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In a football tournament, each team plays exactly twice against each of the others. There are no draws, a win earns two points, and a loss earns nothing. It turns out that only one team won the tournament with 26 points, and there are two teams tied for last with 20 points each. Determine the number of teams, and provi... | Let $n$ be the number of teams.
Each team wins a number of matches between 10 and 13. Since there are $n(n-1)$ matches, each team wins on average $n-1$ matches. The winning team and the two last teams in a tie win on average 11 matches, while the other teams win on average between 11 and 12 matches. Therefore, the ave... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given a point $P$ and a circle $\mathcal{C}$ in the plane, the distance from $P$ to $\mathcal{C}$ is defined as the minimum length $PM$ between $P$ and a point $M$ on the circle $\mathcal{C}$. For example, if $P$ lies on the circle, then the distance from $P$ to $\mathcal{C}$ is zero, and if $P$ is the center of the ci... | Let $\mathcal{C}$ be a circle passing at equal distance from $A, B, C, D$. The four points cannot all be inside the circle, otherwise they would be equidistant from the center, and thus would be concyclic. Similarly, they cannot all be outside.
Determine the number of circles $\mathcal{C}$ passing at equal distance fr... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
(French-Slovak Competition 1996) Find all strictly positive integers $x, y, p$ such that $p^{x}-y^{p}=1$ with $p$ prime. | Let's rewrite the equation in the form $y^{p}+1^{p}=p^{x}$. If $y=1$, we see that $p=2$ and $x=1$. If $p=2$, it readily follows that necessarily $x, y=1$. Therefore, we assume $p$ is odd so that $y+1$ divides $y^{p}+1$.
If the conditions of Zsigmondy's theorem are satisfied, there exists a prime factor of $y^{p}+1$ th... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all strictly positive numbers $a, p, n$ such that $p^{a}-1=$ $2^{n}(p-1)$, where $p$ is a prime number. | It is clear that $p>2$. Suppose by contradiction that $a=$ $uv$ is composite. Then according to Zsigmondy's theorem, $p^{u}-1$ has a prime factor $q$ that does not divide $p-1$. But $p^{u}-1$ divides $p^{a}-1=2^{n}(p-1)$. Therefore, $q=2$. However, $p-1$ is even, which is a contradiction. Therefore, $a$ is prime.
If $... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine (without doing heavy calculations) the value of
$$
\binom{6}{0}-\binom{6}{1}+\binom{6}{2}-\binom{6}{3}+\binom{6}{4}-\binom{6}{5}+\binom{6}{6}
$$
Generalize by replacing 6 with $n$. | By taking $n=6, a=1$ and $b=-1$ in the proposition, since $(-1)^{k}$ is 1 if $k$ is even and -1 if $k$ is odd, we obtain that the first expression is $(1-1)^{6}=0$. More generally, we obtain for any $n$ that
$$
\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\ldots+(-1)^{n-1}\binom{n}{n-1}+(-1)^{n}\binom{n}{n}=0
$... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $x$ and $y$ be two non-zero numbers such that $x^{2} + x y + y^{2} = 0$ ($x$ and $y$ are complex numbers, but that's not too important). Find the value of
$$
\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}
$$
for $k \geq 0$. We deduce that the s... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all real numbers $x, y, z$ such that
$$
x+y+z=3, \quad x^{2}+y^{2}+z^{2}=3, \quad x^{3}+y^{3}+z^{3}=3
$$
Distinction between polynomial and polynomial function Here, we explain why it is necessary to make this distinction, starting by defining a polynomial in another way. Here, $\mathbb{K}=\mathbb{Q}, \mathbb{R}... | We write Newton's relations:
$$
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=\sqrt{4+\sqrt{5-a}}, b=\sqrt{4+\sqrt{5+b}}, c=\sqrt{4-\sqrt{5-c}}$ and $d=\sqrt{4-\sqrt{5+d}}$.
Calculate $a b c d$. | Since we are dealing with polynomials, let's honor them by getting rid of those nasty roots. $a, c, -b, -d$ are roots of $X^{4}-8 X^{2}+X+11$. So $a b c d = b d \times(-a) \times(-c)=11$ and we are done... no! To be sure we have the four roots of the polynomial (the product of the opposites of which gives the constant ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $I$ be the center of the circle inscribed in triangle $ABC$. Suppose that: $AB = AC + CI$. Determine the value of the ratio $\frac{\widehat{ACB}}{\widehat{ABC}}$.
Recall that the bisect... | 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 $A B = A C + C I$, we need to construct a segment $C J$ on the line $(A C)$ such that: $C J = C I$ and $A J = A C + C I$ (so $A$ and $J$ are on opposite sides of $C$). The relation ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
. Determine the units digit of $1789^{1789}$.
. | . On a: $1789 \equiv 9(\bmod 10) \Rightarrow 1789^{1789} \equiv 9^{1789}(\bmod 10)$.
Let's look at the powers of 9 modulo 10. We have: $9^{0} \equiv 1(\bmod 10), 9^{1} \equiv 9$ $(\bmod 10), 9^{2} \equiv 1(\bmod 10), 9^{3} \equiv 9(\bmod 10)$, etc. Then for any natural number $k$, $9^{2 k} \equiv 1(\bmod 10)$ (indeed,... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. Determine the units digit of $2^{49}$.
. | . On a $: 2^{1} \equiv 2(\bmod 10), 2^{2} \equiv 4(\bmod 10), 2^{3} \equiv 8(\bmod 10)$, $2^{4} \equiv 6(\bmod 10)$ and $2^{5} \equiv 2(\bmod 10)$.
We can then prove that $\forall k \in \mathbb{N}^{*}, 2^{4 k} \equiv 6(\bmod 10)$. Let's reason by induction on $k \in \mathbb{N}^{*}$. For $k=1$, we have: $2^{4 k}=2^{4} ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $f$ be the function that associates with a positive integer the sum of its digits when it is written in base 10. For example, $f(537)=15$. Calculate $f\left(f\left(f\left(4444^{4444}\right)\right)\right)$. | Let $n \in \mathbb{N}$. We notice that $f(n) \equiv n[9]$. With $n=4444^{4444}$ and using an easy induction, we get $f(f(f(n))) \equiv 4444^{4444} \equiv 7^{4444}[9]$. Since $7^{3} \equiv 1[9]$, we have $7^{4444} \equiv\left(7^{3}\right)^{1481} \times 7 \equiv 1^{1481} \times 7 \equiv 7[9]$. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the gcd of all numbers of the form
$$
(a-b)(b-c)(c-d)(d-a)(b-d)(a-c)
$$
where $a, b, c, d$ range over $\mathbb{Z}$. | If we calculate the product in question for $a=0, b=1, c=2, d=3$, we find -12, so the gcd in question must divide $12=3 \times 4$.
Conversely, we will show that this gcd is 12, by showing that all numbers of this form are divisible by 3 and by 4.
