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Exercise 2. Let $\mathrm{ABC}$ be a triangle, $\mathrm{O}$ a point inside this triangle. The line parallel to $(\mathrm{BC})$ passing through $O$ intersects $[C A]$ at $D$, the line parallel to $(C A)$ passing through $O$ intersects $[A B]$ at $E$, the line parallel to $(A B)$ passing through $\mathrm{O}$ intersects $[... | ## Solution to Exercise 2 First Solution:
Let $|A B C|$ denote the area of triangle $A B C$. We observe that
$$
\frac{\mathrm{BF}}{\mathrm{BC}}=\frac{|\mathrm{ABF}|}{|\mathrm{ABC}|}=\frac{|\mathrm{ABO}|}{|\mathrm{ABC}|}
$$
because triangles $A B F$ and $A B C$ with bases $B F$ and $B C$ have the same height, and $A ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Exercise 8. Calculate
$$
\sqrt{7+\sqrt{1+\sqrt{7+\sqrt{1+\sqrt{7+\sqrt{1+\sqrt{7+\sqrt{3+\sqrt{1}}}}}}}}}
$$
Only a numerical answer is expected here. | Solution to exercise 8 We notice the following chain of simplifications: $\sqrt{3+\sqrt{1}}=2$ then $\sqrt{7+\sqrt{3+\sqrt{1}}}=\sqrt{7+2}=3$ then $\sqrt{1+\sqrt{7+\sqrt{3+\sqrt{1}}}}=\sqrt{1+3}=2$, and so on...
Step by step, we deduce that any expression of the type $\sqrt{1+\sqrt{7+\sqrt{1+\sqrt{\cdots}}}}$ equals 2... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 4. A number appears on a computer screen. It is known that if $x$ appears on the screen, the number $x^{2}-2 x+1$ appears right after. If the first number to appear is 2, what is the 2020th number to appear? | Solution to Exercise 4 Let's start by looking at the first values displayed by the computer:
$\triangleright$ The first number is 2.
$\triangleright$ The second number is $2^{2}-2 \times 2+1=1$
$\triangleright$ The third number is $1^{2}-2 \times 1+1=1-2+1=0$
$\triangleright$ The fourth number is $0^{2}-2 \times 0+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Exercise 6.
1) Alice wants to color the integers between 2 and 8 (inclusive) using $k$ colors. She wishes that if $m$ and $n$ are integers between 2 and 8 such that $m$ is a multiple of $n$ and $m \neq n$, then $m$ and $n$ are of different colors. Determine the smallest integer $k$ for which Alice can color the int... | Solution alternative $n^{\circ} 1$ We show, as in the previous case, that for the first question, at least 3 colors are needed and at least 4 in the second.
We propose here to generalize the previous coloring: we construct a coloring with the optimal number of colors to color the integers from 2 to $r$, with $r \geqsl... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
## Exercise 11.
1) Alice wants to color the integers between 2 and 8 (inclusive) using $k$ colors. She wishes that if $m$ and $n$ are integers between 2 and 8 such that $m$ is a multiple of $n$ and $m \neq n$, then $m$ and $n$ are of different colors. Determine the smallest integer $k$ for which Alice can color the in... | Solution alternative $n^{\circ} 1$ We show, as in the previous case, that for the first question, at least 3 colors are needed and at least 4 in the second.
We propose here to generalize the previous coloring: we construct a coloring with the optimal number of colors to color the integers from 2 to $r$, with $r \geqsl... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 2. In a deck of cards consisting only of red cards and black cards, there are 2 times as many black cards as red cards. If 4 black cards are added, there are then 3 times as many black cards as red cards. How many cards did the deck have before adding the 4 black cards?
Only a numerical answer is expected her... | Solution to Exercise 2 Let $r$ be the number of red cards and $n$ be the number of black cards in the initial deck. By hypothesis, we have $n=2 r$. After adding 4 black cards, there are $n+4$ black cards in the deck, and by hypothesis, we have $n+4=3 r$. Thus, $4=3 r-n=3 r-2 r=r$ and $n=2 \times 4=8$. Initially, the de... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 9. In a deck of cards consisting only of red cards and black cards, there are 2 times as many black cards as red cards. If 4 black cards are added, there are then 3 times as many black cards as red cards. How many cards did the deck have before adding the 4 black cards?
Only a numerical answer is expected her... | Solution to Exercise 9 Let $r$ be the number of red cards and $n$ be the number of black cards in the initial deck. By hypothesis, we have $n=2 r$. After adding 4 black cards, there are $n+4$ black cards in the deck, and by hypothesis, we have $n+4=3 r$. Thus, $4=3 r-n=3 r-2 r=r$ and $n=2 \times 4=8$. Initially, the de... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 4. Consider a large square grid with side length 10, divided into small squares with side length 1. Two small squares are said to be neighbors if they share a common side. On each of the small squares, a positive real number is inscribed. Furthermore, 5 frogs move on the grid and can cover one small square eac... | Solution to Exercise 4: We will show that the largest possible value of $k$ is 6. First, let's verify that 6 is indeed achievable. For this, we isolate a $2 \times 3$ rectangle in the top left corner of the grid. On all the cells that are not in this rectangle, we write the number 0. On the cells of the rectangle, we w... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 1. Fred and Sarah are the eldest of the same large family. Fred has twice as few brothers as sisters, while Sarah has as many sisters as brothers.
How many children are there in this family? | Solution to Exercise 1 Let $\mathrm{f}$ be the number of girls and $\mathrm{g}$ be the number of boys in this family.
Fred therefore has $g-1$ brothers and $f$ sisters, and the text states that $f=2(g-1)$. On the other hand, Sarah has $\mathrm{f}-1$ sisters and $\mathrm{g}$ brothers, and the statement tells us that $\... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Exercise 6. For any strictly positive integer $m$, we call the first digit of $m$ the leftmost digit in its decimal representation. Let $n$ be a strictly positive integer. Suppose that the two integers $2^{n}$ and $5^{n}$ have the same first digit. Show that this same first digit is 3. | Solution to Exercise 6 Let $a$ be the first digit common to the decimal representation of $2^{n}$ and $5^{n}$.
verified.
The hypothesis of the statement can then be translated as follows: we... | 3 | Number Theory | proof | Yes | Yes | olympiads | false |
Exercise 7. A set of $n$ non-zero and distinct real numbers is said to be regular if, when these numbers are written on the board in ascending order, the difference between two adjacent numbers is always the same, regardless of the two adjacent numbers chosen. For example, the set $\{4,18,-3,11\}$ is regular because if... | ## Solution to Exercise 7
Answer: $n=4$
In this exercise, we aim to demonstrate that 4 is the largest integer satisfying the property stated in the problem. The solution necessarily consists of two parts. First, we show that if a super-regular set has size $k$, then $k \leqslant 4$; this step is called the analysis. ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 12. For any strictly positive integer $m$, we call the first digit of $m$ the leftmost digit in its decimal representation. Let $n$ be a strictly positive integer. Suppose that the two integers $2^{n}$ and $5^{n}$ have the same first digit. Show that this same first digit is 3. | Solution to Exercise 12 Let $a$ be the first digit common to the decimal representation of $2^{n}$ and $5^{n}$.
verified.
