Mathematics · Professor Pi

Every card in IMC 2018

The whole deck, in order — so you can read it through before your child ever sees it.

1 KS3-MATH-IMC-0049

One million written as a power of ten

10⁶

HintA thousand thousands.

Why10³ × 10³. To get a feel for the size: a million seconds is about eleven and a half days. Money questions with hundreds of millions of items are a small number times a power of ten — multiply the small numbers, then attach the zeros and count them.

2 KS3-MATH-IMC-0050

In an isosceles triangle, which two angles are equal?

The two base angles

HintFold along the line of symmetry — which two corners land on each other?

WhyMark both base angles the moment you see two equal sides; in a chain of isosceles triangles each apex angle is an exterior angle of the next.

3 KS3-MATH-IMC-0051

A question asks approximately what fraction 31 out of 243 is, and the options are far apart. What is the fast route?

Round both numbers first

HintThe word "approximately" means an exact division is a waste of time.

WhyRound to numbers that divide nicely — 30 out of 240 is one eighth — and only look harder if two options are close. Non-calculator papers reward a well-chosen rounding far more than long division.

4 KS3-MATH-IMC-0052

You multiply both sides of an inequality by a positive number. What happens to the inequality sign?

It stays the same

HintTest it: 2 < 3, multiply both by 5, then by −5.

WhyMultiplying or dividing by a positive number keeps an inequality true in the same direction, so a double inequality with fractions is best cleared by multiplying every part by a common denominator. Only a negative multiplier reverses the signs.

5 KS3-MATH-IMC-0053

How does a² − b² factorise?

(a + b)(a − b)

HintTry a = 5, b = 3: 25 − 9 = 16 = 8 × 2 — what are 8 and 2 made of?

WhyThe difference of two squares. It tells you which whole numbers are a difference of squares: every odd number is (since 2k + 1 = (k + 1)² − k²), every multiple of 4 is, but a number that is twice an odd number never is — the two brackets would have to be one odd and one even.

6 KS3-MATH-IMC-0054

A regular hexagon is cut from its centre into equilateral triangles. How many are there? (type the number only)

6

HintEvery side is also a radius.

WhyThe three long diagonals of a regular hexagon meet at the centre and split it into six equilateral triangles, each with the hexagon's side as all three of its edges.

7 KS3-MATH-IMC-0055

One ninth, written as a decimal, is the single digit ____ recurring.

1

HintDivide one by nine and watch the same remainder come back every time.

WhyFrom 1/9 = 0.111… you get every ninth: 2/9 = 0.222…, and so on up to 9/9 = 0.999…, which equals 1. Dividing by 90 shifts the recurring block one place: 1/90 = 0.0111….

8 KS3-MATH-IMC-0056

The three-digit number with digits a, b, c from hundreds to units

100a + 10b + c

HintEach column is worth ten times the next one along.

WhyReversing a three-digit number changes it by a multiple of 99, and the middle digit never matters.

9 KS3-MATH-IMC-0057

To test whether a number is divisible by 8, you look at only its last how many digits? (type the number only)

3

HintFind the smallest power of ten that 8 divides — that fixes the tail length.

WhyBecause 1000 = 8 × 125, a number is a multiple of 8 exactly when the number formed by its last three digits is. Combine it with the digit-sum test for 9 and you have a test for 72, since 8 and 9 share no factor.

10 KS3-MATH-IMC-0058

Solving x² = 5x, a student divides both sides by x and gets x = 5. What has been lost?

The solution x = 0

HintDividing by something that might be nothing is never safe.

WhyMove everything to one side and factorise instead: x² − 5x = x(x − 5) = 0 gives both x = 0 and x = 5. Counting "how many values of x" questions hinge on not throwing a root away.

11 KS3-MATH-IMC-0059

A square with side s has diagonals of length s____.

√2

HintPythagoras on either half of the square.

WhyHalf a square is a right-angled isosceles triangle, so the diagonal is √(s² + s²). A square of side 3 placed corner-first spans 3√2; a product of two such spans loses the surd because (√2)² = 2.

12 KS3-MATH-IMC-0060

The cubes of 5, 6 and 7 are ____, ____ and ____ respectively.

125; 216; 343

HintFive squared times five, six squared times six, seven squared times seven.

WhyThe cubes up to ten — 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000 — are worth knowing by sight, and the tenth is exactly a thousand. Questions about sums of consecutive cubes, or about which numbers below a limit are cubes, are then lists of a handful of values rather than calculations.

13 KS3-MATH-IMC-0199

What quick test tells you a whole number is divisible by 3?

Its digit sum is divisible by 3

HintSum the individual figures, not the whole value itself.

Why141 = 3 × 47 fails to be prime because 1 + 4 + 1 = 6 divides by 3 — the digit-sum test spots non-primes fast without needing a division. The same idea, checked against 9 instead of 3, is the test for divisibility by 9.

