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 1 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 $\dfrac{2026^2 - 2024^2}{2025}$?
Answer B. Difference of two squares: $2026^2 - 2024^2 = (2026-2024)(2026+2024) = 2 \times 4050$. Dividing by $2025$ gives $2 \times 2 = 4$. Squaring the numbers out is slow and error-prone.
2[1]
Five different positive integers have mean $10$. What is the largest possible value of the largest of them?
Answer D. The five numbers add up to $5 \times 10 = 50$. To make the largest as big as possible, make the other four as small as possible while still different: $1+2+3+4 = 10$. The largest is then $50 - 10 = 40$. The answer $44$ comes from wrongly allowing $0$, which is not positive.
3[1]
What is the angle between the hour hand and the minute hand of a clock at 7:20?
Answer C. The minute hand is at $20 \times 6^\circ = 120^\circ$ from 12. The hour hand moves $0.5^\circ$ per minute, so it is at $7 \times 30^\circ + 20 \times 0.5^\circ = 220^\circ$. The angle is $220^\circ - 120^\circ = 100^\circ$. Forgetting that the hour hand has moved past the 7 gives $90^\circ$.
4[1]
A rectangle measures $6$ by $8$. What is the perpendicular distance from one vertex to the diagonal that does not pass through it?
Answer A. The diagonal has length $\sqrt{6^2+8^2} = 10$. The triangle formed by two sides and the diagonal has area $\tfrac12 \times 6 \times 8 = 24$, and also $\tfrac12 \times 10 \times h$. So $h = 48/10 = 4.8$. The value $5$ is half the diagonal, the distance from a vertex to the centre, not to the diagonal.
5[1]
Two fair six-sided dice are rolled. What is the probability that the product of the two scores is a multiple of $3$?
Answer E. Work with the complement. The product is not a multiple of 3 only when neither die shows 3 or 6, which has probability $\left(\tfrac46\right)^2 = \tfrac49$. So the answer is $1 - \tfrac49 = \tfrac59$. Adding $\tfrac13 + \tfrac13$ double-counts the rolls where both dice are multiples of 3.
6[1]
Real numbers $a$, $b$, $c$ satisfy $2^a = 3$, $3^b = 5$ and $5^c = 8$. What is the value of $abc$?
Answer C. Chain the powers: $2^{ab} = (2^a)^b = 3^b = 5$, then $2^{abc} = (2^{ab})^c = 5^c = 8 = 2^3$. So $abc = 3$. No logarithm values are needed.
7[1]
For how many integers $n$ with $1 \le n \le 100$ is $n^2 + n + 1$ divisible by $7$?
Answer A. Check the residues of $n$ modulo 7: $n^2+n+1 \equiv 0$ only for $n \equiv 2$ ($4+2+1 = 7$) and $n \equiv 4$ ($16+4+1 = 21$). Between 1 and 100 there are 15 numbers $\equiv 2$ (2, 9, ..., 100) and 14 numbers $\equiv 4$ (4, 11, ..., 95). Total $15 + 14 = 29$. Counting about $\tfrac27$ of 100 gives roughly 29 but not exactly; the end points must be checked.
8[1]
A triangle has sides $5$, $5$ and $6$. What is the radius of its circumcircle?
Answer D. The height onto the side of length 6 is $\sqrt{5^2 - 3^2} = 4$, so the area is $\tfrac12 \times 6 \times 4 = 12$. Then $R = \dfrac{abc}{4 \times \text{area}} = \dfrac{5 \times 5 \times 6}{48} = \dfrac{25}{8}$. The answer $3$ is half the longest side, which would be correct only for a right-angled triangle.
9[1]
What is the remainder when $10^{100}$ is divided by $13$?
Answer B. Since $10^3 = 1000 = 13 \times 77 - 1$, we have $10^3 \equiv -1$ and so $10^6 \equiv 1 \pmod{13}$. Now $100 = 6 \times 16 + 4$, so $10^{100} \equiv 10^4 = 10^3 \times 10 \equiv -10 \equiv 3 \pmod{13}$.
10[1]
A point $P$ lies inside a square $ABCD$ (vertices in that order) with $PA = 3$, $PB = 4$ and $PC = 5$. What is $PD$?
Answer E. Put $A = (0,0)$, $B = (s,0)$, $C = (s,s)$, $D = (0,s)$ and $P = (x,y)$. Expanding shows $PA^2 + PC^2 = PB^2 + PD^2$ (both equal $x^2+y^2+(s-x)^2+(s-y)^2$). So $PD^2 = 9 + 25 - 16 = 18$ and $PD = 3\sqrt2$. Pairing the wrong vertices (adjacent instead of opposite) gives $\sqrt{17}$ or similar wrong values.
