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(BXMO 2014)
() Let $a, b, c$ and $d$ be strictly positive integers. Determine the smallest value that the expression:
$$
S=\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+c+d}{b}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor
$$
can take. | With integer parts, there is not much to do: we apply the inequality $\lfloor x\rfloor>x-1$ to find, by rearranging the terms
$$
S>\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{d}{b}+\frac{b}{d}\right)+\le... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b>1$ be odd such that $a+b=2^{l}$. Find the $k \in \mathbb{N}^{*}$ such that $k^{2} \mid a^{k}+b^{k}$. | Let's start by eliminating the case $k=1$ which works. Already $k$ is odd, because if $k$ were even, we would have a contradiction since $a^{k}+b^{k} \equiv 2[4]$. Next, take $p$ to be the minimal prime divisor of $k$ (possible since we have eliminated the case $k=1$). Since it is necessary that $a$ and $b$ are coprime... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider 2021 lines in the plane, no two of which are parallel and no three of which are concurrent. Let E be the set of their intersection points. We want to assign a color to each point in E such that any two points on the same line, and whose connecting segment contains no other point of E, are of different colors.
... | First, note that it is necessary to have at least 3 colors to obtain such a coloring. Indeed, it is easily shown by induction on the number of lines that the configuration contains at least one triangle formed by the lines (this is true for $n=3$, and if we add a line, either it leaves the triangle intact or it separat... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(IMO 1995, P3) Find all integers $n>3$ for which there exist $n$ points $A_{1}, \ldots, A_{n}$ in the plane and real numbers $r_{1}, \ldots, r_{n}$ such that:
1. no three of the points are ever collinear.
2. for all $i, j, k$, the area of the triangle $A_{i} A_{j} A_{k}$ is equal to $r_{i}+r_{j}+r_{k}$. | If we have four points $A_{i}, A_{j}, A_{k}, A_{l}$ that form a convex quadrilateral in this order, denoting $\mathcal{A}$ as the area, we have
$$
\mathcal{A}\left(A_{i} A_{j} A_{k} A_{l}\right)=\mathcal{A}\left(A_{i} A_{j} A_{k}\right)+\mathcal{A}\left(A_{k} A_{l} A_{i}\right)=\mathcal{A}\left(A_{j} A_{k} A_{l}\right... | 4 | Geometry | proof | Yes | Yes | olympiads | false |
In a room, 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 students. Each stu... | The exercise requires determining 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 chosen by Aline.
Let ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a room, 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 students. Each stu... | The exercise requires determining 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 chosen by Aline.
Let ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Knowing that $a b c=8$, with $a, b, c$ positive, what is the minimum value of $a+b+c$? | By the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality): $\frac{a+b+c}{3} \geqslant \sqrt[3]{a b c}$.
Therefore, $a+b+c \geqslant 3 \cdot \sqrt[3]{8}=6$. This bound is indeed achieved with $a=b=c=2$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What are the integers $k$ such that for all real numbers $a, b, c$,
$$
(a+b+c)(a b+b c+c a)+k a b c=(a+b)(b+c)(c+a)
$$ | Suppose the equation is verified for an integer $k$. By taking $a=b=c=1$, we get $3 \times 3 + k = 2^{3}$, so $k = 8 - 9 = -1$.
It remains to verify that $(a+b+c)(ab+bc+ca) - abc = (a+b)(b+c)(c+a)$ for all real numbers $a, b, c$. Now,
$$
(a+b+c)(ab+bc+ca) - abc = a^2b + a^2c + abc + b^2c + b^2a + abc + c^2a + c^2b + ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $x>0$. Show that $x+\frac{1}{x} \geqslant 2$. Find the cases of equality. | Let $c=x, d=\frac{1}{x}$ and apply proposition 4. We obtain $c+d=x+\frac{1}{x} \geqslant 2 \sqrt{c d}=2 \sqrt{\frac{x}{x}}=$ 2. We have equality if and only if $c=d$ that is $x=\frac{1}{x}$ if and only if $x^{2}=1$. The case of equality is thus achieved for $x=1$. | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
Alice wants to color the integers between 2 and 31 (inclusive) using $k$ colors. She wishes that if $m$ and $n$ are integers between 2 and 31 such that $m$ is a multiple of $n$ and $m \neq n$, then $m$ and $n$ are of different colors. Determine the smallest integer $k$ for which Alice can color the integers $2,3, \ldot... | We show that $k \geqslant 4$. Consider the numbers 2, 4, 8, and 16. 2 divides the other 3, so 2 is not the same color as any of the others. Similarly, 4 divides 8 and 16, so it does not share a color with any of the others. And 8 divides 16, so they each have a different color. We have thus $k \geqslant 4$. Constructio... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
How many integers must one select from the set $\{1,2, \ldots, 20\}$ to ensure that this selection includes two integers $a$ and $\mathrm{b}$ such that $a-b=2$? | Consider the drawers of the form $\{1,3\},\{2,4\},\{5,7\},\{6,8\},\{9,11\},\{10,12\},\{13,15\}$, $\{14,16\},\{17,19\},\{18,20\}$. By choosing 11 integers, by the pigeonhole principle, there will be two that are in the same drawer, and thus have a difference of 2. This quantity is minimal because the set $\{1,2,5,6,9,10... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $n$ be an integer greater than 1.
Show that
$$
\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}=0
$$ | We have the binomial formula
$$
\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}=(1-1)^{n}=0^{n}=0
$$ | 0 | Combinatorics | proof | Yes | Yes | olympiads | false |
Let $a>0$. Show that
$$
a+\frac{1}{a} \geq 2
$$
When do we have equality? | By applying the inequality of Theorem 1 to $\sqrt{a}$ and $\frac{1}{\sqrt{a}}$, we have $\frac{a+\frac{1}{a}}{2} \geq 1$, which gives us the desired inequality.
We have equality for $\sqrt{a}=\frac{1}{\sqrt{a}}$, that is, $a=1$. | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
What is the last digit of $2019^{2020^{2021}}$? | $2020^{2021}$ is even, and since $2019 \equiv-1(\bmod 10)$, we get $2019^{2020^{2021}} \equiv 1(\bmod 10)$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the last digit of $2023^{2024^{2025}}$? | We notice that $3^{2} \equiv 9 \equiv -1(\bmod 10)$, so $3^{4} \equiv 1(\bmod 10)$. Moreover, $2024^{2025}$ is a multiple of 4, so we have once again $2023^{2024^{2025}} \equiv 1(\bmod 10)$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the remainder of the Euclidean division of $2022^{2022^{2022}}$ by 11? | By Fermat's Little Theorem, we have \(2022^{10} \equiv 1 \pmod{11}\). Thus, if we know the value of \(2022^{2022} \mod 10\), we would be very happy.