An integer $n$ divides the difference between two of the integers among... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. Let $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ be the sequence such that $: \forall n \in \mathbb{N}^{*}, u_{n}=\frac{2 n+\cos (n)}{n \sin \left(\frac{1}{n}\right)+\sqrt{(n+1)(n+2)}}$.
Determine the limit of the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$. | . We write: $\forall n \in \mathbb{N}^{*}, u_{n}=\frac{2+\frac{\cos (n)}{n}}{\sin \left(\frac{1}{n}\right)+\sqrt{\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}}$.
Since we have: $\lim _{n \rightarrow+\infty} \frac{\cos (n)}{n}=0, \lim _{n \rightarrow+\infty} \sin \left(\frac{1}{n}\right)=0$ and $\lim _{n \right... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
. Let $\left(u_{n}\right)$ be the sequence defined on $\mathbb{N}$ by $: \forall n \in \mathbb{N}, u_{n}=\frac{n+(-1)^{n}}{n+2}$.
Show that the sequence $\left(u_{n}\right)$ converges and determine its limit. | . On has: $\forall n \in \mathbb{N},-1 \leq(-1)^{n} \leq 1$ thus $\forall n \in \mathbb{N}, \frac{n-1}{n+2} \leq$ $u_{n} \leq \frac{n+1}{n+2}$, hence, by the squeeze theorem, we deduce that the sequence $\left(u_{n}\right)$ tends to 1.
Theorem 4. Let $\left(u_{n}\right)$ and $\left(v_{n}\right)$ be two sequences of re... | 1 | Calculus | proof | Yes | Yes | olympiads | false |
. Consider the sequence $\left(u_{n}\right)$ defined by $u_{0}=3$ and for all $n \in \mathbb{N}^{*}$, $u_{n}=\frac{u_{n-1}+2 n^{2}-2}{n^{2}}$
Show that the sequence $\left(u_{n}\right)$ converges to 2. | . The trick in this kind of exercise (very open!) is to calculate the first terms of the sequence $\left(u_{n}\right)$ to conjecture its direction of variation. Then, according to the conjectured direction of variation, we can try to bound the sequence $\left(u_{n}\right)$ from below or above! We start by showing by in... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
. Let $\left(u_{n}\right)$ be the sequence defined recursively by $u_{0}=1$ and $\forall n \in \mathbb{N}$, $u_{n+1}=\frac{u_{n}}{u_{n}^{2}+1}$.
Show that the sequence $\left(u_{n}\right)$ converges and determine its limit. | . After calculating $u_{1}$ and $u_{2}$, we conjecture that the sequence $\left(u_{n}\right)$ is decreasing and that it is bounded below by 0.
Showing that $\left(u_{n}\right)$ is bounded below by 0 is an immediate induction. To show that $\left(u_{n}\right)$ is decreasing, we calculate: $\forall n \in \mathbb{N}, u_{... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
Let $F=\left\{(x, y, z) \in \mathbb{R}^{3}: x+y+z=0\right\}$ be a subspace of $\mathbb{R}^{3}$. Determine a basis of $F$ and then the dimension of $F$. | . We write: $F=\left\{(x, y, z) \in \mathbb{R}^{3}: x=-y-z, y=y, z=z\right\}=$ $\{(-y-z, y, z), y, z \in \mathbb{R}\}=\{(-y, y, 0)+(-z, 0, z), y, z \in \mathbb{R}\}=\operatorname{Vect}\{(-1,1,0),(-1,0,1)\}$. This writing allows us to deduce that the family $\{(-1,1,0),(-1,0,1)\}$ is a generating set of $F$. We also not... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
We consider the sequence $u_{n}$ as follows: $u_{0}=2010^{2010}$, and $u_{n+1}=u_{n}+7$ if $u_{n}$ is odd, and $\frac{u_{n}}{2}$ if it is even. What is the smallest integer that this sequence will reach? | First, it is evident that the $u_{n}$ are always strictly positive. Next, let's look at the possible minima: if $u_{n}$ is even, then $u_{n+1}=\frac{u_{n}}{2}$ is strictly less than $u_{n}$. If $u_{n}$ is odd and greater than 7, then $u_{n+2}=\frac{u_{n}+7}{2}$ is strictly less than $u_{n}$. It remains to consider 1, 3... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all positive integers $n$ such that $2^{n}+3$ is a perfect square. The same question with $2^{n}+1$. | 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 $2^{n}+3 \equiv 3 \pmod{4}$, but a square is congruent to 0 or 1 modulo 4, so for $n \geq 2, 2^{n}+3$ is not a perfect square.
Suppose that $2^{n}+1=x^{2}$. Then $(x-1)(x+1)=2^{n}$. Therefore, $... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the minimum of the function
$$
f(x)=\frac{\cos (x)^{3}}{\sin (x)}+\frac{\sin (x)^{3}}{\cos (x)}
$$
on $] 0, \pi / 2[$.
Theorem 1.70 (Tchebycheff's Inequality). With the same notations as before, if the $a_{i}$ and the $b_{i}$ are arranged in the same order, then
$$
\sum a_{i} b_{i} \geq \frac{1}{n}\left(\sum a... | Let's use this exercise to recall how to determine the minimum of a function $f$ is $m$. This involves two steps: first, show that $f(x) \geq m$ for all $x$, and then find an $x_{0}$ where this minimum is achieved. Consider the function of interest,
$f(x)=\frac{\cos \left(x^{3}\right)}{\sin (x)}+\frac{\sin (x)^{3}}{\co... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Let $a, b, c, d$ be such that $a b c d=1$. Show that
$$
a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d \geq 10
$$ | We use the arithmetic-geometric inequality:
$$
a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d \geq 10 \sqrt[10]{a^{5} b^{5} c^{5} d^{5}}=10
$$ | 10 | Inequalities | proof | Yes | Yes | olympiads | false |
How many Friday the 13ths can there be in a year (non-leap year)? | In the days of the year, the 13th of each month occupies the ranks: 13, 44, $72, 103, 133, 164, 194, 225, 256, 286, 317, 347$. To determine which day of the week each day corresponds to, it is sufficient to take these values modulo $7: 6,2,2,5,0,3,5,1,4,6,2,4$. If day 1 of the year (January 1st) is a Friday, then there... | 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 (the 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 |
(JBMO 1998 - Problem 2), discussed in class
Let $A B C D E$ be a convex pentagon such that $A B=A E=C D=B C+D E=1$ and $\widehat{A B C}=\widehat{D E A}=90^{\circ}$. Calculate the area of $A B C D E$. | Let $P$ be the point on the ray $[D E)$ such that $D P=1$. Since $\widehat{A E P}=\widehat{A B C}=$ $90^{\circ}$, $A B=A E=1$ and $B C=1-D E=E P$, the triangles $A E P$ and $A B C$ are directly isometric. It follows that $A P=A C$ and that the polygons $A B C D E$ and $A C D P$ have the same area.