11[1]
How many subsets of $\{1, 2, \ldots, 10\}$, including the empty set, contain no two consecutive integers?
Answer C. Let $s_n$ be the number of such subsets of $\{1, \ldots, n\}$. A subset either leaves out $n$ ($s_{n-1}$ ways) or contains $n$ and so leaves out $n-1$ ($s_{n-2}$ ways). So $s_n = s_{n-1} + s_{n-2}$ with $s_1 = 2$, $s_2 = 3$, giving $5, 8, 13, 21, 34, 55, 89, 144$ for $n = 3, \ldots, 10$. The answer $89$ is $s_9$, one step short.
12[1]
How many ordered pairs $(x, y)$ of positive integers satisfy $\dfrac1x + \dfrac1y = \dfrac1{12}$?
Answer A. Clear fractions: $12y + 12x = xy$, so $(x-12)(y-12) = 144$. Both $x, y > 12$ (each fraction is less than $\tfrac1{12}$), so $x - 12$ can be any positive divisor of $144 = 2^4 \times 3^2$, and $y$ is then fixed. There are $(4+1)(2+1) = 15$ divisors, so 15 ordered pairs. Counting unordered pairs instead would give 8.
13[1]
How many positive integers less than $1000$ have digits that add up to $10$?
Answer E. Write each number with three digits $a, b, c$ (leading zeros allowed). We need $a + b + c = 10$ with each digit at most 9. Without the upper limit there are $\binom{12}{2} = 66$ solutions (stars and bars). The only bad ones have a digit equal to 10 and the others 0: 3 of them. So $66 - 3 = 63$. Forgetting the limit of 9 gives 66; forgetting one- and two-digit numbers gives fewer.
14[1]
How many real numbers $x$ satisfy $x = 10\sin x$ (with $x$ in radians)?
Answer B. Solutions satisfy $|x| \le 10$, and the equation is symmetric under $x \to -x$. $x = 0$ is one solution. For $x > 0$ compare the line $y = x/10$ with $y = \sin x$: one crossing in $(0, \pi)$, none in $(\pi, 2\pi)$ where $\sin x \lt 0$, two in $(2\pi, 3\pi)$ because the hump reaches 1 while the line is below 1 there, and none after that since $\sin x \lt 0$ on $(3\pi, 10]$ (note $10 \lt 4\pi$). That is 3 positive solutions, 3 negative and $x = 0$: 7 in total.
15[1]
How many of the integers $1, 2, \ldots, 1000$ can be written as $\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 3x \rfloor$ for some real number $x$?
Answer D. Write $x = k + f$ with $k$ an integer and $0 \le f \lt 1$. The sum is $6k + \lfloor 2f \rfloor + \lfloor 3f \rfloor$, and $\lfloor 2f \rfloor + \lfloor 3f \rfloor$ takes the values $0, 1, 2, 3$ (for $f$ in $[0,\tfrac13)$, $[\tfrac13,\tfrac12)$, $[\tfrac12,\tfrac23)$, $[\tfrac23,1)$). So the values are exactly the integers that leave remainder 0, 1, 2 or 3 on division by 6. Since $1000 = 6 \times 166 + 4$: the 166 full blocks give $4 \times 166 = 664$, and $997, 998, 999$ (remainders 1, 2, 3) add 3, but $1000$ (remainder 4) does not. Total 667.
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.
16A difference of squares that never hits 2026[10 marks]
This question is about integer solutions of equations of the form $x^2 - 3y^2 = N$, where $x$ and $y$ are integers (positive, negative or zero).
(a) Suppose $x^2 - 3y^2 = 2026$. Prove that $x$ and $y$ are both odd.
[3]
Mark scheme
Answer: $x$, $y$ have the same parity; both even gives a multiple of 4, but $2026 \equiv 2 \pmod 4$.
Since $x^2 - 3y^2 \equiv x^2 + y^2 \equiv x + y \pmod 2$ and 2026 is even, $x$ and $y$ have the same parity. [1]
If both are even then $x^2 - 3y^2$ is a multiple of 4. [1]
But $2026 = 4 \times 506 + 2$ is not a multiple of 4, so both are odd. [1]
Accept any equivalent case check (for example a table of $x^2 - 3y^2 \bmod 4$). Simply stating "2026 is even so both are odd" without ruling out the both-even case earns at most [1].
(b) Prove that the square of every odd integer leaves remainder 1 when divided by 8.
[3]
Mark scheme
Answer: $(2k+1)^2 = 4k(k+1) + 1$ and $k(k+1)$ is even.
