Modelled on the shape of a typical first national round in mathematics: a no-calculator multiple-choice contest section that gets harder, followed by two short proof problems of the kind set in a first proof round; the medal bands are only a rough guide.
90minutes35marksNo calculator
Time allowed: 90 minutes. No calculator, no formula sheet.
Section A: 15 multiple-choice questions, one mark each, no penalty for wrong answers. The questions get harder as you go.
Section B: 2 proof problems, 10 marks each. Write full solutions: a correct answer with no reasoning earns little credit.
Rough working is not marked. Diagrams are not drawn to scale.
Total: 35 marks.
The clock starts when you press Start. Multiple-choice answers are marked for you; for written answers you get the mark scheme and give yourself the marks. Your answers stay on this device.
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Feeder round mock paper 2 TheLucidSTEM · Maths olympiad feeder round mock · 90 minutes · 35 marks · No calculator Name: ______________________ Date: ____________ · MARK SCHEME
Section A: multiple choice
Suggested time 40 minutes. One mark each, no penalty for wrong answers. Choose one option for each question.
1[1]
What is the value of $1 - 2 + 3 - 4 + 5 - 6 + \cdots + 99 - 100$?
Answer B. Group the terms in pairs: $(1-2) + (3-4) + \cdots + (99-100)$. There are 50 pairs and each is $-1$, so the total is $-50$. The answer $50$ has the right size but the wrong sign: in every pair the even number is the larger one.
2[1]
Each interior angle of a regular polygon is $156^\circ$. How many sides does the polygon have?
Answer D. Each exterior angle is $180^\circ - 156^\circ = 24^\circ$, and the exterior angles of any convex polygon add up to $360^\circ$. So there are $360/24 = 15$ sides. Check: a 15-gon has interior angle sum $13 \times 180^\circ = 2340^\circ$, and $2340/15 = 156$. The answer $12$ would need interior angles of $150^\circ$.
3[1]
A fair six-sided die is rolled three times. What is the probability that the three scores are in strictly increasing order (first $\lt$ second $\lt$ third)?
Answer A. There are $6^3 = 216$ equally likely outcomes. Each choice of three different scores, $\binom63 = 20$ ways, gives exactly one strictly increasing order. So the probability is $\tfrac{20}{216} = \tfrac{5}{54}$. The tempting $\tfrac16$ (one order out of $3! = 6$) forgets that the three scores need not be different; $\tfrac{7}{27}$ counts non-decreasing orders such as $2, 2, 5$.
4[1]
What is the sum of the digits of $10^{25} - 25$?
Answer E. $10^{25}$ is 1 followed by 25 zeros, so $10^{25} - 25 = 99\ldots975$, a 25-digit number made of 23 nines followed by 7 and 5 (compare $10^4 - 25 = 9975$). The digit sum is $23 \times 9 + 7 + 5 = 207 + 12 = 219$. The answer $207$ counts only the nines.
5[1]
A rectangle is divided into a $4 \times 5$ grid of unit squares (4 rows, 5 columns). How many rectangles of any size, including squares, are drawn on the grid with sides along the grid lines?
Answer C. A rectangle is fixed by choosing two of the 5 horizontal lines and two of the 6 vertical lines: $\binom52 \times \binom62 = 10 \times 15 = 150$. The answer $60 = \binom42\binom52$ chooses from the rows and columns of squares instead of the lines, which misses every rectangle one square wide or tall.
6[1]
A square is drawn inside a semicircle of radius $5$ so that one side of the square lies on the diameter and the other two vertices lie on the arc. What is the area of the square?
Answer C. By symmetry the square is centred on the centre $O$ of the diameter. If the side is $s$, a top vertex is $\tfrac{s}{2}$ across and $s$ up from $O$, and it lies on the arc, so $\left(\tfrac s2\right)^2 + s^2 = 25$. Then $\tfrac54 s^2 = 25$ and the area is $s^2 = 20$. The answer $50$ is the square inscribed in the whole circle (diagonal 10), a different set-up.
7[1]
For how many integers $k$ does the equation $x^2 + kx + 36 = 0$ have two different integer roots?
