Grade 9-10 Mathematics Contest Solutions and Notes

Overview of 9–10th Grade Mathematics Contest Problems

• Taget Audience / Grade Level: 9th and 10th-grade high school students (Feladatok 9–10. oszt!1lyososok r!1sz!1re).

• Topic Coverage:

  • Solid Geometry and Algebra (Volume and Cubic Equations)

  • Combinatorics and Grid Colorings (Binary Matrix Constraints)

  • Number Grids and Submatrix Sum Constraints

  • Graph Theory and Planar Topology (Street Intersections)

  • Number Theory and Consecutive Digit Properties

Problem 15: Volumes of Two Cubic Water Tanks

• Problem Statement:

  • Two cubic water tanks are given, where the larger tank's inner edge is $1\,m/s^2$ or $1\,\text{dm}$ longer than that of the smaller tank.

  • Initially, the larger tank is completely filled with water, while the smaller tank is empty.

  • Water is poured from the larger tank into the smaller tank until the smaller tank is completely full.

  • After filling the smaller tank, $217\,\text{liters}$ ($\text{dm}^3$) of water remain in the larger tank.

  • The task is to determine the capacity (in liters) of the smaller tank.

• Multiple-Choice Options:

  • A) $125$

  • B) $133$

  • C) $243$

  • D) $512$

  • E) $729$

• Mathematical Formulation & Step-by-Step Solution:

  • Let $x\,\text{dm}$ represent the inner edge length of the smaller cubic tank.

  • Then, the inner edge length of the larger cubic tank is $(x + 1)\,\text{dm}$.

  • The volume of a cube with edge length $a$ is given by:     $V = a^3$

  • Volume of the smaller tank:     $V_1 = x^3\,\text{dm}^3 = x^3\,\text{liters}$

  • Volume of the larger tank:     $V_2 = (x + 1)^3\,\text{dm}^3 = (x + 1)^3\,\text{liters}$

  • The volume of remaining water in the larger tank after transferring $V_1$ liters to the smaller tank is:     $V_{\text{remaining}} = V_2 - V_1 = (x + 1)^3 - x^3$

  • Setting this remaining volume equal to $217\,\text{liters}$:     $(x + 1)^3 - x^3 = 217$

  • Expanding $(x + 1)^3$ using the binomial expansion formula $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:     $x^3 + 3x^2 + 3x + 1 - x^3 = 217$

  • Simplifying the algebraic equation:     $3x^2 + 3x + 1 = 217$     $3x^2 + 3x - 216 = 0$

  • Dividing the entire equation by $3$:     $x^2 + x - 72 = 0$

  • Factoring the quadratic polynomial:     $(x + 9)(x - 8) = 0$

  • Roots of the equation:     $x = -9 \quad \text{or} \quad x = 8$

  • Since an edge length must be a strictly positive real number ($x > 0$), the negative root $x = -9$ is discarded.

  • Therefore, the inner edge length of the smaller tank is:     $x = 8\,\text{dm}$

  • Calculating the volume of the smaller tank:     $V_1 = x^3 = 8^3 = 512\,\text{liters}$

• Final Answer:

  • The smaller tank has a capacity of $512\,\text{liters}$.

  • Correct Option: D) 512

Problem 16: Binary Grid Coloring with Row and Column Constraints

• Problem Statement:

  • A $4 \times 4$ grid has each of its 16 cells painted either white or black.

  • Next to each row and below each column, the exact number of required black cells for that row or column is specified:

  • Row 1 constraint: $2$ black cells

  • Row 2 constraint: $0$ black cells

  • Row 3 constraint: $1$ black cell

  • Row 4 constraint: $1$ black cell

  • Column 1 constraint: $2$ black cells

  • Column 2 constraint: $0$ black cells

  • Column 3 constraint: $1$ black cell

  • Column 4 constraint: $1$ black cell

  • The task is to find the total number of distinct valid grid colorings that satisfy all row and column constraints simultaneously.

• Multiple-Choice Options:

  • A) $0$

  • B) $1$

  • C) $3$

  • D) $5$

  • E) $9$

• Detailed Combinatorial Analysis:

  • Denoting the cell in row $r$ and column $c$ as $(r, c)$, where $r, c \in \{1, 2, 3, 4\}$.