The hypothesis of the statement can then be translated as follows: ... | 3 | Number Theory | proof | Yes | Yes | olympiads | false |
Exercise 13. A set of $n$ non-zero and distinct real numbers is said to be regular if, when these numbers are written on the board in ascending order, the difference between two adjacent numbers is always the same, regardless of the two adjacent numbers chosen. For example, the set $\{4,18,-3,11\}$ is regular because i... | ## Solution to Exercise 13
Answer: $n=4$
In this exercise, we aim to demonstrate that 4 is the largest integer satisfying the property stated in the problem. The solution necessarily includes two parts. First, we show that if a super-regular set has size $k$, then $k \leqslant 4$; this step is called the analysis. Se... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 14. Let $A B C D$ be a rectangle and $M$ the midpoint of the segment $[C D]$. A line parallel to $(A B)$ intersects the segments $[A D],[A M],[B M],[B C]$ at points $P, Q, R$ and $S$, respectively. The line $(D R)$ intersects the segment $[A M]$ at $X$ and the segment $[B C]$ at $Y$. If $D X=6$ and $X R=4$, wh... | ## Solution alternative $n^{\circ} 1$
We propose a solution using Thales' theorem in two other "butterflies."
We introduce $E$ as the intersection point of the lines $(A M)$ and $(B C)$, ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Exercise 6. Determine all integers $n \geqslant 3$ such that it is possible to place $n$ distinct real numbers on a circle so that each of these numbers is the product of its two neighbors. | Solution to Exercise 6: Let $a_{1}, a_{2}, \ldots, a_{n}$ be these numbers, with the convention that $a_{n+1}=a_{1}, a_{n+2}=$ $a_{2}$, etc. By hypothesis, we have $a_{i} a_{i+2}=a_{i+1}$ and $a_{i+1} a_{i+3}=a_{i+2}$ for all $i$. Multiplying these two equalities, we get $a_{i} a_{i+1} a_{i+2} a_{i+3}=a_{i+1} a_{i+2}$,... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Exercise 8. Determine the number of natural numbers $n>2015$ such that $n$ is divisible by $n-2015$. | Solution to exercise 8 Let's write $n=2015+k$. We are looking for the number of positive integers $k$ such that $2015+k$ is divisible by $k$. This is equivalent to $k$ dividing $2015=5 \times 13 \times 31$. The divisors of 2015 are numbers of the form $5^{a} 13^{b} 31^{c}$ with $a, b, c$ equal to 0 or 1: there are 8 of... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 5. In a country, there are 10 cities. Any two cities are always directly connected by exactly one one-way road. The roads never cross, with some passing over others using bridges.
Unfortunately, the officials at the Ministry of Traffic Direction, apparently distracted, have oriented the roads in such a way th... | Solution to exercise 5 a) Starting from any city $V_{1}$. We visit successive cities $V_{2}, V_{3}$, etc. Since we can never return to a city already visited, after a maximum of 10 steps, we are inevitably stuck in a city.
b) Starting from any city $V_{1}$. We visit successive cities $V_{2}, V_{3}$, etc., but only by ... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
Exercise 10. 22 cards bearing the numbers $1,2, \ldots, 22$ are on the table. Alcindor and Benoît take turns removing a card of their choice from the table, until there are none left. Alcindor goes first. Then each calculates the units digit of the sum of their cards. The winner is the one with the highest result.
Doe... | Solution to Exercise 10: Let's show that Alcindor can win for sure.
First, we note that the outcome of the game is not affected if we replace the values of the cards with their units digit. All cards are then in pairs, except for the cards 1 and 2, which are in triples.
Alcindor chooses the following strategy: he sta... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 10. Using the numbers from 1 to 22 exactly once each, Antoine writes 11 fractions: for example, he can write the fractions $\frac{10}{2}, \frac{4}{3}, \frac{15}{5}, \frac{7}{6}, \frac{8}{9}, \frac{11}{19}, \frac{12}{14}, \frac{13}{17}, \frac{22}{21}, \frac{18}{16}$, and $\frac{20}{1}$.
Antoine wishes to have ... | Solution to Exercise 10 First, it is possible to form 10 whole numbers with the fractions
$$
\left\{\frac{22}{11}, \frac{20}{10}, \frac{18}{9}, \frac{16}{8}, \frac{14}{7}, \frac{12}{6}, \frac{21}{3}, \frac{19}{1}, \frac{15}{5}, \frac{4}{2}, \frac{17}{13}\right\}
$$
Furthermore, it is impossible to have 11 whole fract... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 2. Over a certain number of days, it was observed that on each day it rained in the morning, it was sunny in the afternoon, and that on each day it rained in the afternoon, it had been sunny in the morning.
During the observation period, it rained on 15 days, and it was sunny 8 mornings and 13 afternoons.
Ho... | ## Solution to Exercise 2
According to the statement, it did not rain for a full day during the period in question. Since it rained for 15 half-days and was sunny for 21 half-days, this gives a total of 36 half-days, and allows us to assert that we are in fact considering a period of 18 days.
Since it was sunny on 8 ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Exercise 8. Let a be a strictly positive integer, such that $a^{3}$ has 5 times more positive divisors than a.
How many positive divisors does $a$ have? | Solution to Exercise 8
Since the divisors of $1^{3}$ are exactly the same as those of 1, we must have $a \geqslant 2$. Let then $a=p_{1}^{\alpha_{1}} \cdots p_{k}^{\alpha_{k}}$ be the prime factorization of $a$, with $\alpha_{i} \geqslant 1$ for all $i$. The number of positive divisors of $a$ is therefore $\left(\alph... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 5. In a classroom, there are ten students. Aline writes ten consecutive integers on the board. Each student chooses one of the ten integers written on the board, such that any two students always choose two different integers. Each student then calculates the sum of the nine integers chosen by the other nine s... | Solution to Exercise 5 The exercise asks to determine the maximum value of a certain quantity. The reasoning therefore consists of two steps, called analysis and synthesis.
Step 1, the analysis: We show that the number of students receiving a gift is necessarily less than or equal to four, regardless of the numbers ch... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 11. In a classroom, there are ten students. Aline writes ten consecutive integers on the board. Each student chooses one of the ten integers written on the board, such that any two students always choose two different integers. Each student then calculates the sum of the nine integers chosen by the other nine ... | Solution to Exercise 11 The exercise asks to determine the maximum value of a certain quantity. The reasoning therefore consists of two steps, called analysis and synthesis.