We look at the powers of 2 modulo 10: they have a period of 4: 2, 4, 6, 8, 2, ... Since 2022 is congruent to \(2 \pmod{4}\), we get \(2022^{2022} \equiv 2^{2022} \equiv 4... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The integers $a$ and $b$ have the following property: for any natural number $n$, the integer $2^{n} a+b$ is a perfect square. Show that $a$ is zero. | For all $n \in \mathbb{N}$ we write $x_{n}^{2}=2^{n} a+b$ with $x_{n} \in \mathbb{N}$. Then $x_{n+2}^{2}=2^{n+2} a+b$ and $\left(2 x_{n}\right)^{2}=$ $4 \cdot\left(2^{n} a+b\right)=2^{n+2} a+4 b$ are perfect squares and we have: $\left(2 x_{n}\right)^{2}-x_{n+2}^{2}=3 b$.
But as soon as $\left(2 x_{n}\right)>2|b|$ the... | 0 | Number Theory | proof | Yes | Yes | olympiads | false |
$P$ is a polynomial of degree 4. $P(0)=P(1)=1, P(2)=4, P(3)=9, P(4)=16$. Calculate $P(-1)$ | $P$ looks a lot like $X^{2}$, so we will study $P(X)-X^{2}$. This polynomial has 4 roots $(1,2,3,4)$. Thus $P(X)-X^{2}=(X-1)(X-2)(X-3)(X-4) Q(X)$, where $Q$ is a polynomial. Let's study the degrees:
$\operatorname{deg}\left(P(X)-X^{2}\right)=4=\operatorname{deg}((X-1)(X-2)(X-3)(X-4) Q(X))=4+\operatorname{deg}(Q) \Long... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Consider 2022 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 c... | 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 |
(A1 IMO 1996)
Let $a, b, c>0$ such that $a b c=1$. Show that
$$
\sum_{c y c} \frac{a b}{a b+a^{5}+b^{5}} \leq 1
$$ | By IAG, we have
$$
\sum_{c y c} \frac{a b}{a b+a^{5}+b^{5}} \leq \sum_{c y c} \frac{a b}{a b+a^{2} b^{2}(a+b)}=\sum_{c y c} \frac{c}{c+a+b}=1
$$ | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
(Infinite Solitaire)(^) You all know the rules of solitaire: there are marbles on a board, and you eliminate marbles by jumping over them with another marble, etc. Now consider the plane and place marbles on all integer points of the negative half-plane. The goal of the game is to place a marble as high as possible in ... | It is easy to raise a ball to height 1, 2, and 3, and with a bit of patience, one can raise a ball to height 4, but that is the limit. Let $\varphi$ be the golden ratio, which satisfies $\varphi^{2}=\varphi+1$:
$$
\varphi=\frac{1+\sqrt{5}}{2}
$$
Then, we need to sum over all points with coordinates $(x, y)$ on which ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Find all odd integers $n \geqslant 1$ such that $n$ divides $3^{n}+1$.
untranslated text remains unchanged. | Let $n>1$ such that $n$ divides $3^{n}+1$. Another way to say this is that $3^{n} \equiv-1(\bmod n)$, so we are dealing with a purely multiplicative problem. We would like to use Fermat's theorem, but it is difficult to exploit because we have poor control over $\varphi(n)$. Let $p$ be a prime factor of $n$, which is o... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find $k$ such that $(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)+k a b c$.
. | It is enough to develop the two members: $(a+b)(b+c)(c+a)=(a b+$ $\left.a c+b^{2}+b c\right)(c+a)=a^{2} b+a^{2} c+a b^{2}+a c^{2}+b^{2} c+b c^{2}+2 a b c,(a+b+c)(a b+b c+c a)+k a b c=$ $a^{2} b+a^{2} c+a b^{2}+a c^{2}+b^{2} c+b c^{2}+3 a b c+k a b c$ so $k=-1$.
## - Notable identities of degree 2 -
Proposition 1.3 (N... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$$
\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\ldots+(-1)^{k}\binom{n}{k}+\ldots+(-1)^{n}\binom{n}{n}=?
$$ | As for the previous exercise, it involves recognizing a binomial of Newton but this time with $a=1$ and $b=-1$. We therefore find $(1-1)^{\mathfrak{n}}=0$.
## Corollary 1.14.
$$
\binom{n}{k}=\binom{n}{n-k}
$$
Proposition 1.15 (Pascal's Formula).
$$
\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}
$$
Proof. We want to ... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $x, y$ and $z$ be three strictly positive real numbers. Show that
$$
\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{z}{y}+\frac{y}{x} \geqslant 6
$$ | We group a term and its inverse, and apply the previous exercise, to obtain $\frac{x}{y}+\frac{y}{x}+\frac{z}{x}+\frac{x}{z}+\frac{z}{y}+\frac{y}{z} \geqslant 2+2+2=6$ | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
Study the sequence defined by $\mathfrak{u}_{0}=0$ and $\mathfrak{u}_{n+1}=\sqrt{12+\mathfrak{u}_{n}}$ for all $\mathfrak{n} \in \mathbb{N}$.