Answer B. If the roots are integers $r$ and $s$ then $rs = 36$ and $k = -(r + s)$. The factor pairs with $r \ne s$ are $\{1, 36\}, \{2, 18\}, \{3, 12\}, \{4, 9\}$ and their negatives, giving $k = \mp 37, \mp 20, \mp 15, \mp 13$: 8 values. The tempting $10$ also counts $\{6, 6\}$ and $\{-6, -6\}$ ($k = \mp 12$), but those give a repeated root, not two different roots; $4$ forgets the negative pairs.
8[1]
A bag holds 3 red and 2 blue balls. Balls are taken out one at a time, at random and without replacement, until both blue balls have been taken out. What is the probability that exactly 4 balls are taken out?
Answer E. Imagine all 5 balls taken out in a random order. The positions of the two blue balls are a random pair from $\binom52 = 10$ equally likely pairs. Exactly 4 draws are needed when the later blue ball is in position 4: the pairs $\{1,4\}, \{2,4\}, \{3,4\}$. So the probability is $\tfrac3{10}$. The tempting $\tfrac15$ treats the position of the last blue ball as equally likely to be any of the 5 places, but later places are more likely.
9[1]
How many of the integers $1, 2, 3, \ldots, 2026$ are neither a perfect square nor a perfect cube?
Answer B. Squares up to 2026: $1^2, \ldots, 45^2 = 2025$, so 45. Cubes: $1^3, \ldots, 12^3 = 1728$ ($13^3 = 2197$ is too big), so 12. Numbers that are both are sixth powers: $1, 64, 729$, so 3. By inclusion-exclusion, $45 + 12 - 3 = 54$ numbers are a square or a cube, leaving $2026 - 54 = 1972$. The tempting $1969 = 2026 - 57$ subtracts the sixth powers twice.
10[1]
Triangle $ABC$ has $AB = AC$ and $\angle BAC = 36^\circ$, and $BC = 1$. What is the length $AC$?
Answer D. The base angles are $72^\circ$. Let the bisector of $\angle B$ meet $AC$ at $D$. Then $\angle DBA = 36^\circ = \angle A$, so $DA = DB$, and $\angle BDC = 72^\circ = \angle C$, so $DB = BC = 1$. Triangle $BCD$ (angles $36^\circ, 72^\circ, 72^\circ$) is similar to $ABC$, so $\dfrac{CD}{BC} = \dfrac{BC}{AC}$. With $AC = x$ and $CD = x - 1$: $x(x - 1) = 1$, so $x = \tfrac{1 + \sqrt5}{2}$. The value $\tfrac{\sqrt5 - 1}{2}$ is $CD$, the short piece, not $AC$.
11[1]
Two numbers $X$ and $Y$ are chosen independently and uniformly at random from the interval $[0, 1]$. What is the probability that $|X - Y| \lt \tfrac13$?
Answer C. Plot $(X, Y)$ as a point in the unit square. The region $|X - Y| \ge \tfrac13$ is two corner triangles, each with legs $\tfrac23$, so total area $2 \times \tfrac12 \times \left(\tfrac23\right)^2 = \tfrac49$. The required probability is $1 - \tfrac49 = \tfrac59$. The tempting $\tfrac23$ adds a band of width $\tfrac13$ on each side of the diagonal without noticing that the band is cut off at the corners of the square.
12[1]
All the diagonals of a convex 12-sided polygon are drawn, and no three of them pass through the same point inside the polygon. How many points inside the polygon are crossing points of two diagonals?
Answer E. Any 4 vertices form a convex quadrilateral whose two diagonals cross at exactly one interior point, and every crossing point comes from exactly one such set of 4 vertices (the ends of the two diagonals through it). So the count is $\binom{12}{4} = 495$. The answer $54$ is the number of diagonals, $\tfrac{12 \times 9}{2}$, not the number of crossings.
13[1]
What is the smallest value of $(x - 1)(x - 2)(x - 3)(x - 4)$ as $x$ ranges over all real numbers?
Answer A. Pair the outer and inner factors: $(x-1)(x-4) = x^2 - 5x + 4$ and $(x-2)(x-3) = x^2 - 5x + 6$. With $t = x^2 - 5x + 5$ the product is $(t - 1)(t + 1) = t^2 - 1 \ge -1$, with equality when $t = 0$, which happens at the real values $x = \tfrac{5 \pm \sqrt5}{2}$. So the minimum is $-1$. The tempting $\tfrac{9}{16}$ is the value at the centre of symmetry $x = \tfrac52$, which is a local maximum, not the minimum.
