Bariz - Problem

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Tags: IJNAMO, algebra, system of equations, geometry, 3D geometry, pyramid, sphere

11.11.2021 19:55

P6. Determine all integer pairs $(x, y)$ satisfying the following system of equations. \[ \begin{cases} x + y - 6 &= \sqrt{2x + y + 1} \\ x^2 - x &= 3y + 5 \end{cases} \] P7. Determine the sum of all (positive) integers $n \leq 2019$ such that $1^2 + 2^2 + 3^2 + \cdots + n^2$ is an odd number and $1^1 + 2^2 + 3^3 + \cdots + n^n$ is also an odd number. P8. Two quadrilateral-based pyramids where the length of all its edges are the same, have their bases coincide, forming a new 3D figure called "8-plane" (octahedron). If the volume of such "8-plane" (octahedron) is $a^3\sqrt{2}$ cm$^3$, determine the volume of the largest sphere that can be fit inside such "8-plane" (octahedron). P9. Six-digit numbers $\overline{ABCDEF}$ with distinct digits are arranged from the digits 1, 2, 3, 4, 5, 6, 7, 8 with the rule that the sum of the first three numbers and the sum of the last three numbers are the same. Determine the probability that such arranged number has the property that either the first or last three digits (might be both) form an arithmetic sequence or a geometric sequence.

HIDE: Remarks (Answer spoiled) It's a bit ambiguous whether the first or last three digits mentioned should be in that order, or not. If it should be in that order, the answer to this problem would be $\frac{1}{9}$, whereas if not, it would be $\frac{1}{3}$. Some of us agree that the correct interpretation should be the latter (which means that it's not in order) and the answer should be $\frac{1}{3}$. However since this is an essay problem, your interpretation can be written in your solution as well and it's left to the judges' discretion to accept your interpretation, or not. This problem is very bashy.

P10. $X_n$ denotes the number which is arranged by the digit $X$ written (concatenated) $n$ times. As an example, $2_{(3)} = 222$ and $5_{(2)} = 55$. For $A, B, C \in \{1, 2, \ldots, 9\}$ and $1 \leq n \leq 2019$, determine the number of ordered quadruples $(A, B, C, n)$ satisfying: \[ A_{(2n)} = 2 \left ( B_{(n)} \right ) + \left ( C_{(n)} \right )^2. \]

Mathological03

11.11.2021 20:06

somebodyyouusedtoknow wrote: P7. Determine the sum of all (positive) integers $n \leq 2019$ such that $1^2 + 2^2 + 3^2 + \cdots + n^2$ is an odd number and $1^1 + 2^2 + 3^3 + \cdots + n^n$ is also an odd number. Dumbest question I've seen this month. $a^i \equiv a \pmod{2}$ for any $i \ge 2$ because $2 \mid a(a - 1) \mid a^i - a$. Therefore we just need \[ 1 + 2 + \dots + n \]to be an odd number, which means $\frac{n(n + 1)}{2}$ has to be odd, i.e. either $n \equiv 1 \pmod{4}$ or $n \equiv 2 \pmod{4}$. The rest is boring.

somebodyyouusedtoknow

11.11.2021 21:37

@above Well yeah but calm down there sir :v

oralayhan

11.11.2021 22:52

for the 6. question

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oralayhan

11.11.2021 23:14

There is definitely a more elegant solution.

BackToSchool

12.11.2021 19:25

oralayhan wrote: for the 6. question Excellent observation at first glance leads to perfect problem solving. The key to solve this problem is to realize that the variable $y$ can be eliminated by substitution.

oralayhan

12.11.2021 21:32

Thanks $BacktoSchool$. I will follow you. I would appreciate it if you could share a short elegant solution.

BackToSchool

27.11.2021 01:13

somebodyyouusedtoknow wrote: P10. $X_n$ denotes the number which is arranged by the digit $X$ written (concatenated) $n$ times. As an example, $2_{(3)} = 222$ and $5_{(2)} = 55$. For $A, B, C \in \{1, 2, \ldots, 9\}$ and $1 \leq n \leq 2019$, determine the number of ordered quadruples $(A, B, C, n)$ satisfying: \[ A_{(2n)} = 2 \left ( B_{(n)} \right ) + \left ( C_{(n)} \right )^2. \] \begin{align*} & \text{ } A_{(2n)} = 2 \left ( B_{(n)} \right ) + \left ( C_{(n)} \right )^2 \\ \implies & \underbrace {AAA \cdots AA}_{2n} = 2\cdot \underbrace {BBB \cdots BB}_{n} + ( \underbrace {CCC \cdots CC} _ {n} )^2 \\ \implies & \frac{A}{9}\cdot (10^{2n}-1) = \frac{2B}{9}\cdot (10^{n}-1) + \left (\frac{C}{9}\cdot (10^{n}-1) \right)^2 \\ \implies & (9A-C^2) \cdot 10^{2n} + (2C^2-18B) \cdot 10^{n} +(18B - 9A - C^2) = 0 \\ \implies & \begin{cases} C^2 = 9A \\ C^2 = 9B \\ C^2 = 18B - 9A \end{cases} \\ \implies & (A, B, C) \in \{(1, 1, 3), (4, 4, 6), (9, 9, 9)\}, n \in \{1, 2, 3, \cdots, 2019\} \end{align*}Thus, the number of ordered quadruples $(A, B, C, n)$ shall be $3 \times 2019 = \boxed {6057}$.

somebodyyouusedtoknow

09.05.2023 12:31

@above you left out 6 solutions, the correct answer is $\boxed{6063}$, which is the case when $11A = C^2 + 2B$ when $n=1$. which are $(A,B,C) = (1,5,1), (2,9,2), (2,3,4), (3,4,5), (5,3,7), (6,1,8)$.