Regarding the second type of equations:
Definition 2.11. Let $\left(u_{n}\right)$ be a real sequence. We say it satisfies a recurrence equation of order $\mathbf{k}$ (linear) i... | We observe, by a quick induction, that the sequence is bounded by 4. We also show by induction on $n$ that $\mathfrak{u}_{n+1}>\mathfrak{u}_{n}:$ the initialization holds with $\mathfrak{u}_{1}=\sqrt{12}>0=\mathfrak{u}_{0}$. If we assume $a_{n+1}>a_{n}$, it follows that $a_{n+2}=\sqrt{12+a_{n+1}} \geq a_{n+1}$, which c... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $x$ and $y$ be two non-zero numbers such that $x^{2} + x y + y^{2} = 0$ (where $x$ and $y$ are complex numbers, but that's not too important). Find the value of
$$
\left(\frac{x}{x+y}\right)^{2013} + \left(\frac{y}{x+y}\right)^{2013}
$$ | Clearly, $x /(x+y)+y /(x+y)=1$ and
$$
\frac{x}{x+y} \cdot \frac{y}{x+y}=\frac{x y}{x^{2}+2 x y+y^{2}}=1
$$
Thus, $x /(x+y)$ and $y /(x+y)$ are the roots of $t^{2}-t+1=0$, so the sums $S_{k}=(x /(x+y))^{k}+(y /(x+y))^{k}$ satisfy $S_{0}=2, S_{1}=1$ and
$$
S_{k+2}=S_{k+1}-S_{k}
$$
for $k \geqslant 0$. We deduce that ... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all real numbers $x, y, z$ such that
$$
x+y+z=3, \quad x^{2}+y^{2}+z^{2}=3, \quad x^{3}+y^{3}+z^{3}=3
$$
## - Polynomials with integer coefficients -
We now present some specific properties of polynomials with integer coefficients:
* We have already seen that if $P, Q \in \mathbb{Z}[X]$ and $\operatorname{deg}... | We write Newton's relations:
$$
S_{1}-\sigma_{1}=0, \quad S_{2}-\sigma_{1} S_{1}+2 \sigma_{2}=0, \quad S_{3}-\sigma_{1} S_{2}+\sigma_{2} S_{1}-3 \sigma_{3}=0
$$
Thus, $\sigma_{1}=3, \sigma_{2}=3, \sigma_{3}=1$. From this, we deduce that $x, y, z$ are roots of $t^{3}-3 t^{2}+3 t-1=0$. Now, $t^{3}-3 t^{2}+3 t-1=(t-1)^{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b, c, d$ be the four roots of $X^{4}-X^{3}-X^{2}-1$. Calculate $P(a)+P(b)+P(c)+$ $P(d)$, where $P(X)=X^{6}-X^{5}-X^{4}-X^{3}-X$. | First, note that $P(X)=X^{2}\left(X^{4}-X^{3}-X^{2}-1\right)-X^{3}+X^{2}-X$. Using the recurrence formula for symmetric polynomials, it suffices to compute $-S_{3}+S_{2}-S_{1}$. Now, $\sigma_{1}=1, \sigma_{2}=-1$ and $\sigma_{3}=0$. Therefore, $S_{1}=\sigma_{1}=1, S_{2}=\sigma_{1} S_{1}-2 \sigma_{2}=3$ and $\mathrm{S}_... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $A, B, C, D$ and $E$ be five points in the plane in general position (that is, such that no three of them are collinear, and no four of them are concyclic). A separator is defined as a circle passing through three of these points, such that one of the two remaining points is located inside the circle, and the other... | We quickly realize that if we try to tackle the problem naively, we will soon end up with perfectly sordid case studies. So we stop for a minute to think about how to simplify this case study as much as possible. To do this, we would like to get rid of as many circles as possible, which encourages us to invert through ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b, c, d$ be real numbers defined by
$$
a=\sqrt{4-\sqrt{5-a}}, b=\sqrt{4+\sqrt{5-b}}, c=\sqrt{4-\sqrt{5+c}}, d=\sqrt{4+\sqrt{5+d}}
$$
Calculate their product.
## 2 Solutions | It would be wonderful if the four reals $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ were roots of the same polynomial, whose constant term would be their product. Fortunately, this is almost true. Indeed, by squaring each equality twice (taking care to isolate the root each time), we find that $a, b, -c, -d$ are r... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all natural numbers $n$ such that $n$ divides $2^{n}-1$. | The number 1 works. Let $p$ be the smallest prime divisor of $n \neq 1$. Since $2^{n} \equiv 1 \pmod{n}$, we also have $2^{n} \equiv 1 \pmod{p}$, and by Fermat's Little Theorem, $2^{p-1} \equiv 1 \pmod{p}$, so the order of 2 modulo $p$ divides the GCD of $n$ and $p-1$, hence it is 1 because $p-1$ is smaller than the sm... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine all strictly positive integers that are coprime with all numbers of the form $2^{n}+3^{n}+6^{n}-1$, for $n$ a natural number. | We are looking for prime numbers that do not divide any of the $2^{n}+3^{n}+6^{n}-1$ for $n$ a natural integer.
2 is not one of them, because it divides $2+3+6-1$.
Let $p \geqslant 5$ be ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $p$ be a prime number. Under what condition does there exist an integer $k$ such that $k^{2} \equiv-1$ $(\bmod p) ?$ | Finding a necessary and sufficient condition requires mathematical tools that you do not have, but you can at least find a necessary condition. If $k^{2} \equiv -1 \pmod{p}$, then $k^{4} \equiv 1 \pmod{p}$, so the order of $k$ - the smallest exponent $n$ such that $k^n \equiv 1 \pmod{p}$ - divides 4. If $-1 \neq 1 \pmo... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many triplets $(a, b, c)$ of integers satisfy the equation $a+b+c=6$ as well as the conditions $-1 \leqslant a \leqslant 2$ and $1 \leqslant b, c \leqslant 4$? | We start by generalizing the problem by looking for the generating series of the sequence of the number of solutions to the equation $a+b+c=n$ with the same conditions on $a, b, c$ (of course, this will actually be a polynomial). Since $-1 \leqslant a \leqslant 2$, the contribution of the variable $a$ is a factor $x^{-... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Calculate $\sum_{i=1}^{n} \prod_{j \neq i} \frac{x-a_{j}}{a_{i}-a_{j}}$ for $a_{1}, \ldots, a_{n} \in \mathbb{R}$ pairwise distinct. | This polynomial is of degree at most $n-1$ and is equal to 1 at $a_{i}$, hence it is identically equal to 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For which positive integers $n$ does the number $n^{2}+1$ divide $n+1$? | The integers considered are strictly positive. Thus, if $\mathfrak{n}^{2}+1$ divides $n+1$, then $n^{2}+1 \leqslant n+1$, hence $n^{2} \leqslant n$. And since $n>0, n \leqslant 1$. Finally, $n$ being an integer, the only candidate left is $n=1$, and we verify that it works: 2 divides 2. It's good!