Step 1, the analysis: We show that the number of students receiving a gift is necessarily less than or equal to four, regardless of the numbers c... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Exercise 1. Seven people dine together every Saturday evening around a round table.
How many times is it possible to dine if everyone wants to have two new neighbors each time?
What is the result for eight people? | Solution: Under the conditions stated, at each dinner, a given person has two new neighbors. Therefore, after the second dinner, they will have had four different neighbors, six different neighbors after the third, and eight different neighbors after the fourth, which is not possible: if there are seven people in total... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 5. Seventeen people dine every Saturday evening around a round table.
How many times is it possible to dine if everyone wants to have two new neighbors each time?
What is the result for eighteen people?
Note: This exercise is the same as Exercise 1 (middle school students had to solve 1, high school student... | Solution: The first thing to notice is the following. Let $n$ be the number of guests (17 or 18), and $k$ be the number of dinners; we are interested in a particular guest among the $n$. Knowing that he has two neighbors at each dinner, and that he must never meet the same person more than once, he will therefore meet ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Exercise 4. Determine the number of natural numbers that are multiples of 8 and 50, and that are divisors of 10000. | Solution to exercise 4 An integer is a multiple of 200. It all comes down to finding the number of divisors of $10000 / 200=50=2 \times 5^{2}$. There are 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4
Find all integers $n \geqslant 0$ such that $20 n+2$ divides $2023 n+210$.
## $\S$ | Solution $\mathbf{n}^{\circ} 1$
Let $n \geqslant 0$ be any solution. The integer $20 n+2$ divides
$$
20 \times(2023 n+210)-2023 \times(20 n+2)=4200-4046=154=2 \times 7 \times 11
$$
whose divisors are $1,2,7,11,14,22,77$ and 154. Thus, $20 n+2$ is 2 or 22, so $n$ is 0 or 1.
Conversely, $n=0$ is a solution, because 2... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Find the smallest integer $n \geqslant 1$ such that the equation
$$
a^{2}+b^{2}+c^{2}-n d^{2}=0
$$
admits the only integer solution $a=b=c=d=0$.
## Solutions | Solution 1 First, we check that no integer $n \leqslant 6$ works. Indeed, the quintuplets $(a, b, c, d, n)$ equal to
$$
(1,0,0,1,1),(1,1,0,1,2),(1,1,1,1,3),(2,0,0,1,4),(2,1,0,1,5) \text { and }(2,1,1,1,6)
$$
are solutions to the equation.
We will now show that $n=7$ is the integer we are looking for. Indeed, let $(a... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate the sums: $a=99+999, b=99+999+9999$ and $c=99+999+9999+99999$.
2. Consider the number $N$ defined as the sum:
$$
N=99+999+9999+\cdots+9999999 \ldots 999
$$
The first term of this sum is written with two nines; then we add the numbers written with three, four nines, etc. The last term of the sum is writte... | 1. The answers are $a=1098, b=11097$ and $c=111096$.
2. Let $N=99+999+\cdots+\overbrace{99 \ldots 9}^{100}$. This sum has 99 terms. We write each term differently:
$$
99=100-1,999=1000-1 \ldots \text {, and } \overbrace{99 \ldots 9}^{100}=1 \overbrace{00 \ldots 0}^{100}-1
$$
$$
\begin{aligned}
N & =(100-1)+(1000-1)+\... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $I$ be the center of the circle inscribed in triangle $A B C$. Suppose that: $C A+A I=$ $B C$. Determine the value of the ratio $\frac{\widehat{B A C}}{\widehat{C B A}}$ | When two segments are not end-to-end on the same line, we cannot say much about the sum of their lengths. Therefore, to use the hypothesis \( CA + AI = BC \), we need to construct a segment \( AJ \) on the line \((CA)\) such that: \( AJ = AI \) and \( CJ = CA + AJ \) (so \( C \) and \( J \) are on opposite sides of \( ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
How many Friday the 13ths can there be in a year? (What is the maximum? What is the minimum?) And if the year is a leap year? | We only consider the case of a non-leap year (the case of a leap year is treated in exactly the same way). Having a Friday the 13th is equivalent to having a Monday the 1st, so we will count the Mondays the 1st. Furthermore, $31 \equiv 3(\bmod 7)$, thus we obtain the shift in the week between January 1st and February 1... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The first term of a sequence is equal to $3^{2012}$. Each subsequent term is equal to the sum of the digits of the previous term. Find the 10th term. | From the correction of the previous exercise, we know that all terms are divisible by 9. On the other hand, $a_{1}=3^{2012}=9^{1006}<10^{1006}$, so the number of digits of $a_{1}$ is less than 1005, and since each digit is less than 10, we have $a_{2} \leq 9 \cdot 1005 < 10000$. Similarly, $a_{3}<40$, $a_{4}<13$, and $... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n$ such that $2^{n}+3$ is a perfect square. The same question with $2^{n}+1$. | A square is never congruent to 3 modulo 4 (if you don't believe me, try it and see!), so $2^{n}+3$ cannot be a square if $n \geq 2$. And if $n=1$ then $2^{1}+3=5$ is not a square and if $n=0$ then $2^{0}+2=3$ is not a square. The second question is less easy. We are looking for $n$ and $x$, two integers, such that $2^{... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many parallelograms with sides 1 and 2, and angles $60^{\circ}$ and $120^{\circ}$, can be placed at most in a regular hexagon with side 3? | Let $s$ be the area of an equilateral triangle with side 1. A regular hexagon can be divided into 6 equilateral triangles with sides of 3, thus each having an area of $9s$: its total area is $54s$. Now, each parallelogram has an area of $4s$: since $\frac{54s}{4s} > 13$, one might think it is possible to place 13 paral... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find all integer solutions $x$ of the equation: $x^{3}+(x+1)^{3}+(x+2)^{3}=(x+3)^{3}$ | By developing the left-hand side, we have: $3 x^{3}+9 x^{2}+15 x+9=(x+3)^{3}$. It follows that $x+3$ is divisible by 3, and thus $x$ is also divisible by 3, and we can set: $x=3 t$, which leads us to: $81 t^{3}+81 t^{2}+45 x+9=27\left(t^{3}+3 t^{2}+3 t+1\right)$, or, after simplifications:
$3 t^{3}-2 t-1=0$. 1 is an ob... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
a) Show that for all $a \in \mathbb{N}$, the equation $x^{2}+y^{2}+z^{2}=8 a+7$ has no solutions in $\mathbb{Q}^{3}$.
b) Find all natural numbers $n$ such that $7^{n}+8$ is a perfect square. | a) To reduce to an equation with integer unknowns, we set $x=\frac{X}{T}, y=\frac{Y}{T}$ and $z=\frac{Z}{T}$, where $X, Y, Z \in \mathbb{Z}$, and where $T \in \mathbb{N}^{*}$ is the minimal integer allowing this writing (in other words, $T$ is the LCM of the denominators of $x, y$, and $z$ written in irreducible form).... | 0 | Number Theory | proof | Yes | Yes | olympiads | false |
Nine cells of a $10 \times 10$ diagram are infected. At each step, a cell becomes infected if it was already infected or if it had at least two infected neighbors (among the 4 adjacent cells).