  • Constraint Reduction:

  • Row 2 has constraint $0$: all cells in Row 2—$(2,1), (2,2), (2,3), (2,4)$—must be white.

  • Column 2 has constraint $0$: all cells in Column 2—$(1,2), (2,2), (3,2), (4,2)$—must be white.

  • This reduces the problem to choosing black cells within a $3 \times 3$ grid formed by Rows$\{1, 3, 4\}$ and Columns$\{1, 3, 4\}$.

  • Reduced Constraints:

  • Row 1 needs $2$ black cells from$\{(1,1), (1,3), (1,4)\}$.

  • Row 3 needs $1$ black cell from$\{(3,1), (3,3), (3,4)\}$.

  • Row 4 needs $1$ black cell from$\{(4,1), (4,3), (4,4)\}$.

  • Column 1 needs $2$ black cells from$\{(1,1), (3,1), (4,1)\}$.

  • Column 3 needs $1$ black cell from$\{(1,3), (3,3), (4,3)\}$.

  • Column 4 needs $1$ black cell from$\{(1,4), (3,4), (4,4)\}$.

  • Case-by-Case Enumeration:

  • Case 1: Cell $(1,1)$ is BLACK.

    • Column 1 already has $1$ black cell at $(1,1)$, so exactly $1$ additional black cell is needed among $(3,1)$ and $(4,1)$.

    • Row 1 already has $1$ black cell at $(1,1)$, so exactly $1$ additional black cell is needed among $(1,3)$ and $(1,4)$.

    • Subcase 1a: Cell $(1,3)$ is BLACK (and $(1,4)$ is WHITE).

    • Since Column 3 requires exactly $1$ black cell, $(1,3)$ fulfills Column 3 completely. Thus, $(3,3)$ and $(4,3)$ must be WHITE.

    • Now, Row 3 must place its $1$ black cell in either $(3,1)$ or $(3,4)$.

    • Row 4 must place its $1$ black cell in either $(4,1)$ or $(4,4)$.

    • Since Column 1 needs $1$ more black cell and Column 4 needs $1$ black cell:

      • Option 1a-i: Set $(3,1)$ BLACK and $(4,4)$ BLACK (with $(3,4)$ and $(4,1)$ WHITE).

      • Valid coloring 1: Black cells at$\{(1,1), (1,3), (3,1), (4,4)\}$.

      • Option 1a-ii: Set $(4,1)$ BLACK and $(3,4)$ BLACK (with $(3,1)$ and $(4,4)$ WHITE).

      • Valid coloring 2: Black cells at$\{(1,1), (1,3), (4,1), (3,4)\}$.

    • Subcase 1b: Cell $(1,4)$ is BLACK (and $(1,3)$ is WHITE).

    • Since Column 4 requires exactly $1$ black cell, $(1,4)$ fulfills Column 4 completely. Thus, $(3,4)$ and $(4,4)$ must be WHITE.

    • Now, Column 3 needs its $1$ black cell at either $(3,3)$ or $(4,3)$.

    • Option 1b-i: Set $(3,3)$ BLACK.

      • Row 3 is now satisfied, so $(3,1)$ must be WHITE.

      • To satisfy Column 1's requirement of $2$ black cells, $(4,1)$ must be BLACK.

      • Valid coloring 3: Black cells at$\{(1,1), (1,4), (3,3), (4,1)\}$.

    • Option 1b-ii: Set $(4,3)$ BLACK.

      • Row 4 is now satisfied, so $(4,1)$ must be WHITE.

      • To satisfy Column 1's requirement of $2$ black cells, $(3,1)$ must be BLACK.

      • Valid coloring 4: Black cells at$\{(1,1), (1,4), (4,3), (3,1)\}$.

  • Case 2: Cell $(1,1)$ is WHITE.

    • Since Row 1 needs $2$ black cells among$\{(1,1), (1,3), (1,4)\}$ and $(1,1)$ is white, BOTH $(1,3)$ and $(1,4)$ MUST BE BLACK.

    • Since $(1,3)$ is black, Column 3 is completely filled (its target is 1), so $(3,3)$ and $(4,3)$ must be WHITE.

    • Since $(1,4)$ is black, Column 4 is completely filled (its target is 1), so $(3,4)$ and $(4,4)$ must be WHITE.

    • Look at Row 3: $(3,3)$ and $(3,4)$ are white, but Row 3 requires $1$ black cell. Therefore, $(3,1)$ MUST BE BLACK.