The following exerci... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all values of $n \in \mathbb{N}$ such that $3^{2 n}-2^{n}$ is prime. | The expression we are discussing in terms of primality can be rewritten as $3^{2 n}-2^{n}=\left(3^{2}\right)^{n}-2^{n}=9^{n}-2^{n}=(9-2)\left(9^{n-1}+9^{n-2} \cdot 2+9^{n-3} \cdot 2^{2}+\cdots+2^{n-1}\right)$.
For $n \geqslant 2$, the right factor is at least $9^{1}$, so it is different from 1. The left factor is itse... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all positive integers $a$ such that $a^{2}+2a$ is a perfect square. | There are two cases. If $a>0$, we notice that $a^{2}<a^{2}+2 a<$ $a^{2}+2 a+1=(a+1)^{2}$. Therefore, $a^{2}+2 a$ cannot be a perfect square. If $a=0$, we find $a^{2}+2 a=0$ which is a perfect square. This is therefore the only one.
## 2 Arithmetic Course
This session discussed congruences. The main points covered dur... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Calculate the sum of the digits of the sum of the digits of the sum of the digits of $A:=4444^{4444}$. | Let $S(n)$ denote the sum of the digits of an integer $n$. Then our trick says that for all $n, S(n) \equiv n[9]$. Thus, $S(S(S(A))) \equiv A[9]$. Let's calculate $A$ modulo 9. We have $4444 \equiv 4+4+4+4 \equiv -2$ [9], so we need to calculate $(-2)^{4444}$ modulo 9. We notice that $(-2)^{3} \equiv 1[9]$. We deduce t... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Show that the fraction $\frac{12 n+1}{30 n+2}$ is always irreducible. | According to the Euclidean algorithm, $\operatorname{gcd}(30 n+2,12 n+1)=\operatorname{gcd}(12 n+$ $1,6 n)=\operatorname{gcd}(6 n, 1)=1$. | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
Let $A=2012^{2012}$, B the sum of the digits of $A$, C the sum of the digits of $B$, and D the sum of the digits of $C$.
What is the value of $\mathrm{D}$? | On a $2012 \equiv 5(\bmod 9)$ hence $2012^{2012} \equiv 5^{2012}(\bmod 9)$. We therefore study the powers of 5 modulo $9: 5^{2} \equiv 7$ so $5^{3} \equiv-1$ and $5^{6} \equiv 1$. Now, we can write $2012=$ $2+6 \times 335$, hence:
$$
2012^{2012} \equiv 5^{2012} \equiv 5^{2} \times\left(5^{6}\right)^{335} \equiv 5^{2} ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all non-negative integers $x, y, z$ such that $x^{3}+4 y^{3}=16 z^{3}+4 x y z$.
## 2 Solutions | Let $x, y, z$ be a solution. We notice that $4 y^{3}, 16 z^{3}$, and $4 x y z$ are even, so $x^{3}$ must be even, and thus $x$ must be even. We set $x^{\prime}=x / 2$. The equation becomes $8 x^{13}+4 y^{3}=16 z^{2}+8 x^{\prime} y z$, or equivalently $2 x^{\prime 3}+y^{3}=4 z^{2}+2 x^{\prime} y z$. We notice that $y$ i... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $\mathrm{a}<\mathrm{b}<\mathrm{c}<\mathrm{d}$ be odd natural integers such that $\mathrm{ad}=\mathrm{bc}, \mathrm{a}+\mathrm{d}$ and $\mathrm{b}+\mathrm{c}$ are powers of 2. Show that $a=1$.
## 2 Solutions | The nerve of the war is to find adequate factorizations, to have divisibility relations.
We set $a+d=2^{k}$ and $b+c=2^{l}$. We then have $a\left(2^{k}-a\right)=b\left(2^{l}-b\right)$, an identity in which two squares appear. One must immediately think of making their difference appear: $b^{2}-a^{2}=(b-a)(b+a)=2^{l} b... | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
Draw a few trees. How many edges at least need to be removed from the graph $\mathrm{K}_{4}$ to make it a tree?
## - Degree of a vertex, Handshaking Lemma -
The degree of a vertex is the number of edges emanating from that vertex. If we sum the degrees of all vertices, each edge is counted twice, since it always conn... | At least one side must be removed because if we keep all four sides, it forms a cycle. Next, if we keep the two diagonals, we need to remove two more sides. Indeed, if we keep two adjacent sides, they will form a triangle with one of the diagonals, which results in a cycle. If we keep two opposite sides, they form a cy... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $E=\{a, b, c, d\}$ be a set. How many functions $f$ from $E$ to $E$ satisfy, for all $x$ in $E, f(f(f(x)))=x$? | The identity, defined for all $x$ by $f(x)=x$, is obviously a solution. If one of the elements has an image distinct from itself, for example $f(x)=y$, what can $f(y)$ be equal to? If we had $f(y)=x$, we would have: $f(f(f(x)))=f(f(y))=f(x)=y$, the relation from the hypothesis would not be satisfied. Similarly, if we h... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(i) Find all integers $n \geq 1$ such that $n$ divides $2^{n}-1$.
(ii) Find all odd integers $n \geq 1$ such that $n$ divides $3^{n}+1$. | (i) Let $n>1$ such that $n$ divides $2^{n}-1$. It is clear that $n$ is odd. Let $p$ be the smallest prime factor of $n$, which is therefore odd. Then $2^{n} \equiv 1 \bmod p$. Let $\omega$ be the order of 2 modulo $p$. Then $\omega$ divides $n$. On the other hand, by Fermat's Little Theorem, $2^{p-1} \equiv 1 \bmod p$.... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the $a, n \geq 1$ such that $\left((a+1)^{n}-a^{n}\right) / n$ is an integer. | Suppose $n>2$. Let $p$ be the smallest prime factor of $n$. Then $p$ divides $(a+1)^{n}-a^{n}$. In other words, $((a+1) / a)^{n} \equiv 1 \bmod p$. Let $\omega$ be the order of $(a+1) / a$ modulo $p$. Then $\omega$ divides $n$. On the other hand, by Fermat's little theorem, $((a+1) / a)^{p-1} \equiv 1 \bmod p$ so that ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $a$ be a strictly positive integer. Suppose that $4\left(a^{n}+1\right)$ is the cube of an integer for all positive integers $n$. Find $a$.
Soit $a$ un entier strictement positif. On suppose que $4\left(a^{n}+1\right)$ est le cube d'un entier pour tout entier positif $n$. Trouver $a$.
