twice the repeated digit — which is a quick way to rule out candidates. Above 99 the pattern breaks (110, 121, 132 …) and the alternating-digit test takes over: 1 − 2 + 1 = 0 for 121, so it is a multiple of 11.

18:07

HintGo to the next whole hour first, then add what is left.

WhyCount to the next whole hour first, then the rest: from 16:52, eight minutes reaches 17:00, and the remaining 67 minutes take you to 18:07. Clock arithmetic carries at 60, not 100 — delays and doubled journey times are just repeated versions of the same step.

Distance divided by time

HintWhich quantity would you divide to compare a fast runner with a slow one?

WhySpeed = distance ÷ time, so a 400 m lap in 50 s is 8 m/s. The same relationship runs backwards: at a steady rate, distance covered in a known time gives the rate, and the rate tells you when a journey began.

(n − 2) × 180°

HintChop the shape into triangles from one corner and count them.

WhyA quadrilateral gives 360°, a pentagon 540°, a hexagon 720°. Questions about marked angles on the outside of a shape are usually this formula in disguise: each marked angle is 180° minus an interior angle, so the outside total is a fixed number.

8

HintRemember 2 counts and 1 does not — then list them.

WhyThey are 2, 3, 5, 7, 11, 13, 17 and 19. With 23 and 29 that is the ten primes below 30 — the working set for prime-square puzzles, where each side must total the same and every entry must be prime.

64

HintThe index has to be even and a multiple of three at once; then take the smallest base you can.

WhySquares need an even index and cubes an index that is a multiple of 3, so a number that is both is a sixth power: 2⁶ = 64 comes first, then 3⁶ = 729. Only 27 and 64 are two-digit cubes, and 64 is the one that is also a square. Turning "search the list" into index arithmetic is what makes the later Challenge questions quick.

99; 1000; 36

HintWhere the columns of place value open and close.

WhyBoundary numbers appear in "largest multiple of … below" questions and in digit-sum arguments: knowing that four digits can sum to at most 36 tells you, for instance, that a four-digit number whose digit sum is a multiple of 20 has digit sum exactly 20.

Add the same amount to both numbers, so one becomes round, then subtract

HintNudge each value by an identical gap until the number you’re taking away is a neat one.

WhyThis avoids borrowing across zeros in a column method. For 4002 − 1998, add 2 to each side: 4004 − 2000 = 2004, matching 4002 − 1998 exactly. The same idea works the other way — round the number being subtracted UP to a multiple of 10 or 100, then correct the answer by the same amount you added.

Multiply together different combinations of its prime factors

HintThink about which small building blocks divide the number exactly, then group them in every possible way.

WhyBreak the number into primes first — say 60 = 2×2×3×5. Every factor is a product using each prime up to its available power: one, some or none of the 2s, times either 3 or not, times either 5 or not. That systematic approach finds all twelve factors of 60 without guesswork, and multiplying the (power+1) values together — 3 × 2 × 2 = 12 — even tells you how many factors to expect before you list a single one.

Divide both by their highest common factor

HintFind the biggest number that goes into the top and the bottom exactly, then shrink each by that amount.

WhyFor 18 over 48, both numbers divide by 6 (their HCF), giving 3 over 8 straight away — no need to cancel by 2, then 3, then again. Spotting the HCF first, rather than dividing by small factors one at a time, saves steps in any fraction question under time pressure.

Its digit product would be 0, and dividing by 0 isn’t allowed

HintThink about what happens when you try to split any number by zero.

WhyDigit products come up often in these puzzles: for a two-digit number with digits a and b, the digit product is a × b. If either digit is 0 the product is 0, and no number is divisible by 0 — so the largest candidate worth checking is one where neither digit is 0. Working from 99 downward and testing only such numbers narrows the search fast.

equal; 180

HintThink about the triangle’s line of symmetry, and remember every triangle’s angles total the same fixed amount.