Furthermore, since $... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A permutation of $\{1,2, \ldots, n\}$ is drawn at random. What is the average number of fixed points | Let $X$ be the set of pairs $(x, \sigma)$ where $x$ is a fixed point of the permutation $\sigma$. For every $x$, there are $(n-1)!$ permutations that fix $x$, so $|X|=n \times(n-1)!=n!$
On the other hand, we have $|X|=\sum_{\sigma}|Fix(\sigma)|$, so the average number of fixed points is equal to 1. | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
How many parallelograms can be placed at most in a regular hexagon with a side length of 3? | Let's call \( s \) the area of an equilateral triangle with side 1. The regular hexagon can be divided into 6 equilateral triangles with sides 3, thus each having an area of \( 9s \): its total area is \( 54s \). Now, each parallelogram has an area of \( 4s \): since \( \frac{54s}{4s} > 13 \), we should be able to plac... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let ABC be a triangle with side lengths 3, 4, 5. Calculate the radius of the inscribed circle. | If you didn't already know, by drawing the $3,4,5$ triangle you will realize that it is a right triangle:
To calculate the radius $r$ of the inscribed circle, we will calculate the area $\m... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $0 \leqslant a, b, c \leqslant 1$. Show that $\frac{a}{b c+1}+\frac{b}{a c+1}+\frac{c}{a b+1} \leqslant 2$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | The function $f(a, b, c)$ on the left-hand side is convex in each variable, so it reaches its maximum at one of the 8 points $\left(\epsilon_{1}, \epsilon_{2}, \epsilon_{3}\right)$ with $\epsilon_{i} \in\{0,1\}$ (indeed, we use the fact that if a function $g:[u, v] \rightarrow \mathbb{R}$ is convex, then $g$ reaches it... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Let $0 \leqslant a, b, c \leqslant 1$. Show that $\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leqslant 1$. | The function $f(a, b, c)$ on the left-hand side is convex in each variable, so it attains its maximum at one of the 8 points $\left(\epsilon_{1}, \epsilon_{2}, \epsilon_{3}\right)$ with $\epsilon_{i} \in\{0,1\}$. We verify that the maximum is 1. | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
We write the numbers $1,2,3, \ldots, 100000$ on a sheet of paper, then replace each number by the sum of its digits, and so on, until each number consists of only one digit. What is the most frequent digit in the resulting list?
## - Counting and Inclusion-Exclusion -
PRINCIPLE OF INCLUSION-EXCLUSION - We denote by $... | Here, it is noted that for any integer $n$, the sum of the digits of $n$ is congruent to $n$ modulo 9. The resulting sequence is therefore periodic: $1,2,3,4,5,6,7,8,9,1,2,3,4, \ldots$ and it ends with a 1, so it is the one that appears most often. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
How many pairs $n$, $m$ exist such that $n \wedge m=50$ and $n \vee m=75$?
And such that $n \wedge m=50$ and $n \vee m=600$? | For the first one, the answer is 0! Indeed, a bit of reflection shows that we must always have $n \wedge m \mid n \vee m$, but $50 \nmid 75$.
For the second, we start by decomposing into prime factors: $50=2 \times 5^{2}$ and $600=2^{3} \times 3 \times 5^{2}$. We also know that if $n$, $m$ are solutions, then $n m=50 ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the gcd of all numbers of the form
$$
(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)
$$
where $a, b, c, d$ are integers. | We will show that the gcd sought is 12. Indeed, by the pigeonhole principle, among $a, b, c, d$ there are two numbers congruent modulo 3, and thus their difference is divisible by 3. Now if we look modulo 4, there are two possibilities: either there are two numbers among $a, b, c, d$ congruent modulo 4, and at that mom... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find an integer $x$ such that $x \equiv 2(\bmod 3), x \equiv 3(\bmod 4)$ and $x \equiv 1(\bmod 5)$. | First, the Chinese Remainder Theorem indicates that such an $x$ does indeed exist, as 3, 4, and 5 are pairwise coprime. Furthermore, it also indicates that there exists a unique natural number $a < 3 \times 4 \times 5 = 60$ such that $x \equiv 2(\bmod 3)$, $x \equiv 3(\bmod 4)$, and $x \equiv 1(\bmod 5)$ if and only if... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the number of ten-thousands in $A=5^{5^{5^{5^{5}}}}$.
Determine the number of ten-thousands in $A=5^{5^{5^{5^{5}}}}$. | We are going to determine the remainder of the Euclidean division of A by $10000=$ $2^{5} \times 5^{5}$. We start by dividing by $2^{5}$. Since $\varphi\left(2^{5}\right)=16$, it suffices to determine the remainder of the Euclidean division of $5^{5^{5^{5}}}$ by 16 and finally, since $\varphi(16)=8$, it is necessary to... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let A be the sum of the digits of 2012 ${ }^{2012}$. We define B as the sum of the digits of A, and $\mathrm{C}$ as the sum of the digits of $B$. Determine $C$. | We start by crudely bounding C. Since $2012^{2012} \leqslant 10000^{2012}$, it has fewer than $4 \times 2012 + 1 < 10000$ digits, so $A < 9 \times 10000 = 90000$. Similarly, $B < 9 \times 5 = 45$ and finally $C \leqslant 13$.
Now we look modulo 9 because we know that the operation of taking the sum of the digits leave... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(i) Find all integers $n \geqslant 1$ such that $n$ divides $2^{n}-1$.
(ii) Find all odd integers $n \geqslant 1$ such that $n$ divides $3^{n}+1$.
## - Solution - | (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 |
For which values of $m \in \mathbb{R}$ does the equation:
$$
\left(x^{2}-2 m x-4\left(m^{2}+1\right)\right)\left(x^{2}-4 x-2 m\left(m^{2}+1\right)\right)=0
$$
have exactly three distinct roots? | The first quadratic $x^{2}-2 m x-4\left(m^{2}+1\right)$ always admits two distinct real roots. Therefore, either the second quadratic must admit a double root, or it must share a root with the first.
It admits a double root when $-2 m\left(m^{2}+1\right)=4$, which simplifies to $\mathfrak{m}^{3}+m+2=0$. This polynomia... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $\alpha, \beta, \gamma$ be the three roots of $x^{3}-x-1$, calculate: $\frac{1-\alpha}{1+\alpha}+\frac{1-\beta}{1+\beta}+\frac{1-\gamma}{1+\gamma}$ | A safe method, even if in this case we can find a faster one, is to look for the equation whose roots are $\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}, \frac{1-\gamma}{1+\gamma}$ 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=\fr... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $\left(a_{n}\right)$ be the sequence defined by $a_{1}=1$ and $a_{n+1}=1+a_{1} a_{2} \ldots a_{n}$. Determine $\sum_{n=1}^{+\infty} \frac{1}{a_{n}}$. | It is clear that for all $n \geqslant 2$, we have $a_{n+1}-1=a_{n}\left(a_{n}-1\right)$, and by inverting this relation, we get $\frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}}$, in other words $\frac{1}{a_{n}}=\frac{1}{a_{n}-1}-\frac{1}{a_{n+1}-1}$. By summing, we obtain $\sum_{k=1}^{n} \frac{1}{a_{k}}=1+\frac{1... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine all integers $n \geqslant 1$ such that $n$ divides $2^{n}-1$.