(Note: The second line is ke... | It is clear that $a=1$ works. Let's show that it is the only one. Suppose therefore $a>1$. Choose $n=2 m$ and note that $a^{2}+1$ is not a power of 2 since it is congruent to 1 or 2 modulo 4. Let $p$ be an odd prime such that $p$ divides $a^{2}+1$. Then, according to LTE:
$$
v_{p}\left(4\left(a^{n}+1\right)\right)=v_{... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all functions $f$ from $\mathbb{R}_{+}^{*}$ to $\mathbb{R}_{+}^{*}$ satisfying the following condition:
$$
\forall x, y \in \mathbb{R}_{+}^{*}, \quad f(x+f(y))=f(x+y)+f(y)
$$
## - Correction - | Let $g$ be the application $f$ - Id.
The equation in the statement can be rewritten after simplification in the following form:
$$
\forall x, y \in \mathbb{R}_{+}^{*}, \quad g(x+y+g(y))=g(x+y)+y
$$
First, note that this equation implies that $g$ is injective. Indeed, if $y_{1}$ and $y_{2}$ are strictly positive real... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
Seven people dine around a round table every Saturday evening. How many times is it possible to have dinner if everyone wants to have two new neighbors each time? What is the result for eight people? | 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, each can... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Seventeen people dine around a round table every Saturday evening. 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 were to solve 1, high school students 5), but in a less particular case. The proposed solution below is very general, but difficult for an average high school student to access. This exercise was therefore intended to identify the students best equip... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
. The sequence $\left(x_{n}\right)$ is defined recursively by $x_{0}=1, x_{1}=1$, and:
$$
x_{n+2}=\frac{1+x_{n+1}}{x_{n}}
$$
for all $n \geqslant 0$. Calculate $x_{2007}$. | . We start by calculating the first terms of the sequence $\left(x_{n}\right)$. We find:
$$
x_{2}=2 \quad ; \quad x_{3}=3 \quad ; \quad x_{4}=2 \quad ; \quad x_{5}=1 \quad ; \quad x_{6}=1
$$
and we find two consecutive terms equal to 1. From this point on, the calculations repeat, which proves that the sequence $\lef... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 as many fract... | First, it is possible to make 10 integers 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 integer fractions. Indeed, there are 3 p... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b, c$ be positive real numbers such that $a b c=8$. Find the minimum value that the expression $a+b+c$ can take. | The form of this exercise is a bit different from the previous ones, but we should not be destabilized by it. To find the minimum value that $a+b+c$ can take, we need to lower-bound this expression, that is, find an inequality of the form $a+b+c \geqslant x$ where $x$ will be a real number, and then find a case of equa... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Find the inverse of 11 modulo 54 | ## 5
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(Exercise 3.9 from the course notes)
Find the inverse of 37 modulo 53
Proposition 11 (Inverse of a product).
Proposition 12 (Proper inverse). | Sometimes, trial and error is enough, but here we proceed with certainty. We start by performing the Euclidean algorithm
\[
\begin{aligned}
& 53=37+16 \\
& 37=16 * 2+5 \\
& 16=5 * 3+1
\end{aligned}
\]
From this, we deduce a Bézout's identity
\[
\begin{array}{r}
16-5 * 3=1 \\
16-(37-16 * 2) * 3=1 \\
53-37-(37-(53-37)... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A person has between 0 and 300,000 hairs on their head. The urban area of Marseille contains 2,000,000 inhabitants. How many Marseillais have the same number of hairs at a minimum? | Here, we need to apply the generalized pigeonhole principle. The socks are the Marseillais, and the possible number of hairs are the drawers. There are 300001 drawers for 2000000 Marseillais. By the generalized pigeonhole principle, $300001 \times 6 + 199994 = 2000000$, so there are at least 7 people who have the same ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,
$$
f(x+y)=x f(x)+y f(y)
$$ | We guess (and at the same time verify) that the null function is a solution. Let's show that it is the only one. By setting $y=f(0)$ we have $f(x)=x f(x)$. In other words, for all $x \in \mathbb{R}$,
$$
(x-1) f(x)=0 \text {. }
$$
Thus, by substituting $x$ with any value different from 1, we obtain that the function i... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What are the $p$ such that $p, p+2, p+4$ are all primes? | Necessarily $p, p+2$ or $p+4$ is a multiple of 3, hence equal to 3. Since $p \geqslant 2$, we have $p=3$. Conversely, $p=3$ is a solution.
## Lemma 3.
Let $n \geqslant 2$, and let $p$ be the smallest divisor of $n$ that is greater than or equal to 2. Then $p$ is prime.
Proof. Let $d$ be a positive divisor of $p$. If... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Calculate the remainder of the Euclidean division of $2022^{2023^{2024}}$ by 19. | On a $2022 \equiv 8[19]$.
Moreover, we observe that $8^{6} \equiv 1[19]$.
With $q$ and $r$ such that $2023^{2024}=6 q+r$ we therefore have
$$
\begin{aligned}
2022^{2023^{2024}} & \equiv 8^{2023^{2024}} \\
& \equiv 8^{q 6} 8^{r} \\
& \equiv 8^{r}
\end{aligned}
$$
it is therefore sufficient to determine the remainder... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(AHSME 1990, P9)
Consider a cube whose edges are colored either red or black such that each face of the cube has at least one black edge. What is the minimum number of black edges? | Of course, we will solve this problem in two steps:
- We need at least 3 black edges. Indeed, each edge is shared between 2 faces, and since there are 6 faces, we need at least $\frac{6}{2}=3$ black edges.
- 3 is indeed sufficient: it is enough that no pair of edges shares a vertex. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the last digit of $2023^{2024^{00252^{206^{2027}}}}$ | On a $2023^{2024^{2005^{20252^{2027}}}} \equiv 3^{2024^{20255^{206^{2027}}}} \equiv 3^{(4 \cdot 506)^{n>0}} \equiv\left(3^{4}\right)^{506^{n}} \equiv 1[10]$ with $n=2025^{2026^{2027}}$ | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the set of positive integers $n$ such that $5^{n}+4$ is a perfect square
untranslated text kept the same format and line breaks. | Let $a>0$ such that $5^{n}+4=a^{2}$, which gives us $5^{n}=(a-2)(a+2)$. Therefore, the two terms on the right must be powers of 5. However, they are not congruent modulo 5, so one of them must be a zero power. Thus, $a-2=1 \Rightarrow a=3$. We then have $5^{n}+4=9 \Rightarrow n=1$. The only solution is therefore $n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the real numbers $x, y$ and $z$ such that
$$
\left\{\begin{array}{l}
x+y+z=3 \\
x^{2}+y^{2}+z^{2}=3 \\
x^{3}+y^{3}+z^{3}=3
\end{array}\right.
$$ | First, by trial and error, we find that $x=y=z=1$ works. We will now prove that this is the only solution. To do this, 1 the idea is to obtain several polynomials of the degrees that interest us, in order to get expressions that we can use.