WhyThe two equal sides create a mirror line, so the angles facing those sides must match — the base-angle property of an isosceles triangle. Combine it with the fact that every triangle’s angles sum to 180°, and one algebraic equation is often enough to pin down every angle in a diagram, even when only one angle is given as a number.

The original starting number, divided by that round’s group size — never by what’s left from the previous round

HintNothing gets used up here — the whole first batch is still there to work with, again and again.

WhyPicture 80 items that are reused rather than used up: available singly at first, then in pairs, then in fours. Because all 80 survive every round, the count each time is 80, then 80 ÷ 2 = 40, then 80 ÷ 4 = 20, giving 80 + 40 + 20 = 140 in total. The trap is chaining instead — taking the 40 from round two and doing 40 ÷ 4 = 10 for round three, giving 130 — which treats the items as if they’d been used up rather than reused. Always check the original number divides exactly by every round’s group size, then add the rounds.

Large side = length + width; small side = length − width

HintThink about how the rectangle’s two dimensions combine to span the outer shape, and how their gap leaves room for the shape tucked inside.

WhyThis is a classic “framed square” motif: if the rectangle has length a and width b, the outer square’s side is a + b and the inner square’s side is a − b, so the outer minus inner area is (a+b) squared minus (a−b) squared = 4ab — exactly the combined area of the four rectangles. Knowing either square’s side instantly gives you a + b or a − b, often enough to answer without ever finding a and b separately.

2

HintAdd the three largest digits possible and see how many tens that makes.

WhyThe biggest column sum from three digits (each at most 9) is 9 + 9 + 9 = 27, which is 2 tens and 7 units — so a carry of 2 into the next column is the ceiling, whatever the actual digits are. This bound is exactly what lets you eliminate impossible cases quickly in a column-addition puzzle with unknown digits.

Systematic listing — organised casework that covers every case once

HintWrite out each valid possibility in an orderly way rather than searching for a shortcut pattern.

WhyWhen a counting question has too few possibilities for a combinatorics formula to be worth the setup, and too many to just eyeball, working through every case in an organised order — often building up from the smallest value of one variable — avoids missing or double-counting a case. It is slower than a formula but far more reliable when extra conditions are attached.

Write every quantity as a multiple of one base variable, then use the difference to solve for it

HintPick the smallest amount as your unknown, express the rest in terms of it, and plug in the given gap.

WhySay A is twice B, and C is three times A, with C exactly 15 more than B. Let B = n: then A = 2n and C = 6n, and 6n − n = 15 gives n = 3, so B = 3, A = 6, C = 18. Chaining every quantity off a single variable — rather than juggling several separate unknowns — turns a wordy ratio problem into one simple equation.

A year’s digit sum stays small while age (current year minus birth year) changes quickly

HintCompare how fast one side of the equation can grow versus how little the other side can move.

WhyA birth year in the 2000s has digit sum at most 2 + 0 + 9 + 9 = 20, so the age it must match is capped near 20 too — while the actual age (this year minus the birth year) climbs by exactly 1 every year. Writing “age = current year minus birth year” and “age = digit sum of birth year” as two expressions for the same age turns the puzzle into one equation with only a handful of candidate years to try.

reverse

HintStart from the end and work your way backward, undoing addition with subtraction and multiplication with division.

WhySuppose a sequence of steps — say, doubling a number then subtracting 5 — turns some starting value into a known result. Working backwards, first undo the “subtract 5” by adding 5, then undo the doubling by halving. Repeating this through every step recovers the very first amount, even when several such steps are chained together — exactly the approach for problems where you’re told about a final share but asked for the original total.

The sum of the two interior angles not next to it

HintAdd together the two corners of the triangle that are farthest from the extended side.

WhyExtend one side of a triangle to make an exterior angle: because angles on a straight line total 180° and the triangle’s own angles total 180°, the exterior angle must equal 180° minus the one adjacent interior angle — exactly the sum of the other two interior angles. Worth knowing whenever a diagram marks angles OUTSIDE a triangle, especially in multi-triangle diagrams built from lines through a shared point.