## - Polynomials in $\mathbb{Z} / p \mathbb{Z}-$
Unless otherwise stated, throughout the following, $p$ will denote a prime number, and we will work in the field $\mathbb{Z} / \mathrm{p} \mathbb{Z}$.
If $A$ is an integral domain, many results th... | $n=1$ works, and we will show that it is the only solution. Let $n \geqslant 2$, and suppose that $n \mid 2^{n}-1$. It is clear that $n$ is odd, since $2^{n}-1$ is. Let $p$ be the smallest prime divisor of $n$, and set $n=k p$. We have $p \mid\left(2^{n}-1\right)$ and by Fermat's Little Theorem, $2^{n}=\left(2^{k}\righ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Calculate $\left(\frac{37}{97}\right)$.
To translate the text into English, I have kept the format and line breaks as in the original text. | Let's apply quadratic reciprocity:
$$
\begin{array}{rlrl}
\left(\frac{37}{97}\right) & =\left(\frac{97}{37}\right) & & \text{since }\left(\frac{37}{97}\right)\left(\frac{97}{37}\right)=1 \\
& =\left(\frac{23}{37}\right) & & \text{since only the class of } 97 \text{ mod } 37 \text{ matters } \\
& =\left(\frac{37}{23}\r... | -1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(given to Jean-Denis Zafar). Let $m$ and $n$ be two integers that are coprime. Calculate the GCD of $5^{m}+7^{m}$ and $5^{n}+7^{n}$. | . We denote $S_{k}=5^{k}+7^{k}$. We will show that the GCD of $S_{n}$ and $S_{m}$ is 12 if $m$ and $n$ are odd, and 2 otherwise. We denote $\delta$ as the GCD we wish to calculate. For $m=n=1$, we directly have $\delta=12$. We assume $m>n$ and set $m=n+a$; $a$ and $n$ are coprime. Let $U_{a}=7^{a}-5^{a}$. Then $S_{m}-5... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(solved by François Caddet, Marc Coiffier, and Jean-Alix David). Let $\left(a_{n}\right)$ be defined by $a_{1}=3$, $a_{2}=2$, and for $n \geqslant 1, a_{n+2}$ is the remainder of the Euclidean division of $a_{n}+a_{n+1}$ by 100. Calculate the remainder of the Euclidean division of:
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{200... | . Consider the sequence $\left(b_{n}\right)$ such that $0 \leqslant b_{n} \leqslant 3$ and $b_{n} \equiv a_{n}(\bmod 4)$. Since 4 divides 100, $\left(b_{n}\right)$ satisfies the recurrence relation $b_{n+2} \equiv b_{n}+b_{n+1}(\bmod 4)$. We notice that the sequence $\left(b_{n}\right)$ is periodic, with the sequence $... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
. In the decimal writing of $A$, the digits appear in (strictly) increasing order from left to right. What is the sum of the digits of $9 A$? | . Let $A=\overline{a_{1} a_{2} \cdots a_{k}}$ be the decimal representation of $A$. By performing the subtraction
$$
\begin{aligned}
& \begin{array}{cccccc}
a_{1} & a_{2} & a_{3} & \cdots & a_{k} & 0
\end{array} \\
& \begin{array}{llllll}
- & a_{1} & a_{2} & \cdots & a_{k-1} & a_{k} \\
\hline
\end{array}
\end{aligned}... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find $k$ such that, for all real $a, b$ and $c$:
$$
(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)+k a b c
$$ | $k=-1$ works. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find $k$ such that, for all $n$, the following quantity is a square:
$$
4 n^{2}+k n+9
$$ | $k=12$ works. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Calculate the greatest common divisor of 15 and 70.
## - Prime numbers and the Sieve of Eratosthenes - Group A only -
Some integers in particular play a role of primary importance: these are the integers whose GCD is 1.
Definition 2.2.11.
Let $a$ and $b$ be two integers. We say that $a$ and $b$ are coprime if $a \w... | It is enough to apply Euclid's algorithm: it is easily verified that $70=15 \times 4+10$, then
that $15=10 \times 1+5$, and finally that $10=5 \times 2+0$. This shows that $70 \wedge 15=5$, which was not really surprising anyway. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Among the pairs of integers $(a, b)$ such that $1 \leqslant a \leqslant b \leqslant 5$, how many are there for which $a$ and $b$ are coprime?
Moreover, there are also other numbers, much better known to the general public, which have remarkable divisibility properties: these are the prime numbers.
## Definition 2.2.1... | It suffices to list the positive divisors of each integer between 1 and 5: each integer $n \in\{1,2,3,5\}$ is only divisible by 1 and itself, whereas 4 is divisible by 1, 2, and 4. Consequently, if $1 \leqslant a \leqslant b \leqslant 5$, the integer $a \wedge b$ is different from 1 if and only if $a=b \geqslant 2$ or ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
( JBMO 2018 )
$n$ numbers are written on a board, satisfying the following properties:
- all the numbers have 3 digits
- none of them contain a 0
- two different numbers never have the same hundreds digit, or the same tens digit, or the same units digit
- the sum of the digits of each number is 9
What is the maximum... | We already observe that $n \leqslant 7$, because we cannot use digits larger than 7 and we must use $n$ different unit digits. The sum of the $3 n$ digits used must be equal to $9 n$. The sum of the unit digits, for example, is at least: $1+\ldots+n=\frac{n(n+1)}{2}$. Hence the inequality satisfied by $n$:
$$
9 n \geq... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the limit of the following sequence:
$$
u_{n}=\sum_{i=1}^{n} \frac{1}{F_{i} F_{i+2}}
$$ | We find that $\frac{1}{F_{k} F_{k+2}}=\frac{1}{F_{k} F_{k+1}}-\frac{1}{F_{k+1} F_{k+2}}$. Therefore,
$$
u_{n}=\sum_{i=0}^{n} \frac{1}{F_{i} F_{i+2}}=\sum_{i=0}^{n}\left(\frac{1}{F_{i} F_{i+1}}-\frac{1}{F_{i+1} F_{i+2}}\right)=1-\frac{1}{F_{n+1} F_{n+2}}
$$
by telescoping sum.
Since $F_{k}$ tends to infinity, the lim... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A number is called sympathetic if, for each divisor $d$ of $n, d+2$ is a prime number. Find the maximum number of divisors of a sympathetic number.
## - Solutions - | Let $n$ be a fixed friendly integer, and let $d$ be a divisor of $n$. If $d \equiv 1(\bmod 3)$, then $d+2 \equiv 0$ $(\bmod 3)$, and since $d+2$ is prime, this means that $d+2=3$, so $d=1$.
