JMC 2023

Exactly the same

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

为什么The 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

提示How many times does the small one fit into the large one?

为什么Dividing 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°

提示Every slice at the centre is the same size.

为什么The 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

提示A prime cannot be split between two numbers.

为什么A 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

提示Think about the parity of two neighbouring numbers.

为什么Consecutive 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

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

为什么From 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

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

为什么Count 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

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

为什么Speed = 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°

提示Chop the shape into triangles from one corner and count them.

为什么A 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

提示Remember 2 counts and 1 does not — then list them.

为什么They 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

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

为什么Squares 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

提示Where the columns of place value open and close.

为什么Boundary 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

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

为什么This 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

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

为什么Break 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

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

为什么For 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

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

为什么Digit 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

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

为什么The 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

提示Nothing gets used up here — the whole first batch is still there to work with, again and again.

为什么Picture 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

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

为什么This 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

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

为什么The 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

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

为什么When 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

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

为什么Say 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

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

为什么A 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

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

为什么Suppose 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

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

为什么Extend 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.