14[1]
What is the sum of all positive integers $n$ for which $n + 10$ divides $n^2 + 10$?
Answer D. Since $n^2 + 10 = (n + 10)(n - 10) + 110$, the number $n + 10$ divides $n^2 + 10$ exactly when it divides $110 = 2 \times 5 \times 11$. The divisors of 110 above 10 are $11, 22, 55, 110$, giving $n = 1, 12, 45, 100$ and sum $158$. The tempting $148$ comes from the sign slip $n^2 + 10 = (n+10)(n-10) + 90$, which leads to the divisors of 90 instead.
15[1]
How many triangles have whole-number side lengths and perimeter $24$? (Triangles with the same three side lengths count once.)
Answer A. Write the sides as $a \le b \le c$ with $a + b + c = 24$. The triangle inequality $a + b \gt c$ means $c \lt 12$, and $c \ge 8$ because $c$ is the largest. For each $c$ count the pairs $a \le b \le c$ with $a + b = 24 - c$. $c = 8$: $(8,8)$, 1 pair. $c = 9$: $(6,9), (7,8)$, 2 pairs. $c = 10$: $(4,10), (5,9), (6,8), (7,7)$, 4 pairs. $c = 11$: $(2,11), (3,10), (4,9), (5,8), (6,7)$, 5 pairs. That is $1 + 2 + 4 + 5 = 12$. The tempting $18$ also counts the 6 flat cases with $c = 12$ and $a + b = 12$, which are not triangles.
Section B: proof problems
Suggested time 50 minutes (about 25 minutes per problem). Give complete, clearly explained arguments. Later parts may use earlier parts even if you could not prove them.
16The quadrilateral of midpoints[10 marks]
$ABCD$ is a convex quadrilateral. $P$, $Q$, $R$ and $S$ are the midpoints of the sides $AB$, $BC$, $CD$ and $DA$. The diagonals $AC$ and $BD$ are drawn dashed in the figure, which is not to scale. You may use the midpoint theorem: the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. Areas are written $[\ldots]$, so $[ABCD]$ is the area of $ABCD$.
(a) Prove that $PQRS$ is a parallelogram.
[3]
Mark scheme
Answer: $PQ$ and $SR$ are both parallel to $AC$ and both equal to $\tfrac12 AC$.
In triangle $ABC$, $P$ and $Q$ are midpoints of $AB$ and $BC$, so $PQ \parallel AC$ and $PQ = \tfrac12 AC$. [1]
In triangle $ADC$, $S$ and $R$ are midpoints of $AD$ and $DC$, so $SR \parallel AC$ and $SR = \tfrac12 AC$. [1]
So $PQ$ and $SR$ are parallel and equal, and a quadrilateral with one pair of opposite sides parallel and equal is a parallelogram. [1]
Accept a vector proof: with position vectors, $\vec{PQ} = \tfrac12(\mathbf c - \mathbf a) = \vec{SR}$ [2], hence a parallelogram [1]. Showing only that $PQ \parallel SR$ (without equal length or the second pair of sides) earns at most [2].
(b) Prove that $[PQRS] = \tfrac12 [ABCD]$.
[3]
Mark scheme
Answer: The four corner triangles cut off have total area $\tfrac12 [ABCD]$.
Triangle $APS$ is an enlargement of triangle $ABD$ with scale factor $\tfrac12$ (centre $A$), so $[APS] = \tfrac14 [ABD]$; similarly $[CRQ] = \tfrac14 [CDB]$. Adding, $[APS] + [CRQ] = \tfrac14 [ABCD]$. [1]
In the same way, using diagonal $AC$: $[BQP] + [DSR] = \tfrac14([BCA] + [DAC]) = \tfrac14 [ABCD]$. [1]
Since $ABCD$ is convex, it is made up of $PQRS$ and these four corner triangles without overlap, so $[PQRS] = [ABCD] - \tfrac14[ABCD] - \tfrac14[ABCD] = \tfrac12 [ABCD]$. [1]
Accept equivalent methods (for example coordinates or vectors with the cross product) with full working. Common error: assuming the corner triangles are all equal in area, which is false in general: [0] for that step.
(c) Prove that $PQRS$ is a rectangle if and only if the diagonals $AC$ and $BD$ are perpendicular.
[2]
Mark scheme
Answer: $PQ \parallel AC$ and $QR \parallel BD$, so $\angle PQR = 90^\circ$ exactly when $AC \perp BD$.