    • Look at Row 4: $(4,3)$ and $(4,4)$ are white, but Row 4 requires $1$ black cell. Therefore, $(4,1)$ MUST BE BLACK.

    • Check Column 1: $(3,1)$ and $(4,1)$ are black, giving Column 1 exactly $2$ black cells, which satisfies Column 1's constraint.

    • Valid coloring 5: Black cells at$\{(1,3), (1,4), (3,1), (4,1)\}$.

  • Summary of All Valid Colorings:

  1. $\{(1,1), (1,3), (3,1), (4,4)\}$

  2. $\{(1,1), (1,3), (4,1), (3,4)\}$

  3. $\{(1,1), (1,4), (3,3), (4,1)\}$

  4. $\{(1,1), (1,4), (4,3), (3,1)\}$

  5. $\{(1,3), (1,4), (3,1), (4,1)\}$

  • Total number of valid colorings = $5$.

• Final Answer:

  • There are $5$ distinct valid colorings.

  • Correct Option: D) 5

Problem 17: Completing a 3x3 Grid with Subgrid Sum Constraints

• Problem Statement:

  • Niki wants to complete a $3 \times 3$ grid such that the sum of the four numbers in every $2 \times 2$ subgrid is equal to 10$.\n * The partially filled 3 \times 3 grid is given as follows:\n * Top row: 1$,$b$,$0\n * Middle row: d$,$2$,$f\n * Bottom row: 4$,$h$,$3\n * The task is to determine the sum of the four missing numbers written into the empty positions (b + d + f + h).\n\n* **Multiple-Choice Options:**\n * **A)** 9\n * **B)** 10\n * **C)** 11\n * **D)** 12\n * **E)** 13\n\n* **Algebraic Derivation:**\n * Let the 3 \times 3 matrix be represented as:\n    \begin{pmatrix} 1 & b & 0 \ d & 2 & f \ 4 & h & 3 \end{pmatrix}\n * Set up the system of linear equations for each of the four 2 \times 2 subgrids:\n 1. **Top-Left 2 \times 2 Subgrid:**\n       1 + b + d + 2 = 10 \implies b + d + 3 = 10 \implies b + d = 7\n 2. **Top-Right 2 \times 2 Subgrid:**\n       b + 0 + 2 + f = 10 \implies b + f + 2 = 10 \implies b + f = 8\n 3. **Bottom-Left 2 \times 2 Subgrid:**\n       d + 2 + 4 + h = 10 \implies d + h + 6 = 10 \implies d + h = 4\n 4. **Bottom-Right 2 \times 2 Subgrid:**\n       2 + f + h + 3 = 10 \implies f + h + 5 = 10 \implies f + h = 5\n\n * **Calculating the Sum S = b + d + f + h:**\n * Summing equation (1) and equation (4):\n      S = (b + d) + (f + h) = 7 + 5 = 12\n * Alternatively, summing equation (2) and equation (3):\n      S = (b + f) + (d + h) = 8 + 4 = 12\n\n * **Consistency Check:**\n * From (1) and (2): f - d = (b + f) - (b + d) = 8 - 7 = 1$.

  • From (3) and (4): f - d = (f + h) - (d + h) = 5 - 4 = 1$.\n * The system is fully consistent with one free parameter (e.g., choosing d = 2$yields$b = 5$,$f = 3$,$h = 2$, giving$5 + 2 + 3 + 2 = 12).\n\n* **Final Answer:**\n * The sum of the four missing numbers is 12$.

  • Correct Option: D) 12

Problem 18: Map Sketch and Topological Identification of the Curved Street

• Problem Statement:

  • On a boat trip, Julcsi attempted to draw a map sketch of her village.

  • She drew four streets containing her friends' houses (Anna, Bea, Cili, D!1ori) and seven street intersections.

  • Due to heavy sea waves, she could not draw straight lines.

  • In reality, three of the four streets are perfectly straight lines, while only the fourth street is curved.

  • The task is to identify which friend lives in the curved street.

• Multiple-Choice Options:

  • A) Anna

  • B) Bea

  • C) Cili

  • D) D!1ori

  • E) Cannot be determined (nem lehet meghatározni)

• Geometric and Topological Analysis:

  • Fundamental Geometric Principle:

  • In Euclidean geometry, two distinct straight lines can intersect at most once ($|L_i \cap L_j| \le 1$).