(a) Can the infection spread everywhere?
(b) How many initially infected cells are needed to spread the infection everywhere? | (a) No, this is not possible. Note that the perimeter of the area delimited by the infected cells decreases in a broad sense at each step. Initially, it is at most $9 \times 4=36$, and cannot reach 40.
(b) According to question (a), at least 10 cells must already be infected. We place 10 cells along a diagonal. We ver... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n \geq 1$ such that $3^{n-1}+5^{n-1}$ divides $3^{n}+5^{n}$. | (Algebraic manipulations, using the fact that if $a \mid b$, then $a \mid b+k a$ for any integer $k$)
Let $n \geq 1$ such that $3^{n-1}+5^{n-1} \mid 3^{n}+5^{n}$. Then
$$
3^{n-1}+5^{n-1} \mid 5\left(3^{n-1}+5^{n-1}\right)-3^{n}+5^{n}=2 \cdot 3^{n-1}
$$
But $3^{n-1}+5^{n-1}>2 \cdot 3^{n-1}$ for $n \geq 2$, so $3^{n-1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Calculate the value of
$$
\frac{2014^{4}+4 \times 2013^{4}}{2013^{2}+4027^{2}}-\frac{2012^{4}+4 \times 2013^{4}}{2013^{2}+4025^{2}}
$$ | For better readability, let $n=2013$. Our expression can be rewritten as
$$
\frac{(n+1)^{4}+4 n^{4}}{n^{2}+(2 n+1)^{2}}-\frac{(n-1)^{4}+4 n^{4}}{n^{2}+(2 n-1)^{2}}=\frac{5 n^{4}+4 n^{3}+6 n^{2}+4 n+1}{5 n^{2}+4 n+1}-\frac{5 n^{4}-4 n^{3}+6 n^{2}-4 n+1}{5 n^{2}-4 n+1}
$$
To simplify (because we always need to simplify... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $\alpha, \beta, \gamma$ be the three roots of $x^{3}-x-1$. What is the value of $\frac{1-\alpha}{1+\alpha}+\frac{1-\beta}{1+\beta}+\frac{1-\gamma}{1+\gamma}$? | A safe method, even if in this case, one can find a faster one, is to look for the equation having roots $\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}, \frac{1-\gamma}{1+\gamma}$ and to calculate the sum of the roots of this equation from its coefficients. If $x$ is a root of $x^{3}-x-1$, what equation is $y=\fra... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(IMO 1959) Show that for all $n \in \mathbb{Z}$, the fraction $\frac{21 n+4}{14 n+3}$ is irreducible. | Here is a one-line solution:
$$
(21 n+4,14 n+3)=(7 n+1,14 n+3)=(7 n+1,7 n+2)=(1,2)=1
$$ | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
Let $p$ be a prime number. Show that $\binom{2 p}{p} \equiv 2[p]$. | According to the Vandermonde convolution formula
$$
\binom{2 p}{p}=\sum_{0 \leqslant n \leqslant p}\binom{p}{n}^{2}
$$
Now, $p \left\lvert\,\binom{ p}{n}\right.$ for $1 \leqslant n \leqslant p-1$, so $\binom{2 p}{p} \equiv\binom{p}{0}+\binom{p}{p} \equiv 2$ modulo $p$. | 2 | Number Theory | proof | Yes | Yes | olympiads | false |
Find all $n \in \mathbb{N}^{*}$ such that $n$ divides $2^{n}-1$. | Let $n>1$ such that $n$ divides $2^{n}-1$, and $p$ be the smallest prime divisor of $n$. We denote $d$ as the order of 2 modulo $p$: we know that $d$ divides $p-1$, and on the other hand, from the statement $p \mid 2^{n}-1$ so $d$ divides $n$, thus $d$ divides $\text{GCD}(n, p-1)$.
However, since $p$ is minimal, $p-1$... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all $n \in \mathbb{N}^{*}$ odd such that $n$ divides $3^{n}+1$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | First, $n=1$ is a solution.
Moreover, let $q$ be the smallest prime divisor of $n$ and $d$ the order of 3 modulo $q$: $n$ is odd, so $q>2$. On one hand, $d \mid q-1$ and on the other hand $3^{n} \equiv-1(\bmod q)$, so $3^{2 n} \equiv 1(\bmod n)$, thus $d \mid 2 n$.
Therefore, $d$ divides the GCD of $2 n$ and $q-1$, wh... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Check that 2 is a solution of the equation:
$$
(x+1)^{3}+(x+2)^{3}+(x+3)^{3}=(x+4)^{3}
$$
Does this equation have any other integer solutions? | $3^{3}+4^{3}+5^{3}=27+64+125=216=6^{3}$. Several techniques can be used to solve the equation: setting $x=2+k$, which, when expanded, gives a third-degree equation in $k$ of which one root is zero (this equation does not admit any other real root). But one can also expand directly: $(x+1)^{3}+(x+2)^{3}+(x+3)^{3}-(x+4)^... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Consider 2015 lines in the plane, no two of which are parallel and no three of which are concurrent. Let $E$ be the set of their intersection points.
We want to assign a color to each point in $E$ such that any two points on the same line, whose segment connecting them contains no other point of $E$, are of different ... | The configuration contains at least one triangle. This can be proven, for example, by induction on the number of lines: three lines form a triangle, and if you add a line, it either leaves the triangle intact or splits it into a triangle and a quadrilateral. Consequently, it is impossible to perform a coloring with two... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find all positive integers $n$ such that $2^{n}+3$ is a perfect square. | It is about solving $2^{n}+3=a^{2}$ with $a$ being a natural integer. If $2 \leq n$, then we have $a^{2} \equiv 3 \bmod 4$, which is not possible according to what we have seen earlier. Therefore, the only possibilities are $n=0$ and $n=1$, which, after verification, gives $n=0$ as the only solution.
