Mathematics

JMC 2023

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MathematicsJMC 2023
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Mathematics · JMC 2023Professor Pi
Answer in your head…An L-shape is made by cutting one corner out of a rectangle. How does its perimeter compare with the original rectangle’s?
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★ KS3-MATH-JMC-0109Front

Mathematics · JMC 2023Professor Pi

Exactly the same

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

The 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.

★ KS3-MATH-JMC-0109Back

Mathematics · JMC 2023Professor Pi
Answer in your head…You divide a positive number by a SMALLER positive number. Is the answer bigger or smaller than 1?
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★ KS3-MATH-JMC-0110Front

Mathematics · JMC 2023Professor Pi

Bigger than 1

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

The 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.

★ KS3-MATH-JMC-0110Back

Mathematics · JMC 2023Professor Pi
Answer in your head…Lines from the centre of a regular pentagon to its five corners make five equal angles at the centre. How big is each one?
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★ KS3-MATH-JMC-0111Front

Mathematics · JMC 2023Professor Pi

72°

HintEvery slice at the centre is the same size.

The 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°.

★ KS3-MATH-JMC-0111Back

Mathematics · JMC 2023Professor Pi
Answer in your head…Several whole numbers are multiplied together and the product is a multiple of 7. What can you say about the numbers themselves?
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★ KS3-MATH-JMC-0112Front

Mathematics · JMC 2023Professor Pi

One is a multiple of 7

HintA prime cannot be split between two numbers.

The 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.

★ KS3-MATH-JMC-0112Back

Mathematics · JMC 2023Professor Pi
Fill the gapAnswer in your head…The sum of two consecutive square numbers is always ____.
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★ KS3-MATH-JMC-0113Front

Mathematics · JMC 2023Professor Pi

odd

HintThink about the parity of two neighbouring numbers.

The 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.

★ KS3-MATH-JMC-0113Back

Mathematics · JMC 2023Professor Pi
Fill the gapAnswer in your head…The two-digit multiples of 11 all have a repeated digit, but the smallest three-digit multiple of 11 is ____.
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★ KS3-MATH-JMC-0114Front

Mathematics · JMC 2023Professor Pi

110

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

The 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.

★ KS3-MATH-JMC-0114Back

Mathematics · JMC 2023Professor Pi
Fill the gapAnswer in your head…A bus leaves at 16:52 and the journey takes 75 minutes, so it arrives at ____ — remember an hour has 60 minutes, not 100.
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★ KS3-MATH-JMC-0115Front

Mathematics · JMC 2023Professor Pi

18:07

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

The 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.

★ KS3-MATH-JMC-0115Back

Mathematics · JMC 2023Professor Pi
↔ Asked both waysAnswer in your head…Average speed, as a calculation
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★ KS3-MATH-JMC-0116Front

Mathematics · JMC 2023Professor Pi

Distance divided by time

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

The 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.

★ KS3-MATH-JMC-0116Back

Mathematics · JMC 2023Professor Pi
↔ Asked both waysAnswer in your head…The sum of the interior angles of an n-sided polygon
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★ KS3-MATH-JMC-0117Front

Mathematics · JMC 2023Professor Pi

(n − 2) × 180°

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

The 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.

★ KS3-MATH-JMC-0117Back

Mathematics · JMC 2023Professor Pi
⌨ Type the answerAnswer in your head…How many prime numbers are there below 20? (type the number only)

★ KS3-MATH-JMC-0118Front

Mathematics · JMC 2023Professor Pi

8

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

The 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.

★ KS3-MATH-JMC-0118Back

Mathematics · JMC 2023Professor Pi
⌨ Type the answerAnswer in your head…A whole number bigger than 1 is both a square number and a cube number. What is the smallest it can be? (type the number only)

★ KS3-MATH-JMC-0119Front

Mathematics · JMC 2023Professor Pi

64

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

The 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.