Write the odd integer as $2k+1$, so $(2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$. [1]
$k$ and $k+1$ are consecutive integers, so one is even and $k(k+1)$ is even. [1]
Hence $4k(k+1)$ is a multiple of 8 and the square is $1 \pmod 8$. [1]
Accept checking the four odd residues $1, 3, 5, 7 \pmod 8$ (squares $1, 9, 25, 49$, all $\equiv 1$) for full marks, as long as it is explained why residues are enough. Checking a few examples such as $3^2$ and $5^2$ only proves nothing: [0].
(c) Hence prove that the equation $x^2 - 3y^2 = 2026$ has no integer solutions.
By (a) and (b), $x^2 \equiv y^2 \equiv 1 \pmod 8$, so $x^2 - 3y^2 \equiv 1 - 3 = -2 \equiv 6 \pmod 8$. [1]
But $2026 = 8 \times 253 + 2 \equiv 2 \pmod 8$, a contradiction, so there are no solutions. [1]
Common error: writing $-2 \equiv 2 \pmod 8$. That loses the second mark.
(d) Explain why the same argument does not rule out $x^2 - 3y^2 = 2022$, and find one pair of integers $(x, y)$ with $x^2 - 3y^2 = 2022$.
[2]
Mark scheme
Answer: $2022 \equiv 6 \pmod 8$; for example $(x, y) = (45, 1)$.
$2022 = 8 \times 252 + 6 \equiv 6 \pmod 8$, which matches $x^2 - 3y^2$ for odd $x, y$, so there is no contradiction. [1]
A correct pair with a check, for example $45^2 - 3 \times 1^2 = 2025 - 3 = 2022$. [1]
Accept any correct pair, such as $(\pm 45, \pm 1)$, $(87, 43)$ since $7569 - 5547 = 2022$, $(93, 47)$, or $(303, 173)$. A pair with no check shown still earns the mark if it is correct.
17Equalising numbers round a circle[10 marks]
The positions $0, 1, 2, \ldots, n-1$ are marked in order round a circle, so that each position has two neighbours (position $n-1$ and position $0$ are neighbours). An integer is written at each position. A move is: choose two neighbouring positions and add 1 to both of their numbers.
At the start, the number at position $0$ is $1$ and every other number is $0$.
(a) Take $n = 2026$. Prove that, however many moves are made, the numbers can never all be equal.
[4]
Mark scheme
Answer: The (sum at even positions) minus (sum at odd positions) is always 1, but would be 0 if all were equal.
Split the positions into even and odd ones. Because $n = 2026$ is even, every pair of neighbours (including $2025$ and $0$) is one even and one odd position. [1]
Let $E$ be the sum of the numbers at even positions and $O$ the sum at odd positions. Each move adds 1 to $E$ and 1 to $O$, so $E - O$ never changes. [1]
At the start $E - O = 1 - 0 = 1$. [1]
If all 2026 numbers equalled $k$, then $E = O = 1013k$ and $E - O = 0 \ne 1$, a contradiction. [1]
Alternative full solution [4]: each move raises the total by 2, so the total is always odd [2]; if all numbers equalled $k$ the total would be $2026k$, which is even [2]. Common error: claiming the total itself is invariant (it rises by 2 each move): [0] for the invariant marks.
(b) Take $n = 2025$. Show that the numbers can all be made equal.
[3]
Mark scheme
Answer: Make one move on each of the pairs $(1,2), (3,4), \ldots, (2023, 2024)$: every number becomes 1.
A correct sequence of moves stated, for example one move on each pair $(1,2), (3,4), \ldots, (2023,2024)$, which is 1012 moves. [1]
Explanation that positions $1$ to $2024$ are each increased exactly once and position $0$ is untouched. [1]
Conclusion that every number is then 1. [1]
Accept any other correct construction (for example one that makes every number 3). An argument that only says "the invariant of (a) no longer works" proves nothing about possibility: [0].
(c) Take $n = 2026$ again, but now start with the number $1$ at position $0$, the number $1$ at position $j$ (where $1 \le j \le 2025$), and $0$ everywhere else. For which values of $j$ can the numbers all be made equal? Prove your answer.
[3]
Mark scheme
Answer: Exactly when $j$ is odd.
Using $E - O$ from (a): it starts at $2$ if $j$ is even and at $0$ if $j$ is odd. [1]
All equal needs $E - O = 0$, so $j$ even is impossible. [1]
For $j$ odd: positions $1, \ldots, j-1$ are an even number ($j - 1$) of consecutive positions, and so are positions $j+1, \ldots, 2025$ ($2025 - j$ of them). Pair each block into neighbouring pairs and make one move on each pair; every number becomes 1. [1]
Note: the total-parity argument does not settle (c), because the starting total is 2 for every $j$. Stating the answer "$j$ odd" with no proof earns [0]; a correct construction alone earns [1].
TheLucidSTEM practice paper. Questions are original and written in the style of a first national round; this is not an official olympiad paper.