Solution to the e... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all positive integers $a$ such that $a^{2}+2 a$ is a perfect square. | There are two cases. If $a>0$, we note that $a^{2}<a^{2}+2a<$ $a^{2}+2a+1=(a+1)^{2}$. Therefore, $a^{2}+2a$ cannot be a perfect square. If $a=0$, we find $a^{2}+2a=0$ which is a perfect square. This is thus the only one. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $x$ such that $x^{3}+(x+1)^{3}+(x+2)^{3}=(x+3)^{3}$. | Let's develop both sides of the equality:
$$
x^{3}+x^{3}+3 x^{2}+3 x+1+x^{3}+6 x^{2}+12 x+8=x^{3}+9 x^{2}+27 x+27
$$
, so
$x^{3}-6 x-9=0$. We deduce that $x$ divides 9, so $x$ is $-9,-3,-1,1,3$ or 9. Of all these possibilities, only 3 is correct. The solution to this equation is therefore 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For which positive integers $n$, does the number $n^{2}+1$ divide $n+1$? | If $n^{2}+1$ divides $n+1$, we must have $n^{2}+1 \leq n+1$, or equivalently $n^{2} \leq n$, that is, $n \leq 1$. The only possibility is therefore $n=1$. Conversely, $n=1$ does work. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all natural numbers $n$ such that $2^{n}+3$ is a perfect square. | For $n=0,2^{n}+3=2^{2}$ is a perfect square. We verify that $n=1$ is not a solution. Therefore, we can assume $n \geq 2$ and then $2^{n}+3 \equiv 3[4]$. However, a square is congruent to 0 or 1 modulo 4, so for $n \geq 2, 2^{n}+3$ is not a perfect square.
Proposition 150. If $n$ is an integer, then $n^{3} \equiv -1, 0... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What are the last two digits of $7^{9^{9^{9}}}$? | We look at the remainders of the powers of 7 modulo 100. It is a cycle of length 4. Since $9 \equiv 1(\bmod 4)$, the last two digits are 07. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Show that if $0 \leqslant a, b, c \leqslant 1$ then
$$
\frac{a}{b c+1}+\frac{b}{a c+1}+\frac{c}{a b+1} \leqslant 2
$$ | The left member is twice differentiable in each variable with positive second derivatives. Therefore, the function is convex in each variable. However, the global maximum of a convex function defined on a closed and bounded space is attained at an extremal point. Note that here we only know that the function is convex ... | 2 | Inequalities | proof | Yes | Yes | olympiads | false |
Calculate $\frac{1}{3}$ in $\mathbb{Z}_{5}$.
untranslated text:
Calculer $\frac{1}{3}$ dans $\mathbb{Z}_{5}$. | We find the Bézout coefficients (using, for example, the extended Euclidean algorithm) to progress from decimal to decimal:
$$
1=3 \times 2-5 \times 1
$$
so
$$
\frac{1}{3}=2+5 \times \frac{-1}{3}
$$
We know that $\frac{-1}{3}$ is a $p$-adic integer, so the first decimal of $\frac{1}{3}$ is 2. Similarly,
$$
\begin{... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the maximum number of angles less than $150^{\circ}$ that a non-crossing 2017-sided polygon can have, all of whose angles are strictly less than $180^{\circ}$?
## 3 High School Eliminators: Statements
Translate the text above into English, please keep the original text's line breaks and format, and output the... | Answer: 11. If the angles are $180-x_{i}$, then we have $x_{i}>0$ and $x_{1}+\cdots+x_{2017}=360$, so at most 11 of the $x_{i}$ can be greater than 30. We can verify that conversely there exists such a polygon.
## 5 High School Eliminatory: Solutions | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $x, y, z$ be integers such that for any triangle $ABC$ we have $16 H N^{2} = x BC^{2} + y CA^{2} + z AB^{2}$, where $H$ is the foot of the altitude from $A$, $M$ is the midpoint of $[BC]$, and $N$ is the midpoint of $[AM]$. Determine $x^{2} + y^{2} + z^{2}$. | Answer: 9. Since $A H M$ is a right triangle at $H$, $N$ is the center of the circumcircle of $A H M$ so $H N=A M / 2$. The median theorem then gives $16 H N^{2}=$ $4 A M^{2}=-A C^{2}+2\left(A B^{2}+B C^{2}\right)$, thus $x=2, y=2, z=-1$, and consequently $x^{2}+y^{2}+z^{2}=9$. | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Five numbers are given in ascending order, which are the lengths of the sides of a quadrilateral (not crossed, but not necessarily convex, meaning a diagonal is not necessarily inside the polygon) and one of its diagonals $D$. These numbers are $3, 5, 7, 13$ and 19. What can be the length of the diagonal $D$? | We can rephrase the problem: let $a, b, c, d$ be the lengths of the sides of the quadrilateral, and $e$ be the length of the diagonal that separates, on one side, the sides of lengths $a$ and $b$, and on the other side, the sides of lengths $d$ and $e$. It is necessary and sufficient that the triplets $(a, b, e)$ and $... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A number was written on the board. At each step, we add to it the largest of its digits (for example, if we have written 142, the next number will be 146). What is the maximum number of odd numbers that can be written consecutively by proceeding in this way?
## High school statements | The answer is 5. Suppose we start from an odd number $n$. We denote $n_{i}$ as the $i$-th number written, with $n_{1}=n$. Let $c_{i}$ and $d_{i}$ be the largest digit and the units digit of $n_{i}$, respectively. If $c_{1}$ is odd, then $n_{2}=n_{1}+c_{1}$ is even, and we have written only one odd number. Note also tha... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the last digit of $2017 7^{2017}$? | It is enough to look at the last digit of $7^{2017}$. We notice that the powers of 7 are periodic modulo 10 (i.e., the value of the last digit of $7^{n}, n \geq 1$, is periodic), with a period of 4: the last digit is successively $7,9,3,1,7,9,3,1$, etc. 2016 is a multiple of 4, so the last digit of $7^{2016}$ is a 1. T... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Show that if $a, b, c$ and $d$ are four positive real numbers such that $a b c d=1$, then
$$
a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d \geq 10
$$ | We have a sum of 10 terms, let's try to apply the AM-GM inequality which suggests
Sum of 10 terms $\geq 10 \sqrt[10]{\text{ Product of these } 10 \text{ terms }}$
which gives us here
$$
\begin{aligned}
a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d & \geq 10 \sqrt[10]{a^{2} \cdot b^{2} \cdot c^{2} \cdot d^{2} \cdot... | 10 | Inequalities | proof | Yes | Yes | olympiads | false |
We have 21 pieces of type $\Gamma$ (formed of three small squares). We are allowed to place them on an $8 \times 8$ chessboard (without overlapping, so that each piece covers exactly three squares). An arrangement is said to be maximal if no additional piece can be added while respecting this rule. What is the smallest... | On pave the chessboard with 16 squares $2 \times 2$. Each square contains two cells of the arrangement, so it covers at least 32 cells, which requires a minimum of 11 pieces. Conversely, we can construct such an arrangement:
$ as the coordinates of the midpoint between $\left(x_{1} ; y_{1}\right)$ and $\lef... | We are interested in the midpoint $\left(x_{m} ; y_{m}\right)$ of a segment between $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$. For this midpoint to have integer coordinates, it is necessary that $x_{m}$ AND $y_{m}$ are integers, which means that $x_{1}+x_{2}$ AND $y_{1}+y_{2}$ must be even numbers.
... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(warm-up)
Let $x>0$. Show that
$$
x+\frac{1}{x} \geq 2
$$
For which $x$ do we have equality? | Let $x>0$. We have:
$$
x+\frac{1}{x} \geq 2 \Longleftrightarrow x^{2}-2 x+1 \geq 0 \Longleftrightarrow(x-1)^{2} \geq 0
$$
The last assertion is true, which concludes the proof.
Case of equality:
$$
x+\frac{1}{x}=2 \Longleftrightarrow x^{2}-2 x+1=0 \Longleftrightarrow(x-1)^{2}=0
$$
Therefore, equality is only achie... | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
Let $a$ and $b$ be two distinct roots of the polynomial $X^{3}+3 X^{2}+X+1$. Calculate $a^{2} b+a b^{2}+3 a b$. | Let $c$ be the third root of this polynomial. The relations between coefficients and roots give:
$$
\left\{\begin{aligned}
\sigma_{1}=a+b+c & =-3 \\
\sigma_{2}=a b+b c+c a & =1 \\
\sigma_{3}=a b c \quad & =-1
\end{aligned}\right.
$$
Thus $a^{2} b+a b^{2}+3 a b=a^{2} b+a b^{2}-(a+b+c) a b=-a b c=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $P(x)=x^{3}-3 x-1$. Denote $x_{1}, x_{2}$ and $x_{3}$ as its three real roots. Evaluate the quantity $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$. | A first solution is to use the Gauss identity
$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}=\left(x_{1}+x_{2}+x_{3}\right)\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1} x_{2}-x_{2} x_{1}-x_{1} x_{3}\right)
$$
which leads to
$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=3 \sigma_{3}+\sigma_{1}\left(\sigma_{1}^{2}-3 \sigma_{2}\r... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all rational numbers $x$ such that $x^{3}+(x+1)^{3}+(x+2)^{3}=(x+3)^{3}$. | After simplification, it is a matter of finding the rational roots of the polynomial $P(x)=x^{3}-6 x-9$. We notice that 3 is a root of $P$. By polynomial division, $P(x)=(x-3)\left(x^{2}+3 x+3\right)$. Since $x^{2}+3 x+3$ has no real solutions $(\Delta=-3), P$ has no other rational roots than 3.
Another idea is to exp... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all odd integers $y$ such that there exists an integer $x$ satisfying
$$
x^{2}+2 y^{2}=y x^{2}+y+1
$$
## 2 Solution | We rewrite the equation in the form
$$
(y-1) x^{2}=2 y^{2}-y-1
$$
since $2 y^{2}-y-1=(y-1)(2 y+1)$, therefore if $y \neq 1$, we can divide by $y-1$, and the equation becomes $x^{2}=2 y+1$. Since $y$ is odd, it can be written in the form $y=2 k+1$. In this case, $x^{2}=4 k+3$, but a table of congruences shows that $x^... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine all complex numbers $z$ such that $|z|+|\bar{z}|=|z+\bar{z}|$. | We note $z=a+i b$. We observe that $z+\bar{z}=a+i b+a-i b=2 a$ and the equality becomes $\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}=\sqrt{4 a^{2}}$. However, $2 \sqrt{a^{2}+b^{2}}=\sqrt{4 a^{2}+4 b^{2}} \geq \sqrt{4 a^{2}}$ with equality if and only if $b=0$. The complex solutions are therefore the complexes whose imaginary... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For any real number $x$, we denote $\lfloor x\rfloor$ as the integer part of $x$, that is, the greatest integer less than or equal to $x$. We denote $\{x\}$ as its decimal part, that is, $\{x\}=x-\lfloor x\rfloor$. For example, we have $\lfloor 3.1\rfloor=3$ and $\{3.1\}=3.1-3=0.1$. We also have $\lfloor-2.7\rfloor=-3$... | Let $x$ be a solution to the problem. We denote $n=\lfloor x\rfloor$ and $y=\{x\}$. Then $n$ is an integer and $0 \leqslant y < 1$. We have $n+y=x$ and $n-2019y=n-2019\{x\}=n-2019(x-n)=2019n-x$.
If $n>0$, then $2019 n>n$, so $n-2019>0$, thus $y>1$, which is impossible. If $0<n<2019$, then $0>n-2019$, so $y<0$, which ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider a large square grid with side length 10, divided into small squares with side length 1. Two small squares are said to be neighbors if they share a common side. On each of the small squares, a positive real number is inscribed. Furthermore, 5 frogs move on the grid and can each cover a small square. Two frogs n... | We will show that the largest possible value of $k$ is 6. First, let's verify that 6 is indeed achievable. For this, we isolate a $2 \times 3$ rectangle in the top left corner of the grid. On all the cells that are not in this rectangle, we write the number 0. On the cells of the rectangle, we write the numbers $10, 10... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
- Show that for any strictly positive real number $x: x+\frac{1}{x} \geq 2$. Find the case of equality.
- Show that for all strictly positive real numbers $a$ and $b: \frac{a}{b}+\frac{b}{a} \geq 2$. Find the case of equality.
- Determine the minimum for $a$ and $b$ two non-zero real numbers of:
$$
\frac{a^{6}}{b^{6}}... | - Since $x$ must be strictly positive, the inequality is equivalent to $x^{2}-2 x+1 \geq 0$, which is $(x-1)^{2} \geq 0$ and this is always true, with equality if and only if $x=1$.
- It suffices to apply the previous question to $x=\frac{a}{b}$, and we have equality if and only if $a=b$.
- We have $\frac{a^{6}}{b^{6}}... | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
What are the integers $k$ such that the following identity is always true for any real numbers $a, b$, and $c$?