For example, $(x+y+z)^{2}=\left(x^{2}+y^{2}+z^{2}\right)+2(x y+y z+z x)$, so
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find the non-zero real numbers $x$ such that $x+\frac{1}{x}=2$. | Let $x$ be a real number such that $x+\frac{1}{x}=2$. It suffices to consider the polynomial
$$
P(t)=(t-x)\left(t-\frac{1}{x}\right)=t-2 x+1=(t-1)^{2}
$$
to deduce that $x=1$. Conversely, $x=1$ indeed works. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n>0$ such that $n \mid 2^{n}-1$. | First, $n=1$ is a solution.
Suppose now $n>1$ such that $n \mid 2^{n}-1$ and $p$ the smallest prime divisor of $n$.
$$
2^{n} \equiv 1 \quad(\bmod p)
$$
Thus the order $\omega$ of 2 modulo $p$ divides $n$. Moreover,
$$
\omega \leq \phi(p)<p
$$
so $\omega$ is smaller than any prime divisor of $n$, yet $\omega \mid n... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all odd integers $n>0$ such that $n \mid 3^{n}+1$. | First, $n=1$ is a solution. Now, consider the case $n \geqslant 2$: $n$ therefore has at least one prime divisor, let $p$ be the smallest of them. We have
$$
3^{n} \equiv-1 \quad(\bmod p)
$$
We know how to manipulate the order and Fermat's theorem, and we have just seen in exercise 7 that these manipulations can solv... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b, c$ be three strictly positive real numbers such that $a+b+c+a b+b c+c a+a b c=7$. Show that:
$$
\sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} \geq 6
$$ | Given that the constraint is quite ugly, we try to simplify it. We almost have the product $(1+a)(1+b)(1+c)$. Let's calculate this product: $(1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ca+abc=1+7=8$ by hypothesis. Therefore, we need to try to make $1+a, 1+b, 1+c$ appear in the inequalities. Note that by the QM-AM inequality, $a^{2}+... | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
(treatise)
For what condition on $y \in \mathbb{N}$ is $y^{2}+5 y+12$ a square? | We know that $(y+2)^{2}=y^{2}+4 y+4y^{2}+5 y+12$ if and only if $6 y+9>5 y+12$, if and only if $y>3$.
Therefore, for $y \geq 4, (y+2)^{2}<y^{2}+5 y+12<(y+3)^{2}$, under these conditions $y^{2}+5 y+12$ cannot be a square.
Finally, the only potential solutions are $0 \leq y \leq 3$. We test:
| $y$ | $y^{2}+5 y+12$ | P... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(unprocessed)
Find all positive integers $x$ satisfying $x!=x^{3}-11 x+4$. | Intuitively, the factorial grows faster than the cube, so quite quickly $x^{3}-11 x+40$. Therefore, for $x \geq 5$, we have $x! > x^{3}-11 x+4$, so we cannot have equality. We need to test the values $0 \leq x \leq 4$:
| $x$ | $x!$ | $x^{3}-11 x+4$ | Equality? |
| :---: | :---: | :---: | :---: |
| 0 | 1 | 4 | no |
| 1... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What are the $n$ such that $4^{n}+5^{n}$ is a perfect square? | Let $k, n$ be such that $4^{n}+5^{n}=k^{2}$.
Then $5^{n}=k^{2}-2^{2 n}=\left(k-2^{n}\right)\left(k+2^{n}\right)$. If 5 divides each of the two factors, 5 also divides their difference, namely $2^{n+1}$, which is a contradiction. Thus $k+2^{n}=5^{n}$, and $k-2^{n}=1$. By taking the difference again, we find $2^{n+1}=5^... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
( Inspired by USAMO 2014, P1 )
Let $P \in \mathbb{R}[X]$ be a monic polynomial of degree 2. Suppose that $P(1) \geq P(0)+3$, and that $P$ has two real roots (not necessarily distinct) $x_{1}$ and $x_{2}$. Find the smallest possible value for $\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right)$. | Let's write $P=X^{2}+a X+b$ with $a, b \in \mathbb{R}$. The condition in the statement becomes $1+a+b \geq b+3$ thus $a \geq 2$. Furthermore, by Vieta's formulas, $x_{1}+x_{2}=-a$ and $x_{1} x_{2}=b$. We write:
$$
\begin{aligned}
\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right) & =x_{1}^{2} x_{2}^{2}+x_{1}^{2}+x_{2}^{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A polyhedron has 20 triangular faces. Determine its number of vertices and edges. | Let's perform a double counting between the faces and the edges, using the notations $S, A, F$ from the previous exercise. Let $N$ be the number of pairs $(f, a)$ where $f$ is a face and $a$ is an edge that is a side of $f$. On one side, if we fix $f$, there are three edges that work because all the faces of the polyhe... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the integers $n \geqslant 1$ for which $n$ divides $2^{n}-1$. | First, $n=1$ is a solution. We now look for a potential solution $n \geqslant 2$.
Let $p$ be the smallest prime factor of such an integer $n$. Since $2^{n} \equiv 1(\bmod p)$, we know that $p$ is odd and that $\omega_{p}(2)$ divides $n$. However, $\omega_{p}(2)$ also divides $p-1$. But, by the minimality of $p$, the i... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $p$ be a prime number. Find the integers $k \geqslant 0$ for which $p$ divides $1^{k}+2^{k}+\cdots+p^{k}$. | When $k=0$, our sum is equal to $p$. The integer $k=0$ is therefore a solution, and we now focus on the cases where $k \geqslant 1$.