Now suppose there exist two primes $p$ and $q$, distinct from 3, which both divide $n$. Then $p \equiv q \equiv 2(\bmod 3)$, so $... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
For any natural number $n$, we denote $\mathbf{S}(n)$ as the sum of the digits of $n$. Calculate $\mathbf{S}^{5}\left(2018^{2018^{2018}}\right)$. | It is clear that $A \equiv B \equiv C \equiv D \equiv E(\bmod 9)$. Since 2 has order 6 modulo 9, we then find
$$
2018^{2018} \equiv 2^{2018} \equiv 0 \quad(\bmod 2) \text { and } 2018^{2018} \equiv 2^{2018} \equiv(-1)^{2018} \equiv 1 \quad(\bmod 3)
$$
so $2018^{2018} \equiv 4(\bmod 6)$, which means that
$$
E \equiv ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Seven cases of a chessboard $(8 \times 8)$ are on fire. At each step, a case catches fire if at least two of its neighbors (by a side, not a corner) are on fire. Is it possible for the fire to spread everywhere? If not, how many cases initially on fire are needed at a minimum for the fire to be able to spread everywher... | We can show that the perimeter of the burning area decreases at each step. Therefore, the fire cannot spread to the entire chessboard (perimeter of 32 if the squares have a side of 1) from seven burning squares (maximum perimeter of 28). At least eight squares are needed: if all the squares of a major diagonal are init... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest non-zero natural number $n$ such that 2016 divides $20^{n}-16^{n}$. | We start by factoring 2016 into a product of prime factors: $2016=2^{5} \times 3^{2} \times 7$. We need to find the integers $n \geqslant 1$ such that the integer $20^{n}-16^{n}=2^{2 n}\left(5^{n}-4^{n}\right)$ is divisible by $2^{5}, 3^{2}$, and 7. This is exactly the set of integers such that $2 n \geqslant 5$, $5^{n... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider 2018 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 |
Let $S$ be a finite set of points in the plane, no three of which are collinear. Each point in $S$ is colored either blue or red. Suppose that every triangle whose three vertices are blue contains at least one red point, and every triangle whose three vertices are red contains at least one blue point.
Determine the la... | The answer is 8. We will in fact show that there are at most 4 blue points and at most 4 red points. Suppose there are at least 5 blue points. By performing a rotation if necessary, we can assume that the points have distinct x-coordinates. We denote $\mathcal{B}$ as the set of 5 blue points with the smallest x-coordin... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A number appears on a computer screen. We know 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 2020-th number to appear | Let's start by looking at the first values displayed by the computer:
- The first number is 2.
- The second number is $2^{2}-2 \times 2+1=1$
- The third number is $1^{2}-2 \times 1+1=1-2+1=0$
- The fourth number is $0^{2}-2 \times 0+1=1$
- The fifth number is $1^{2}-2 \times 1+1=1-2+1=0$
- The sixth number is $0^{2}-2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 integers $2,3, \ldo... | In this problem, we are looking for the smallest integer $k$ satisfying a certain property. Suppose we want to show that the smallest integer sought is the integer $c$. There will then be two parts to the proof. On the one hand, it is necessary to show that if an integer $k$ satisfies the property, then $k \geqslant c$... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $x>0$. Show that
$$
x+\frac{1}{x} \geq 2
$$
For which $x$ do we have equality? | This inequality is equivalent to $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} \geq 0$.
Therefore, equality is obtained for $x=\frac{1}{x}$, i.e., for $x=1$. | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
How many positive common divisors do $10^{100}$ and $10^{121}+10^{813}+10$ have? | The number of common divisors is equal to the number of divisors of the GCD of these two numbers. Indeed: a common divisor of the two numbers will be a divisor of the GCD.
$$
\begin{aligned}
10^{813}+10^{121}+10 & =\left(10^{713}+10^{21}\right) \times 10^{100}+10 \\
10^{100} & =10^{99} \times 10+0
\end{aligned}
$$
Th... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the GCD of 1000000000 and 1000 000, 005. | $$
\begin{aligned}
P G C D(1000000000,1000000005) & =P G C D(1000000005,1000000005-1000000000) \\
& =P G C D(1000000005,5) \\
& =5
\end{aligned}
$$
since $5 \mid 1000005$. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
We write down all the numbers from 1 to one million successively. Then we replace each number with the sum of its digits. Then we repeat this process until only single-digit numbers remain. Which digit appears most frequently? | Among the integers from 1 to 999,999, there are exactly 111,111 that are congruent to 1 modulo 9, 111,111 to 2, ..., 111,111 to 9. Since an integer is congruent to the sum of its digits modulo 9, we conclude that the digit 1 will be written 111,111 times, the digit 2 will be written 111,111 times, ..., and the digit 9 ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
65 people are lined up in the central column of a 17-column grid. At each step, each person takes a random step either to the left or to the right (all steps are of the same magnitude).
Show that there is always at least one column that contains at least 8 people. | Color every other column pink, every other column green, starting with pink. Initially, everyone is on a pink column. Then, with each operation, everyone changes color. Thus, everyone is always on the same color at any given time. Therefore, either all the pink columns are empty, or all the green columns are empty. The... | 8 | Combinatorics | proof | Yes | Yes | olympiads | false |
For which natural numbers $n$ is the integer $n^{2}+n+1$ a perfect square? | If $n=0$, we have $n^{2}+n+1=1$ which is a perfect square. Otherwise, we have
$$
n^{2}<n^{2}+n+1<n^{2}+2 n+1=(n+1)^{2}
$$
thus $n^{2}+n+1$ is strictly between two consecutive squares and cannot be one. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all positive integers $n$ such that $n+1$ divides $2n^2 + 5n$. Be sure to verify that the $n$ found are solutions. | - We factorize: $n+1$ divides $2n^2 + 5n = n(2n + 5)$. Since $\text{GCD}(n+1, n) = \text{GCD}(n+1-n, n) = \text{GCD}(1, n) = 1$, by Gauss's lemma, since $n+1$ and $n$ are coprime, $n+1$ divides $2n + 5 = 2(n+1) + 3$, so $n+1$ divides 3. Therefore, $n+1 = 1$ or 3, so $n = 0$ or 2.
Conversely, $n=0$ works because 1 divi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $k \geqslant 1$ such that there exists a pair of integers $(n, m)$ such that $9 n^{6}=2^{k}+$ $5 m^{2}+2$. | Given that we have squares $\left(n^{6}\right.$ and $\left.m^{2}\right)$, we look at the equation modulo 4. To look at it modulo 4, suppose $k \geqslant 2$, the equation becomes $n^{6} \equiv m^{2}+2(\bmod 4)$. Since a square is either 0 or 1, the left-hand side is 0 or 1, and the right-hand side is 2 or 3.