$$
(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)+k a b c
$$ | By developing:
$$
\begin{aligned}
(a+b)(b+c)(c+a) & =a b c+a b a+a c c+a c a+b b c+b b a+b c c+b c a \\
(a+b+c)(a b+b c+c a) & =a a b+a b c+a c a+b a b+b b c+b c a+c a b+c b c+c c a
\end{aligned}
$$
By substituting the two equalities, we get $2 a b c=3 a b c-k a b c$ which gives $k=-1$. The calculation ensures that t... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many integers must one select from the set $\{1,2, \ldots, 20\}$ to ensure that this selection includes two integers $a$ and $b$ such that $a-b=2$? | Consider the drawers of the form $\{1,3\},\{2,4\},\{5,7\},\{6,8\},\{9,11\},\{10,12\},\{13,15\}$, $\{14,16\},\{17,19\},\{18,20\}$. By choosing 11 integers, by the pigeonhole principle, there will be two that are in the same drawer, and thus have a difference of 2. This quantity is indeed minimal because the set $\{1,2,5... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Paris has two million inhabitants. A human being has, at most, 600000 hairs on their head. What is the largest number of Parisians that we can hope to find who have exactly the same number of hairs on their head? | Here, the drawers are the numbers of hairs and the socks will be the Parisians. We then have 2000000 socks for 600001 drawers (we don't forget the bald Parisians): the pigeonhole principle assures us that it is possible to find $\left\lceil\frac{2000000}{600001}\right\rceil=4$ Parisians who have exactly the same number... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A fly and $\mathrm{k}$ spiders move on a $2019 \times 2019$ grid. On their turn, the fly can move 1 square and the $k$ spiders can each move 1 square. What is the minimal $k$ for which the spiders are sure to catch the fly. | We show that 2 spiders suffice: when the fly moves to a coordinate, the spider on that coordinate replicates the movement, the other approaches. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find all non-negative integers $x, y$ such that:
$$
2^{x}=y^{2}+y+1
$$ | The left side of the equation is even if $x>0$. The right side is always odd: indeed $y$ and $y^{2}$ have the same parity, their sum is therefore even, so $1+y+y^{2}$ is always odd.
If $x=0$, then it is necessary and sufficient that $1+y+y^{2}=1$ with $y \geqslant 0$ so the only solution is $x=y=0$.
## Congruences
D... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $\mathrm{a}, \mathrm{b}$ be two positive real numbers such that $a b \geqslant 1$. Show that:
$$
\frac{1}{1+a}+\frac{1}{1+b} \leqslant 1
$$
Determine the case of equality. | $\frac{1}{1+a}+\frac{1}{1+b} \leqslant 1 \Longleftrightarrow \frac{1+b+1+a}{(1+a)(1+b)} \leqslant 1 \Longleftrightarrow 2+a+b \leqslant 1+a+b+a b \Longleftrightarrow 1 \leqslant a b$ The inequality is therefore satisfied by assumption, the case of equality is thus $a b=1$. | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
A permutation of $\{1, \ldots, n\}$ is drawn at random. On average, how many fixed points does it have? | Consider an $n \times n$ table, where the rows correspond to the elements of $\{1, \ldots, n\}$ and the columns to permutations. We mark a 1 in position $(i, s)$ if the element $i$ is a fixed point of the permutation $s$. The number of 1s is counted:
- by rows: each element $i$ is a fixed point of $(n-1)!$ permutation... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
If $\alpha, \beta, \gamma$ are the three roots of $x^{3}-x+1$, determine the value of
$$
\frac{1-\alpha}{1+\alpha}+\frac{1-\beta}{1+\beta}+\frac{1-\gamma}{1+\gamma}
$$ | Given that $x$ is a root of $x^{3}-x+1$, we wonder about the polynomial for which $y=\frac{1-x}{1+x}$ is a root. We find $x=\frac{1-y}{1+y}$ and substitute it into the polynomial. We then find that $y$ is a root of $-y^{3}+y^{2}-7 y-1$. By Viète's formulas, we conclude that the desired sum is 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b$ and $c$ be strictly positive real numbers such that $a^{2}+b^{2}+c^{2}=\frac{1}{2}$. Show that
$$
\frac{1-a^{2}+c^{2}}{c(a+2 b)}+\frac{1-b^{2}+a^{2}}{a(b+2 c)}+\frac{1-c^{2}+b^{2}}{b(c+2 a)} \geqslant 6
$$ | We note that $1-a^{2}+c^{2}=2\left(a^{2}+b^{2}+c^{2}\right)-a^{2}+c^{2}=a^{2}+2 b^{2}+3 c^{2}$ using the hypothesis. On the other hand, by the inequality of means, we have $a c \leqslant \frac{1}{2}\left(a^{2}+c^{2}\right)$ and $2 b c \leqslant b^{2}+c^{2}$, so
$$
c(a+2 b)=a c+2 b c \leqslant \frac{1}{2}\left(a^{2}+c^... | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
(N1 2002) Find the smallest integer $t$ such that there exist strictly positive integers $x_{1}, \ldots, x_{t}$, satisfying $x_{1}^{3}+\ldots+x_{t}^{3}=2002^{2002}$. | Let's show that $t \geq 4$. We are looking for an integer $n$ such that $3 \mid \varphi(n)$, so that cubes take on only a few values modulo $n$. This is the case for 9, for which cubes can only be 0, 1, or -1. We observe that $2002^{2002} \equiv 4 \pmod{9}$, so we have $t \geq 4$. It suffices to find $x_{1}, x_{2}, x_{... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest integer $n$ such that there exist polynomials $f_{1}, f_{2}, \ldots, f_{n}$ such that:
$$
f_{1}(x)^{2}+f_{2}(x)^{2}+\ldots+f_{n}(x)^{2}=x^{2}+7
$$ | We show that the equation $a^{2}+b^{2}+c^{2}=7 d^{2}$ has no non-trivial solutions, which concludes the cases for $n=1,2,3$ by looking at the constant coefficient. Moreover, the $f_{i}$ are affine since otherwise the degree would be too high, so there exists an $i$ and a $t$ such that $f_{i}(t)= \pm t$ for $n=4$, reduc... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A triangle has side lengths: 3, 4, 5. Calculate the radius of the inscribed circle (circle inside the triangle and tangent to all three sides of the triangle). | Let's call $A, B, C$ the vertices of the triangle, $a=BC, b=CA, c=AB$ the lengths of the three sides, $I$ and $r$ the center and the radius of the inscribed circle. The heights from $I$ to the triangles $IAB, IBC$, and $ICA$ are all equal to $r$, so the areas of these three triangles are: area $(IAB)=\frac{rc}{2}$, are... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
We write on the board the integers from 1 to 2013. At each step, we erase two of them and write their difference instead. The number of integers thus decreases by 1. Can the last integer obtained at the two thousand and twelfth step be equal to 0? | This is still an argument of parity that allows us to conclude. Since the difference of two integers has the same parity as their sum, the sum of all integers on the board retains the same parity at each step of the process. Initially, it is: $1+2+\ldots+2011=\frac{2011 \times 2012}{2}$, which is even. This sum will th... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On each cell of a $9 \times 9$ grid, there is a chip. At a certain signal, each chip jumps to one of the cells touching diagonally the one it is in. Find the minimum number of empty cells after this move. | We color the columns of the table alternately in black and white. There are then $9 \times 5=45$ black squares, and $9 \times 4=36$ white squares. During the movement, each chip changes the color of its column. Thus, there are more than 36 chips on the 45 black squares after the signal, and therefore at least 9 empty s... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a deck of cards consisting only of red and black cards, there are 2 times as many black cards as red cards. If 4 black cards are added, there are then 3 times as many black cards as red cards. How many cards were in the deck before adding the 4 black cards?