Let $x$ be an element of $\mathbb{Z} / p \mathbb{Z}$ of order $p-1$. The first $p-1$ powers of $x$ coincide with the invertibles modulo $p$, so that
$$
1^{k}+2^{k}+\cdots+(p-1)^{k}+p^{k... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(BXMO 2014) Let $a, b, c$ and $d$ be strictly positive integers. Determine the smallest value that the expression:
$$
S=\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+c+d}{b}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor
$$
can take. | With integer parts, there is not much to do: we apply the inequality $\lfloor x\rfloor>x-1$ to find, by rearranging the terms
$$
S>\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{d}{b}+\frac{b}{d}\right)+\le... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(IMO SL 2020 A3) Let $a, b, c, d$ be strictly positive real numbers satisfying $(a+c)(b+d)=a c+b d$. Determine the smallest value that
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}
$$
can take. | The hypothesis suggests grouping the terms in $a$ and $c$ on one side, and the terms in $b$ and $d$ on the other. We can thus group as follows (and this preserves a certain symmetry in $a$ and $c$ and in $b$ and $d$):
$$
\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)
$$
To make factors $a c... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all $n>1$ such that $n^{2} \mid 2^{n}+1$. | (P3 IMO 1990)
First, $n$ must be odd, so $2^{n}+1$ must be divisible by 3, which means $n$ must be a multiple of 3. Let's find the 3-adic valuation of $2^{n}+1: v_{3}\left(2^{n}+1\right)=v_{3}(2+1)+v_{3}(n)$. On the other hand,
$v_{3}\left(n^{2}\right)=2 v_{3}(n)$ and by divisibility we should have $2 v_{3}(n) \leqsl... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $N \in \mathbb{N}$.
Discriminant the cat proposes the following game to you:
- You choose a finite sequence (of your choice and kept secret) of integers $\left(a_{0}, \ldots, a_{n}\right) \in \mathbb{N}^{n+1}$
- At each turn of the game, Discriminant gives you a real number $\beta \in \mathbb{N}$ and you respond ... | Discriminant can always win in 1 turn.
It is enough to evaluate at an algebraic number.
For example, to be explicit, at the Liouville number
$$
\sum_{k=1}^{+\infty} \frac{1}{10^{k!}}
$$
(meditate on the link with the previous exercise). | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(British MO 1998)
Solve for $x, y, z \in \mathbb{R}^{+}$ real positive numbers the following system of equations:
$$
\begin{aligned}
& x y + y z + z x = 12 \\
& x y z = 2 + x + y + z
\end{aligned}
$$ | This exercise is a disguised inequality problem. We notice that $(2,2,2)$ is a solution, so we would like to show that we are in the case of equality of an inequality.
Indeed, by AM-GM we have
$$
12 \geq x y+y z+z x \geq 3 \sqrt[3]{x^{2} y^{2} z^{2}}
$$
as well as
$$
x y z=2+x+y+z \geq 2+3 \sqrt[3]{x y z}
$$
We se... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Solve the equation $x^{4}-3 x^{3}+4 x^{2}-3 x+1=0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Let $y=x+1 / x$. We have $0=\frac{x^{4}-3 x^{3}+4 x^{2}-3 x+1}{x^{2}}=$ $\left(x^{2}+\frac{1}{x^{2}}\right)-3\left(x+\frac{1}{x}\right)+4=\left(y^{2}-2\right)-3 y+4=y^{2}-3 y+2$.
The roots of the polynomial $y^{2}-3 y+2$ are 1 and 2, so $y^{2}-3 y+2=(y-1)(y-2)$. Therefore, the equation is equivalent to $y=1$ or $y=2$.... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Solve $1-4^{x}+2^{x+1}=2^{x}+2^{-x}$. | Let $t=2^{x}$. The equation is equivalent to $1-t^{2}+2 t=t+1 / t$, which simplifies to $1-t^{2}+t-1 / t=0$. By multiplying by $t$, we get $t(1-t^{2})+t^{2}-1=0$, or equivalently $0=(1-t^{2})(t-1)=(1-t)(1+t)(t-1)=-(t+1)(t-1)^{2}$.
Therefore, $t=1$ or $t=-1$. Since $t=2^{x}>0$, we necessarily have $t=1$, hence $x=0$.
... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n$ that are coprime with every number of the form $2^{m} + 3^{m} + 6^{m} - 1, m \in \mathbb{N}$. | We will show that only 1 is suitable. For this, it suffices to establish that for any prime $p$, there exists an integer $m$ such that $p \mid 2^{m}+3^{m}+6^{m}-1$. For $p=2$, any $m$ works. For $p=3$, any even $m$ (for example, 2) works. For $p \geq 5, p$ is coprime with $2,3,6$. We leave it to the reader to study the... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(not treated in class) Determine the minimum of $x^{2}+y^{2}+z^{2}$ under the condition $x^{3}+y^{3}+z^{3}-3 x y z=1$. | We calculate that $S_{3}=\sigma_{1} S_{2}-\sigma_{1} \sigma_{2}+3 \sigma_{3}$, so $S_{3}-3 \sigma_{3}=\sigma_{1}\left(S_{2}-\frac{3 S_{2}-\sigma_{1}^{2}}{2}\right)$. We must therefore minimize $S_{2}$ under the condition that $\sigma_{1}\left(3 S_{2}-\sigma_{1}^{2}\right)=2$.
Now, $3 S_{2}=\frac{2}{\sigma_{1}}+\sigma_... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine all integers $n>1$ such that $\frac{2^{n}+1}{n^{2}}$ is an integer. | This exercise is a classic: it is problem 3 of the 1990 International Olympiad, held in Beijing, where France placed 5th with three gold medals, notably thanks to Vincent Lafforgue (younger brother of Fields Medalist Laurent Lafforgue), who achieved perfect scores of 42/42 in 1990 and 1991. It is worth noting, however,... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all odd integers $n \geq 1$ such that $n$ divides $3^{n}+1$.
| Let $n>1$ such that $n$ divides $3^{n}+1$. Let $p$ be the smallest prime factor of $n$, which is therefore odd, and satisfies $p>3$. Then $3^{2 n} \equiv 1 \pmod{p}$. Let $\omega$ be the order of 3 modulo $p$. Then $\omega$ divides $2 n$. On the other hand, by Fermat's Little Theorem, $3^{p-1} \equiv 1 \pmod{p}$. Thus ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Calculate the gcd of $21 n+4$ and $14 n+3$. | We apply the Euclidean algorithm.