It remains... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all natural numbers $n$ such that: $n^{5}-2 n^{4}-7 n^{2}-7 n+3=0$ | Let $n \in \mathbb{N}$. We notice that $n \mid n^{5}-2 n^{4}-7 n^{2}-7 n=-3$ thus $n \in\{-3,-1,1,3\}$. Only 3 works. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine all positive integers $n$ such that:
$$
5^{n-1}+3^{n-1} \mid 5^{n}+3^{n}
$$ | Let $n \in \mathbb{N}$. We have:
$$
5^{n-1}+3^{n-1} \mid\left(5^{n}+3^{n}\right)-5\left(5^{n-1}+3^{n-1}\right)
$$
Then
$$
5^{n-1}+3^{n-1} \mid 3^{n}-5 \times 3^{n-1}
$$
i.e.
$$
5^{n-1}+3^{n-1} \mid 3^{n-1}(3-5)
$$
. We also have $\operatorname{gcd}\left(5^{n-1}+3^{n-1}, 3^{n}-1\right)=1$: let $d$ be a common divi... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What are the integers $p$ such that $p, p+2$ and $p+4$ are all prime? | We look modulo 3:
If $p \equiv 0(\bmod 3):$ then $p=3$.
If $p \equiv 1(\bmod 3): p+2 \equiv 0(\bmod 3)$ and thus $3 \mid p+2$ which is prime i.e. $p+2=3:$ impossible.
If $p \equiv 2(\bmod 3): p+4 \equiv 0(\bmod 3)$ and thus $3 \mid p+4$ which is prime i.e. $p+4=3:$ impossible.
Conversely, $p=3$ works: 3, 5, and 7 a... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Determine the $gcd$ of all numbers of the form $(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$ where $a, b, c, d$ range over the integers.
## Solutions | We denote $P G C D$ as the integer sought.
We only have 4 numbers at our disposal and thus no chance to certify that a prime number $p \geqslant 5$ always divides integers of this form: $\forall p \geqslant 5$, prime, $p \nmid P G C D$. Therefore, we must stick to the factors 2 and 3 that would be present in all these... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the remainder of the Euclidean division of $2018^{2019^{2020}}$ by 11. | We know that $2018^{10} \equiv 1(\bmod 11)$, so we seek to calculate the value of $2019^{2020}$ modulo
10. Since $2019 \equiv -1(\bmod 10)$ and 2020 is even, we have $2019^{2020} \equiv 1(\bmod 10)$. Therefore, $2018^{2019^{2020}} \equiv 2018^{1} \equiv 5(\bmod 11)$. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(CMC 2020) Let $n$ be an integer. We call $C_{n}$ the grid of side $n$ from which one diagonal has been removed. For which integers $n \geqslant 2$ can $C_{n}$ be tiled with tiles consisting of three adjacent squares in the shape of an $\mathrm{L}$ (tiles can be flipped in any direction, but must not overlap or extend ... | It is necessary that the number of cells in $C_{n}$ is a multiple of 3, in other words, $n(n-1)$ is divisible by 3. For this, $n \equiv 0[3]$ or $n \equiv 1[3]$. Then, we notice that if we know how to tile $C_{n}$, then:
- we can tile $C_{n+6}$ if $n$ is even
- we can tile $C_{n+6}$ if $n$ is odd
- we can tile $C_{n+2... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(EGMO 2017) Find the smallest positive integer $k$ for which there exists a $k$-coloring of the positive integers and a function $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ such that
1. $\forall n, m \in \mathbb{N}^{*}$ of the same color, $f(n+m)=f(n)+f(m)$
2. $\exists n, m \in \mathbb{N}^{*}$ such that $f(n+m) \ne... | Obviously, at least two colors are needed to satisfy point (2). Moreover, we notice that 3 colors suffice: we color in blue the integers congruent to 0 modulo 3, in red those congruent to 1, and in yellow those congruent to 2 modulo 3, and the function
$$
f: n \in \mathbb{N}^{*} \mapsto \begin{cases}n & \text { if } n... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find the positive integers $n$ such that $2^{n}+1$ is a perfect square. | We already see that 3 is a solution: let's show that it is the only one.
Let $n$ be such an integer: there exists $a \geq 0$ such that $2^{n}+1=a^{2}$. Clearly $a \neq 0$. But then $2^{n}=(a-1)(a+1)$, so $a+1$ and $a-1$ are positive divisors of a power of 2, hence they are powers of 2. Since they differ by 2, they can... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n$ such that $n^{3}-3 n^{2}+n+2$ is a power of 5. | 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 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $x$ such that $x^{4}+x^{3}+x^{2}+x+1$ is a perfect square. (Bounding) | Let $A=x^{4}+x^{3}+x^{2}+x+1$. We have $x^{2}-x+1>0$ (quadratic with a strictly negative discriminant) and by immediate induction, for $x \geqslant 2, x^{3}>x+1$. We then have: $\left(x^{2}+x\right)^{2}=x^{4}+2 x^{3}+x^{2}>$ $A>x^{4}+2 x^{2}+1=\left(x^{2}+1\right)^{2}$. Therefore, there is no solution for $x \geqslant ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(factorisation) Find all natural numbers $n$ such that $n^{4}+4^{n}$ is prime. | For $n \geqslant 1, n^{4}+4^{n}=n^{4}+4 \cdot\left(2^{n-1}\right)^{2}=\left(n^{2}-n 2^{n}+2 \cdot 4^{n-1}\right)\left(n^{2}+n 2^{n}+2 \cdot 4^{n-1}\right)$ Thus, since $n^{2}+n 2^{n}+2 \cdot 4^{n-1} \geqslant 2$, if $n^{4}+4^{n}$ is prime, then $n^{2}+n 2^{n}+2=n^{4}+4^{n}$, and $n^{2}-n 2^{n}+2 \cdot 4^{n-1}=1$. Hence... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n \geq 1$ such that $n$ divides $2^{n}-1$. | We notice that $n=1$ works. Let $p$ be a prime divisor of $n$. Then $2^{n} \equiv 1 \pmod{p}$ and $2^{p-1} \equiv 1 \pmod{p}$ by Fermat's little theorem. Therefore, considering $p$ minimal, $p \mid 2^{\gcd(n, p-1)}-1=2-1=1$, which is absurd. The answer is therefore 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the multiplicity of the root 1 of $X^{2 n}-n X^{n+1}+n X^{n}-X^{2}$ for $n \in \mathbb{N}_{\geq 2}$? | 1 is at least a root of multiplicity 1. We derive the polynomial and obtain: $2 n X^{2 n-1}-n(n+1) X^{n}+n^{2} X^{n-1}-2 X$. 1 is a root if and only if $n=2$. We derive once more if $n=2$: $12 X^{2}-12 X+4$ for which 1 is not a root. Therefore, 1 is always a simple root except if $n=2$ and then $c'$ is a double root.
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $P=X^{3}-3 X-1$. Calculate the sum of the inverses of the squares of the roots (real or complex) of $P$.