14 KS3-MATH-IMC-0200

To convert a rate given per minute into the same rate per second, what number do you divide by?

60

HintHow many seconds make up one minute?

Why300 downloads counted in a minute is 300 ÷ 60 = 5 downloads per second — the same divide-by-60 step turns any per-minute rate (heartbeats, footsteps, words typed) into a per-second one, and multiplying by 60 reverses it.

15 KS3-MATH-IMC-0201

Among the digits 0 to 9, the perfect squares are ____, ____, ____ and ____.

0; 1; 4; 9

Hint0², 1², 2² and 3² all land back inside a single digit.

WhyThese four values matter whenever a puzzle asks whether some digit, or a product of two digits, is itself square — checking against just 0, 1, 4, 9 is faster than testing a square root each time.

16 KS3-MATH-IMC-0202

A smaller square sits inside a larger square with each vertex touching a side, splitting the border into four triangles. What is true of the four triangles?

They are all congruent

HintEach one is a copy of its neighbour turned through a quarter of a full turn.

WhyBecause they're congruent right-angled triangles, the shaded band between the two squares splits into four equal pieces — so one triangle's area is a quarter of (big square's area − small square's area), without needing individual side lengths.

17 KS3-MATH-IMC-0203

You swap the order of the two numbers in a multiplication. What happens to the product?

Nothing — the product is unchanged

HintSeven threes and three sevens land on the same total.

WhyThis is the commutative property: a × b = b × a for any numbers a and b. A calculator whose keys have been scrambled still gives a correct answer for a calculation only when reordering its two factors truly leaves the result unchanged.

18 KS3-MATH-IMC-0204

Two similar shapes have corresponding side lengths in ratio 1 : k. What is the ratio of their areas?

1 : k²

HintArea is a two-dimensional measurement, so it scales with length squared.

WhyDoubling every side (k = 2) always multiplies area by 4, and tripling sides multiplies area by 9 — true for any pair of similar shapes, from triangles to hexagons, not just squares or circles.

19 KS3-MATH-IMC-0205

When two triangles overlap, what is the greatest possible number of sides the overlapping region can have?

6

HintEach triangle can only contribute at most three of its own edges to the boundary.

WhyThe overlap can be a triangle, quadrilateral, pentagon or hexagon depending on how the two triangles cross — but it can never have seven or more sides, since between them the two triangles only have six edges to contribute.

20 KS3-MATH-IMC-0206

Three sectors are drawn at a triangle's three vertices, each sector's angle equal to that vertex's interior angle. What fraction of a full circle do the three sectors make together?

One half

HintThe three interior angles of any triangle add to 180°.

Why180° is exactly half of the 360° in a full turn, so however the triangle is shaped, three same-radius sectors placed at its corners always combine into half a circle — a shortcut for area problems that dot circles around a triangle's vertices.

21 KS3-MATH-IMC-0207

The exterior angle of a regular polygon with n sides

360°/n

HintAll the exterior angles of any polygon add up to one full turn.

WhyA regular pentagon's exterior angle is 360°/5 = 72°, so its interior angle is 180° − 72° = 108°. Composite figures that glue a regular polygon onto a square or triangle almost always start from this calculation.

22 KS3-MATH-IMC-0208

Two expressions like xy and (x + a)(y + b) are set equal because they represent equal areas. After expanding, why does the xy term disappear from the equation?

It appears on both sides and cancels

HintExpand fully — the same xy term is hiding in every expansion, so it drops out.

WhyOnce the xy term cancels, what's left is a linear equation in x and y — the standard shortcut whenever two rectangles built from linear expressions are given equal area, turning what looks like a quadratic problem into ordinary simultaneous equations.

23 KS3-MATH-IMC-0209

The area of an equilateral triangle with side length l is ____.

(√3/4)l²

HintSplit it into two right-angled halves; the height works out to (√3/2) times the side.

WhyThis comes from area = ½ × base × height, with height = (√3/2)l. A hexagon or star built from equilateral triangles of different sizes becomes a few applications of this one formula rather than fresh area work each time.

24 KS3-MATH-IMC-0210

The largest value in a list, given its smallest value x and range r

x + r

HintRange is defined as the gap between the biggest and smallest values in a set.

WhyWriting the largest value as x + r turns a 'how many lists satisfy…' problem into an algebraic one: fix the smallest value as x, and every other constraint (mean, median, mode) becomes an equation or inequality in x and the other unknowns.

25 KS3-MATH-IMC-0211

What is the exact value of tan 30°?

1/√3

HintBisect an equilateral triangle to get a right-angled triangle with sides 1, √3 and 2.

WhyThat bisected equilateral triangle gives every standard trig ratio at 30° and 60° without a calculator — rationalised, 1/√3 = √3/3. It comes up whenever a problem asks for a perpendicular distance created by a 60° angle.

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