$$
\begin{aligned}
21 n+4 & =(14 n+3) \cdot 1+7 n+1 \\
14 n+3 & =2 \cdot(7 n+1)+1 \\
7 n+1 & =1 \cdot(7 n+1)+0
\end{aligned}
$$
Therefore, this gcd is 1, and this holds for all $n \in \mathbb{N}$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the integers $n \in \mathbb{N}^{*}$ such that the function $x \mapsto x^{n}$ is convex. | The second derivative of the function $x \mapsto x^{n}$ is $x \mapsto n(n-1) x^{n-2}$, which is positive on $\mathbb{R}$ if and only if $n=1$ or if $n$ is even. Consequently, $x \mapsto x^{n}$ is convex if and only if $n=1$ or if $n$ is even. | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Find the largest constant $\mathrm{C}$ such that for all $x, y, z \geq 0,(y z+z x+$ $x y)^{2}(x+y+z) \geq C x y z\left(x^{2}+y^{2}+z^{2}\right)$. | By setting $y=z=1, \frac{(y z+z x+x y)^{2}(x+y+z)}{x y z\left(x^{2}+y^{2}+z^{2}\right)}=\frac{(2 x+1)^{2}(x+2)}{x\left(x^{2}+2\right)}$. This quantity tends to 4 as $x$ tends to $+\infty$, so we will try to prove the inequality with $C=4$.
By expanding, we need to show that $\left(x^{3} y^{2}+x^{2} y^{3}+y^{3} z^{2}+y... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
(*) We call $I$ the set of points in the plane such that their abscissa and ordinate are irrational numbers, and $R$ the set of points whose both coordinates are rational. How many points of $R$ at most can lie on a circle of irrational radius whose center belongs to $I$? | (Solved by Yakob Kahane)
If we have two points on a circle, then the center of said circle lies on the perpendicular bisector of the two points. If the said two points have rational coordinates, their perpendicular bisector has an equation with rational coefficients.
We assume henceforth that there exist $A, B, C$ th... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Consider a $4 \times 4$ grid consisting of 16 cells. What is the minimum number of cells that need to be blackened so that by eliminating any two columns and any two rows, we are sure that at least one black cell remains?
## 2 Solution | We will show that at least 7 squares must be blackened. First, if we color the squares as follows:
then we verify the hypothesis of the statement. Indeed, let's try to erase two rows and two... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the GCD of all numbers of the form $i \times(i+1) \times(i+2)$ with $i \geq 1$ (we are looking for the greatest common divisor of all these numbers). | For $i=1$ we have $i \times(i+1) \times(i+2)=6$. Therefore, the GCD must be less than 6.
But among three consecutive numbers, there is always a multiple of 2 and a multiple of 3, so $i \times(i+1) \times(i+2)$ is always a multiple of 6.
The GCD of the numbers $i \times(i+1) \times(i+2)$ for $i \geq 1$ is exactly 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The act of adding the digits of a number is called addichiffre. For example, when we addichiffre 124, we get $1+2+4=7$.
What do we get when we addichiffre $1998^{1998}$, then addichiffre the result obtained, and so on, three times in a row? | 1998 and $1998^{1998}$ are multiples of 9, and therefore their digit sums are also multiples of 9. By induction, all digit sums obtained by repeated digit sum operations starting from $1998^{1998}$ are multiples of 9.
Moreover:
$$
1998^{1998} \leq 2000^{1998}=(2 \cdot 1000)^{1998}=2^{1998} \cdot 10^{3 \cdot 1998}=2^{... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Twelve candidates for the position of mayor are participating in a televised debate. After a while, one of them declares, "Up until now, we have lied once." A second then says, "Now, it's twice." A third exclaims, "Three times, now," and so on until the twelfth, who asserts that before him, there were twelve lies. The ... | We denote $C_{1}, \ldots, C_{12}$ as the candidates in the order of their speaking. Since at least one of them told the truth, we can consider the smallest integer $k$ such that $C_{k}$ told the truth (i.e., $C_{k}$ is the first to tell the truth). By the minimality of $k$, all those who spoke before him lied. Furtherm... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Consider the graph $\mathcal{G}_{n}$ whose vertices are the integers from 1 to $n$. Two points $a$ and $b$ are connected if and only if $a+b$ is prime. For which $n$ is the graph $\mathcal{G}_{n}$ planar?
## Paths on graphs
Definition 3. A graph is said to be Eulerian if there exists a path that passes through each e... | It is possible to draw $\mathcal{G}_{8}$ in the plane, but arranging $\mathcal{G}_{9}$ can result in the graph of the three houses and three factories, which is not planar.
 | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $a$ and $b$ be strictly positive integers such that $ab$ divides $a^{2}+b^{2}+1$. Calculate the quotient. | As before, we denote $k=\frac{a^{2}+b^{2}+1}{a b}$ and consider a pair $(\alpha, \beta)$ of minimal sum among the pairs $(a, b)$ satisfying this equality. If $\alpha=\beta$, then $\alpha^{2}$ divides $2 \alpha^{2}+1$ so $\alpha^{2}=1$ and $k=3$. Otherwise, we can assume without loss of generality that $\alpha>\beta \ge... | 3 | Number Theory | 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 $r$ be the radius of the inscribed circle. We start by drawing a figure.
The triangle $ABC$ with side lengths $3, 4, 5$ is a right triangle. We can calculate its area $S$ in two diffe... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
. For what value of $k$ do we have the identity (true for all $a, b, c$):
$$
(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)+k a b c ?
$$ | . It suffices to develop each of the two members. All terms are identical except the terms in $a b c$ which appear twice on the left and $3+k$ times on the right. Therefore, $k=-1$. We note that for $a=b=c=1$, we get: $8=9+k$, so the only possible value of $k$ is -1, but this does not dispense with doing the calculatio... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Show that: $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=9$.
## 2 Test Solution
Translate the text above into English, keep the original text's line breaks and format, and directly output the translation result. | The main idea of the exercise is to use the relation $a^{2}-b^{2}=(a-b)(a+b)$ to write: $1=(\sqrt{k+1}+\sqrt{k}) \cdot(\sqrt{k+1}-\sqrt{k})$, or equivalently $\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k+1}-\sqrt{k}$.
In other words, $\frac{1}{\sqrt{1}+\sqrt{2}}=\sqrt{2}-\sqrt{1}, \frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2... | 9 | Algebra | proof | Yes | Yes | olympiads | false |
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