Only a numerical answer is expected here. | Let $r$ be the number of red cards and $n$ be the number of black cards in the initial deck. By hypothesis, we have $n=2 r$. After adding 4 black cards, there are $n+4$ black cards in the deck, and by hypothesis, we have $n+4=3 r$. Thus, $4=3 r-n=3 r-2 r=r$ and $n=2 \cdot 4=8$. Initially, the deck contains $r+n=4+8=12$... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n$ such that $4 n^{4}+1$ is a prime number. | By the Sophie-Germain identity:
$1+4 n^{4}=\left(1-2 n+2 n^{2}\right)\left(1+2 n+2 n^{2}\right)$ Therefore, it is necessary that $1-2 n+2 n^{2}=1$, so $n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find $k$ such that, for all real $a, b, c$,
$$
(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)+k \cdot a b c
$$ | In the development, we find $k=-1$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all even primes. | If a prime is even, then it is divisible by 2, and 2 is a divisor. Therefore, since $2 \neq 1$, it must be 2. Conversely, 2 is indeed prime. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all primes of the form $a^{2}-1$, with $a \geqslant 2$ natural. | 3 is the only prime of this form. Indeed, if a prime $p$ is of the form $a^{2}-1=(a-1)(a+1)$, it is divisible by $a-1$ and $a+1$. Therefore, these must be 1 and $p$. Since they are distinct and $a-1$ is the smallest of them, $a-1=1$ and $p=a^{2}-1=2^{2}-1=3$. Conversely, $2^{2}-1=3$ is indeed prime. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What are the prime numbers $p$ such that $p+2$ and $p+4$ are also prime? | It is known that among three consecutive numbers, one of them will be a multiple of 3. The idea here is the same, and we will reason as we would for consecutive numbers. We know that $p$ can be written in the form $3k$, $3k+1$, or $3k+2$. Let's consider these three cases:
- If $p=3k$, $p$ is divisible by 3 and is prim... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
For which strictly positive integers $n$ do we have $5^{n-1}+3^{n-1} \mid 5^{n}+3^{n}$? | If $5^{n-1}+3^{n-1} \mid 5^{n}+3^{n}$, then $5^{n-1}+3^{n-1} \mid 5^{n}+3^{n}-5\left(5^{n-1}+3^{n-1}\right)=-2 \cdot 3^{n-1}$. Thus, $5^{n-1}+3^{n-1} \leqslant 2 \cdot 3^{n-1}$, hence $5^{n-1} \leqslant 3^{n-1}$ and we deduce that $n=1$. Conversely, for $n=1,2$ divides 8 and $n=1$ is our only solution. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the set of natural numbers $n$ such that $5^{n}+4$ is a square. | We are therefore looking for the $n$ such that the equation
$$
5^{n}+4=a^{2}
$$
This can be rewritten as
$$
5^{n}=a^{2}-4
$$
By recognizing a notable identity, this is equivalent to
$$
(a-2)(a+2)=5^{n}
$$
Consequently, since 5 is prime, we deduce that $a-2$ and $a+2$ are powers of 5, i.e., there exist integers $r... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Show that if $n \geqslant 2$ divides $2^{n}+1$, then $n$ is a multiple of 3. | As $n \geqslant 2$, consider one of its prime divisors $p$. Then $2^{n} \equiv-1[p]$ so $2^{2 n} \equiv 1[p]$. We deduce that $p$ divides $2^{2 n}-1 \wedge 2^{p-1}-1$ by Fermat's little theorem. A classical lemma (which we will reprove later) ensures that $p$ divides $2^{(2 n) \wedge(p-1)}-1$. Suppose then that $p$ is ... | 3 | Number Theory | proof | Yes | Yes | olympiads | false |
Find all integers $n$ such that $n^{3}-3 n^{2}+n+2$ is a power of 5.
## More exercises | We then have: $(n-2)\left(n^{2}-n-1\right)=5^{a}$. Thus, we have: $n-2=5^{x}$ and $n^{2}-n-1=5^{y}$. It follows that $5^{y}-5^{2 x}-3 \cdot 5^{x}=1$. Therefore, if $x, y \geqslant 1$ then $0 \equiv 1[5]$, which is absurd. Therefore, $x=0$ or $y=0$. If $x=0$, then $y=1$, which gives $n=3$. If $y=0$, then $5^{2 x}+3 \cdo... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine the non-zero real numbers $x$ such that $x+\frac{1}{x}=2$. | We could just multiply by $x$ (non-zero) to obtain $(x-1)^{2}=0$ and that's it, but that's not fun, we need to use Viète!
Let's take a polynomial whose roots are $x$ and $\frac{1}{x}$. Their sum is 2 and their product is 1, so they are roots of $X^{2}-2 X+1$, hence $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$:
$$
f\left(x^{2022}+y\right)=f\left(x^{1747}+2 y\right)+f\left(x^{42}\right)
$$ | For complicated exponents, we will try to simplify them. For a given $x$, we have $x^{2022} + y = x^{1747} + 2y$ if and only if $y = x^{2022} - x^{1747}$, so we take this $y$. We then have $f\left(2 x^{2022} - x^{1747}\right) = f\left(2 x^{2022} - x^{1747}\right) + f\left(x^{42}\right)$, from which: $f\left(x^{42}\righ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A sequence $a_{1}, a_{2}, a_{3}, \ldots$ of strictly positive integers satisfies $a_{1}>5$ and $a_{n+1}=5+6+\cdots+a_{n}$ for all strictly positive integers $n$. We assume that, regardless of the value of $a_{1}$, this sequence always contains a multiple of $p$. Show that $p=2$.
Bonus: Show that $p=2$ satisfies the st... | Suppose $p \neq 2$, note that by immediate induction $a_{n} \geq 5$ so the statement is well-posed. We have $a_{n+1}=\frac{a_{n}\left(a_{n}+1\right)}{2}-10$
We would ideally like no term to be divisible by $p$. The simplest case would be for all $a_{n}$ to be congruent to $a_{0}$ modulo $p$, and for $a_{1}$ not to be ... | 2 | Number Theory | proof | Yes | Yes | olympiads | false |
Determine all prime numbers $p$ for which there exists a unique $a$ in $\{1, \ldots, p\}$ such that $a^{3}-3 a+1$ is divisible by $p$.
---
The translation maintains the original formatting and structure of the source text. | Let $p$ satisfy the statement, and $a$ be the unique solution of $a^{3}-3 a+1 \equiv 0(\bmod p)$. Note that $a \neq 0$, so by setting $b \equiv \frac{1}{a}(\bmod p), b^{3}-3 b^{2}+1 \equiv 0$ thus $(b-1)^{3}-3 b+2 \equiv 0$ thus $(b-1)^{3}-3(b-1)-$ $1 \equiv 0$. In particular, $-(b-1)$ is also a root modulo $p$ of $X^{... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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