★ KS3-MATH-JMC-0119Back

Mathematics · JMC 2023Professor Pi
Fill the gapsAnswer in your head…The largest two-digit number is ____, the smallest four-digit number is ____, and the largest digit sum a four-digit number can have is ____.
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★ KS3-MATH-JMC-0120Front

Mathematics · JMC 2023Professor Pi

99; 1000; 36

HintWhere the columns of place value open and close.

The 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.

★ KS3-MATH-JMC-0120Back

Mathematics · JMC 2023Professor Pi
↔ Asked both waysAnswer in your head…The compensation strategy for mental subtraction
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★ KS3-MATH-JMC-0269Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0269Back

Mathematics · JMC 2023Professor Pi
Answer in your head…If you know a number’s prime factorisation, how can you find every one of its factors without testing each candidate?
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★ KS3-MATH-JMC-0270Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0270Back

Mathematics · JMC 2023Professor Pi
Answer in your head…When a fraction’s numerator and denominator share a common factor bigger than 1, what’s the quickest way to simplify it?
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★ KS3-MATH-JMC-0271Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0271Back

Mathematics · JMC 2023Professor Pi
Answer in your head…In a puzzle where a number must be divisible by the product of its own digits, why can that number never contain the digit 0?
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★ KS3-MATH-JMC-0272Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0272Back

Mathematics · JMC 2023Professor Pi
Fill the gapsAnswer in your head…In an isosceles triangle, the two angles opposite the equal sides are always ____ to each other, and no matter how the triangle looks its three angles still sum to ____ degrees.
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★ KS3-MATH-JMC-0273Front

Mathematics · JMC 2023Professor Pi

equal; 180

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

The 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.

★ KS3-MATH-JMC-0273Back

Mathematics · JMC 2023Professor Pi
Answer in your head…In a puzzle where the same items get reused for a bigger swap each round instead of being used up, what should each round’s count be based on?
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★ KS3-MATH-JMC-0274Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0274Back

Mathematics · JMC 2023Professor Pi
Answer in your head…Four congruent rectangles are arranged edge-to-edge to frame a smaller square inside a larger one. How do the two squares’ side lengths relate to the rectangle’s length and width?
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★ KS3-MATH-JMC-0275Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0275Back

Mathematics · JMC 2023Professor Pi
⌨ Type the answerAnswer in your head…When three single-digit numbers are added in one column of a written addition, what is the largest possible carry into the next column? (type the number only)

★ KS3-MATH-JMC-0276Front

Mathematics · JMC 2023Professor Pi

2

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

The 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.

★ KS3-MATH-JMC-0276Back

Mathematics · JMC 2023Professor Pi
↔ Asked both waysAnswer in your head…Counting method for too many outcomes to guess, too few for a formula
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★ KS3-MATH-JMC-0277Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0277Back

Mathematics · JMC 2023Professor Pi
Answer in your head…In a word problem with several quantities linked by ratios, and one absolute difference given between two of them, what’s the reliable solving method?
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★ KS3-MATH-JMC-0278Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0278Back

Mathematics · JMC 2023Professor Pi
Answer in your head…Why do “this person’s age equals the digit sum of their birth year” puzzles have only a few years worth testing?
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★ KS3-MATH-JMC-0279Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0279Back

Mathematics · JMC 2023Professor Pi
Fill the gapAnswer in your head…To find an unknown starting amount from a known final result and a chain of steps, undo each step in ____ order.
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★ KS3-MATH-JMC-0280Front

Mathematics · JMC 2023Professor Pi

reverse

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

The 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.

★ KS3-MATH-JMC-0280Back

Mathematics · JMC 2023Professor Pi
Answer in your head…The exterior angle of a triangle, formed by extending one side, equals what in terms of the triangle’s other two angles?
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★ KS3-MATH-JMC-0281Front

Mathematics · JMC 2023Professor Pi

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.

The 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.

★ KS3-MATH-JMC-0281Back

JMC 2023

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