Soit $P=X^{3}-3 X-1$. Calculer la somme des inverses des carrés des racines (réelles ou complexes) de $P$.
(Note: The second line is kept in French as per the original text, but the first line is translated to... | Let's denote $r_{1}, r_{2}, r_{3}$ as these roots. Then by Viète's formulas:
$$
\begin{aligned}
\frac{1}{r_{1}^{2}}+\frac{1}{r_{2}^{2}}+\frac{1}{r_{3}^{2}} & =\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}\right)^{2}-2 \cdot\left(\frac{1}{r_{2} r_{3}} \frac{1}{r_{1} r_{2}} \frac{1}{r_{1} r_{3}}\right) \\
& =\le... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $P$ be a polynomial with integer coefficients such that: $P(-1)=-4, P(-3)=-40$ and $P(-5)=$
$-156$.
What is the maximum number of integers $x$ for which: $P(P(x))=x^{2}$? | On a : $P(x) \equiv P(x \bmod 3) \bmod 3$ for all integer $x$. For each of the cases $x \equiv 0,1,2 \bmod 3$, we find $P(P(x)) \equiv 2 \bmod 3$ but a square is congruent to 0 or $1 \bmod 3:$ there are therefore none. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine the gcd of all numbers of the form
$$
(b-a)(c-b)(d-c)(d-a)(c-a)(d-b)
$$
where $a, b, c, d$ range over $\mathbb{Z}$.
## Bézout's Theorem and Euclidean Algorithm | Let $D$ be the gcd of all these numbers.
If we take $a=0, b=1, c=2$ and $d=3$, the product is
$$
1 \cdot 1 \cdot 1 \cdot 3 \cdot 2 \cdot 2=12
$$
Therefore, by the definition of gcd, $D \mid 12$.
Now let's check that 12 divides all numbers of this form. Let $a, b, c, d$ be four integers, and $n$ the corresponding pr... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(IMO 2016 P4) A set of natural numbers is said to be fragrant if it contains at least two elements and if each of its elements has a prime factor in common with at least one of the other elements. Let $P(n)=n^{2}+n+1$. Determine the smallest positive integer $b$ for which there exists a positive integer $a$ such that t... | Here it is natural to think that $b$ is likely to be relatively small, otherwise the problem risks being very complicated. We quickly calculate the $P(n)$ for $n$ between 1 and 20, and factorize them into prime factors to understand a bit how the problem works. Let's look at $b=2$: the question comes down to whether $P... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Find all integers $n \geqslant 1$ such that $n$ divides $2^{n}-1$. | First, $n=1$ works. Conversely, suppose we have a solution $n \geqslant 2$. Let $p$ be the smallest prime factor of $n$. Since $2^{n}-1$ is odd, $n$ is also odd, hence $p \geqslant 3$.
Let $\omega$ be the multiplicative order of 2 modulo $p$. Since $\omega$ divides $n$ and divides $p-1$, it also divides $n \wedge(p-1)... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(F-M) Let $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $x+f(x)=f(f(x))$ determine the solutions of $f(f(x))=0$. | We have the injectivity of $f$ and since $f(x)+f^{2}(x)=f^{3}(x)=f(x+f(x))$ we have $f(f(x))=0$ if and only if $f(x)=f(x+f(x))$ if and only if $x=x+f(x)$ if and only if $f(x)=0$ and thus $x+f(x)=f(f(x))=x=0$ so the only potential solution is 0. Furthermore, since $f(0)=f(f(0))$ we have $f(0)=0$ and thus $f(f(0))=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Consider the sequence $\left(a_{n}\right)_{n \in \mathbb{N}}$ defined by $a_{0}=2$ and $a_{n+1}=\sqrt{2 a_{n}-1}$ for all $n \geq 0$.
- Show that $a_{n}$ is irrational for all $n \geq 1$.
- Is the limit of $\left(a_{n}\right)$ irrational (if it exists)? | (solved by Pierre-Andréa Silvente)
First, the sequence is well-defined: indeed, we have $a_{n}>1$ for all integers $n \geqslant 0$. We show this by induction: we have $a_{0}>1$. Let $n$ be a positive integer, suppose $a_{n}>1$, then $a_{n+1}>$ $\sqrt{2 \times 1-1}=1$. Thus, by the principle of induction, $a_{n}>1$ for... | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
Find the polynomials with integer coefficients $P$ such that:
$$
\forall n, n \text { divides } P\left(2^{n}\right)
$$ | (let solved by Brieux Madeline-Derou)
Let $k$ be a positive integer and $q$ an odd prime integer. Set $n=2^{k} q$, then $n$ divides $P\left(2^{n}\right)$. Now $2^{n}=2^{(} 2^{k} \times 2^{q}$. By Fermat's Little Theorem, we have $0 \equiv P\left(2^{n}\right) \equiv P\left(2^{2^{k}} 2^{q}\right) \equiv P\left(2^{2^{k}}... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $n$ be a natural number. Consider $2 n+1$ tickets, each having a number which is a positive integer. Suppose that the sum of the numbers on the $2 n+1$ tickets does not exceed 2330 but that the sum of the numbers on any group of $n$ tickets is always greater than 1165.
Determine the maximum value of $n$. | (solved by Yaël Dillies)
Let $a_{1}<\cdots<a_{2 n+1}$ be the numbers written on the tickets.
We directly have $a_{i} \leqslant a_{j}+(i-j)$ for all $1 \leqslant i \leqslant j \leqslant 2 n+1$. Therefore, for all $i \in \llbracket 1, n+1 \rrbracket$,
$$
a_{i}=\frac{n a_{i}}{n} \leqslant \frac{1}{n} \sum_{j=n+2}^{2 n+... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $\sigma(n)$ be the sum of the prime divisors of $n$ (counted without multiplicity). Find all $n$ such that:
$$
\sigma\left(2^{n}+1\right)=\sigma(n)
$$ | (Solved by Elias Caeiro)
Let us note that by Zsigmondy, if $n \neq 3, 2^{n}+1$ admits a primitive prime factor which is therefore congruent to 1 modulo $2n$. Indeed, if we denote $q$ this factor, $q \neq 2$ and the order of 2 modulo $q$ divides $2n$, but does not divide $n$. In particular, the order can be written as ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A triangle has sides of length 3, 4, and 5. Determine the radius of the inscribed circle (circle inside the triangle and tangent to all three sides). | If we call $I$ the center of the inscribed circle, and $r$ the radius of the inscribed circle, since $r$ is the height of each of the triangles $I A B, I B C$ and $I C A$, the area of these triangles is: $\frac{1}{2} r \times A B$, $\frac{1}{2} r \times B C, \frac{1}{2} r \times C A$, so their sum is $r p$ by calling $... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. The natural number $n$ and the real number $a$ satisfy the relation
$$
n-\left(\frac{a-1}{a+1517}+\frac{\sqrt{|a|-2023}+\sqrt{2023-|a|}}{a-2023}\right)^{2024}=0
$$
Determine the last digit of the number $n$. | ## Solution.