  • Intersection Upper Bound for Straight Streets:

  • If three streets $L_1, L_2, L_3$ were all straight lines, the maximum possible number of intersection points between them would be:       $\binom{3}{2} = 3$

  • Role of the Curved Street:

  • Since the map contains $7$ intersection points in total, and the three straight streets can account for at most $3$ intersections among themselves, the curved street $C$ must intersect the straight streets at least:       $7 - 3 = 4$       intersection points.

  • Pairwise Intersection Rule:

  • Any pair of lines that intersect each other two or more times CANNOT both be straight lines; at least one of them must be the curved street.

  • Tracing the Map Sketch:

  • Following the path of the street attached to the label D!1ori (originating in the upper-right region):

    • D!1ori's street crosses the other streets multiple times, intersecting another single street at two distinct crossing points.

    • Therefore, D!1ori's street fails the straight-line intersection property and must be the unique curved street in the village.

• Final Answer:

  • D!1ori lives in the curved street.

  • Correct Option: D) D!1ori

Problem 19: Longest Sequence of Three-Digit Numbers Containing an Odd Digit

• Problem Statement:

  • Pista wants to write down as many consecutive three-digit positive integers ($100 \le n \le 999$) as possible such that every written number contains at least one odd digit ($1, 3, 5, 7, 9$).

  • The task is to find the maximum number of consecutive integers Pista can write down.

• Multiple-Choice Options:

  • A) $1$

  • B) $10$

  • C) $110$

  • D) $121$

  • E) Other value (m!1s !1rt!1k)

• Detailed Number-Theoretic Analysis:

  • Complementary Condition:

  • A number fails Pista's condition if and only if ALL of its digits are EVEN ($0, 2, 4, 6, 8$).

  • We refer to a number composed entirely of even digits as an "all-even-digit number".

  • Pista's sequence terminates whenever it encounters an all-even-digit number.

  • Thus, maximizing the length of consecutive valid numbers requires finding the largest gap between two consecutive all-even-digit numbers (or between an all-even-digit number and the boundary of three-digit numbers).

  • Structure of All-Even-Digit Numbers:

  • A three-digit number $n = 100a + 10b + c$ is all-even if $a \in \{2, 4, 6, 8\}$ and $b, c \in \{0, 2, 4, 6, 8\}$.

  • Analyzing Ranges with Odd Hundreds Digits:

  • When the hundreds digit $a$ is odd ($a \in \{1, 3, 5, 7, 9\}$), EVERY single number in that entire century contains at least one odd digit (namely, its hundreds digit).

  • For example, every number from $300$ to $399$ has hundreds digit $3$, so all $100$ numbers in this block are valid.

  • Extending Across the Century Boundary:

  • Consider the transition into and through a century with an odd hundreds digit, such as the $300\text{s}$:

    • Look at numbers preceding $300$ in the late $200\text{s}$:

    • The last all-even-digit number in the $200\text{s}$ is $288$ (digits $2, 8, 8$ are all even).

    • Numbers from $289$ to $299$:

      • $289$: contains odd digit 9$.\n * 290$to$299$: tens digit is$9$(odd), so ALL$10 numbers contain an odd digit.\n * Total valid numbers in this stretch: 299 - 289 + 1 = 11$numbers ($289, 290, 291, \dots, 299).\n * Numbers from 300$to$399:\n * Hundreds digit is 3$(odd), so ALL$100$numbers are valid ($300, 301, \dots, 399).\n * Number 400:\n * Digits are 4, 0, 0 (all even). This number violates the condition and terminates the sequence.\n\n * **Calculating Total Sequence Length:**\n * The unbroken sequence of valid numbers ranges from 289$to$399 inclusive.\n * Count of numbers in the sequence:\n      \text{Length} = 399 - 289 + 1 = 111\n\n * **Verification of Maximality:**\n * An identical maximum length of 111$is achieved in other odd-century transitions (e.g.,$889$to$999$, since$1000 is a four-digit number).\n * Since 111$is not listed among options A ($1$), B ($10$), C ($110$), or D ($121), the correct answer is "other value".\n\n* **Final Answer:**\n * The maximum number of consecutive three-digit integers is 111$.

  • Correct Option: E) m!1s !1rt!1k