By (a) $PQ \parallel AC$, and in the same way (triangle $BCD$) $QR \parallel BD$. [1]
A parallelogram is a rectangle exactly when two adjacent sides are perpendicular, and $PQ \perp QR$ holds exactly when $AC \perp BD$; both directions follow. [1]
An argument that proves only one direction earns [1] in total.
(d) Suppose $AC = 6$, $BD = 8$ and $AC \perp BD$. Find the perimeter of $PQRS$ and the area of $ABCD$.
[2]
Mark scheme
Answer: Perimeter $14$; $[ABCD] = 24$.
$PQRS$ is a rectangle with sides $\tfrac12 AC = 3$ and $\tfrac12 BD = 4$, so its perimeter is $2(3 + 4) = 14$. [1]
Its area is $12$, so by (b) $[ABCD] = 24$ (also $\tfrac12 \times 6 \times 8 = 24$ directly). [1]
A perimeter of $28$ (using the full diagonals as sides) loses the first mark.
17A sequence that squares and subtracts[10 marks]
A sequence is defined by $a_1 = 2$ and $a_{n+1} = a_n^2 - a_n + 1$ for $n \ge 1$, so $a_2 = 3$, $a_3 = 7$, $a_4 = 43$. You may use without proof that every term is an integer with $a_n \ge 2$.
(a) Show that $a_{n+1} - 1 = a_n(a_n - 1)$, and deduce that $\dfrac{1}{a_n} = \dfrac{1}{a_n - 1} - \dfrac{1}{a_{n+1} - 1}$ for every $n \ge 1$.
Rearranging gives the result; all denominators are non-zero because $a_n \ge 2$. [1]
Checking the identity for $n = 1, 2, 3$ only, without the general algebra, earns [0] for the last two marks.
(b) Prove that $\dfrac1{a_1} + \dfrac1{a_2} + \cdots + \dfrac1{a_n} = 1 - \dfrac{1}{a_{n+1} - 1}$, and hence that this sum is less than $1$ for every $n$.
[2]
Mark scheme
Answer: The sum telescopes to $\dfrac{1}{a_1 - 1} - \dfrac{1}{a_{n+1} - 1}$ and $a_1 - 1 = 1$.
Since $a_1 - 1 = 1$ this is $1 - \dfrac{1}{a_{n+1} - 1}$, which is less than 1 because $a_{n+1} - 1 \gt 0$. [1]
Accept a proof by induction on $n$ using (a). For example $\tfrac12 + \tfrac13 + \tfrac17 + \tfrac1{43} = \tfrac{1805}{1806}$, but a numerical check alone earns [0].
(c) Prove by induction that $a_{n+1} = a_1 a_2 \cdots a_n + 1$ for every $n \ge 1$.
Substituting the assumption gives $a_{n+2} - 1 = a_{n+1} \cdot a_1 a_2 \cdots a_n = a_1 a_2 \cdots a_{n+1}$, which is the statement for $n + 1$; so it holds for all $n$ by induction. [1]
Accept a direct argument: multiplying $a_{k+1} - 1 = a_k(a_k - 1)$ for $k = 1, \ldots, n$ and cancelling gives $a_{n+1} - 1 = a_1 \cdots a_n (a_1 - 1)$ [3]. The induction mark is lost if the conclusion of the step is not clearly stated.
(d) Prove that any two different terms of the sequence have no common factor greater than 1. Deduce that there are infinitely many prime numbers.
[2]
Mark scheme
Answer: For $m \lt n$, $a_n = (a_1 \cdots a_{n-1}) + 1 \equiv 1 \pmod{a_m}$; each term has its own prime factor.
Take $m \lt n$. By (c), $a_n = a_1 \cdots a_{n-1} + 1$, and $a_m$ is one of the factors of the product, so any common divisor $d$ of $a_m$ and $a_n$ divides $a_n - a_1 \cdots a_{n-1} = 1$. So $d = 1$. [1]
Each $a_n \ge 2$ has at least one prime factor; since no two terms share a factor, these primes are all different, and there are infinitely many terms, so there are infinitely many primes. [1]
Saying the terms are prime is wrong ($a_5 = 1807 = 13 \times 139$): an argument that relies on it earns [0] for the second mark.
TheLucidSTEM practice paper. Questions are original and written in the style of a first national round; this is not an official olympiad paper.