From the existence conditions of $\sqrt{|a|-2023}$ and $\sqrt{2023-|a|}$, we obtain $|a|-2023 \geq 0$ and $2023-|a| \geq 0$, which are equivalent to $|a| \geq 2023$ and $|a| \leq 2023$. From the last two inequalities, we get $|a|=2023$, or $a= \pm 2023$.
Since the denominator of the fraction $\frac{\sqrt... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. Determine all natural numbers $n$, which have exactly four positive divisors, knowing that the sum of all positive divisors of the number $n$ is twice as large as $n$.
| ## Solution:
Since the natural number $n$ has exactly four positive divisors, it follows that the number has the form $n=p \cdot q$, $p \neq q$ or $n=p^{3}$, where $p$ and $q$ are prime numbers.
Assume that $n=p \cdot q$. Then the positive divisors of the number $n$ are $1, p, q$, and $p \cdot q$. From the problem da... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.6. Consider the numbers: $a=2^{2}+4^{2}+6^{2}+\cdots+2020^{2} ; b=1^{2}+3^{2}+5^{2}+\cdots+2019^{2}$; $c=1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+2019 \cdot 2020$. Calculate the value of the expression $E=\frac{a-b}{a+b-c}$. | ## Solution:
$a-b=2^{2}+4^{2}+6^{2}+\cdots+2020^{2}-\left(1^{2}+3^{2}+5^{2}+\cdots+2019^{2}\right)=\left(2^{2}-1^{2}\right)+\left(4^{2}-3^{2}\right)+$ $\left(6^{2}-5^{2}\right)+\ldots .+\left(2020^{2}-2019^{2}\right)=3+7+11+\ldots .+4039=4042 \cdot 505$. $a+b-c=1^{2}+2^{2}+3^{2}+$. . $+2020^{2}-(1 \cdot 2+2 \cdot 3+3 ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12.3. Let the complex number $z=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{1}{4} i$. Show that the value of the ratio $\sqrt{3} \cdot \frac{\operatorname{Re}\left(z^{2024}\right)}{\operatorname{Im}\left(z^{2024}\right)}$ is a rational number. | Solution 1. $z=\frac{1}{2}\left(1-\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)=\frac{1}{2}\left(1-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)=\frac{1}{2}\left(2 \sin ^{2} \frac{\pi}{12}+2 i \sin \frac{\pi}{12} \cos \frac{\pi}{12}\right)=$
$$
\begin{gathered}
=\sin \frac{\pi}{12}\left(\sin \frac{\pi}{12}+i \cos \frac{\pi... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
11.2. Calculate the limit $\lim _{n \rightarrow \infty}\left(\left(1+\frac{1}{2}\right) \cdot\left(1+\frac{1}{2+3}\right) \cdot \ldots \cdot\left(1+\frac{1}{2+3+\ldots+n}\right)\right)$. | Solution.
Let $P_{n}=\left(1+\frac{1}{2}\right) \cdot\left(1+\frac{1}{2+3}\right) \cdots \cdot\left(1+\frac{1}{2+3+\ldots+n}\right)$.
$$
\begin{aligned}
& S_{n}=1+2+\ldots+n=\frac{n(n+1)}{2}, \forall n \in \mathbb{N}^{*} \Rightarrow \\
& S_{n}-1=2+3+\ldots+n=\frac{n(n+1)}{2}-1=\frac{n^{2}+n-2}{2}=\frac{(n-1)(n+2)}{2}... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
12.4. Let $I_{n}=\int_{1}^{n} \frac{[x]}{x^{2}+1} d x, n \in \mathbb{N}, n \geq 2$. Calculate: $\lim _{n \rightarrow \infty} \frac{I_{n}}{\ln n}$. | ## Solution:
According to the Stolz-Cesàro lemma, we obtain
$$
\begin{gathered}
\lim _{n \rightarrow \infty} \frac{I_{n}}{\ln n}=\lim _{n \rightarrow \infty} \frac{I_{n+1}-I_{n}}{\ln (n+1)-\ln n} \\
I_{n+1}-I_{n}=\int_{n}^{n+1} \frac{[x]}{x^{2}+1} d x=n \int_{n}^{n+1} \frac{d x}{x^{2}+1}= \\
=n(\operatorname{arctg}(n... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
8.6. Let the function $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=a x+b, a, b \in \mathbb{R}^{*}$. If $m, n$ are distinct natural numbers such that
$$
\begin{gathered}
n>2, m>2, f(1)+f(2)+\cdots+f(m)=-n \text { and } f(1)+f(2)+\cdots+f(n)=-m \text {, calculate } \\
\frac{f(1)+f(2)+\cdots+f(m+n)}{m+n}
\end{gathered}
$$ | Solution: $f(1)+f(2)+\cdots+f(m)=a \cdot \frac{m(m+1)}{2}+m b$, similarly $f(1)+f(2)+\cdots+f(n)=$ $a \cdot \frac{n(n+1)}{2}+n b \Rightarrow a m^{2}+a m+2 m b+2 n=0$ and $a n^{2}+a n+2 n b+2 m=0 \Rightarrow$ $a m^{2}+a m+2 m b+2 n-a n^{2}-a n-2 n b-2 m=0 \Leftrightarrow a(m+n+1)=2(1-b)$ $(m \neq n)$.
$f(1)+f(2)+\cdots... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Solve in $\mathbb{R}$ the equation $\sqrt{8 x^{2}+10 x-3}-\sqrt{8 x+12}=3+\sqrt{4 x+8}-\sqrt{4 x^{2}+7 x-2}$. | ## Solution.
The equation from the statement is equivalent to the equation $\sqrt{(4 x-1)(2 x+3)}+\sqrt{(4 x-1)(x+2)}-2 \sqrt{2 x+3}-2 \sqrt{x+2}=3$.
We will have $D V A=\left[\frac{1}{4} ;+\infty\right)$.
Next, we write the equation in the form
$\sqrt{4 x-1}(\sqrt{2 x+3}+\sqrt{x+2})-2(\sqrt{2 x+3}+\sqrt{x+2})=3 \L... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4. Let the set $M=\{f: \mathbb{R} \rightarrow \mathbb{R} \mid f(2 f(x)+x)+f(x)=4 x-4\}$.
a) Show that the set $M$ is not empty.
b) Find all numbers $a \in \mathbb{R}$ such that $f(a)=0$ holds for at least one function $f \in M$.
c) Find all numbers $a \in \mathbb{R}$ such that $f(a)=0$ holds for all functions $f ... | Solution. a) Let's show that the function $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=x-1$ belongs to the set $M$. Indeed, $f(2 f(x)+x)+f(x)=f(3 x-2)+x-1=4 x-4$. Therefore, $M \neq \varnothing$.
b) We will show that the set of these values $a$ is not empty, i.e., there exists a zero for at least one function in $M$. F... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
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