be real numbers such that 
,![\[
a + \frac{1}{a} + b + \frac{1}{b} + c + \frac{1}{c} + d + \frac{1}{d} = 0.
\]](http://latex.artofproblemsolving.com/a/9/0/a90c94d82e3650560635616dacaf392d300b8c91.png)
Prove that one of the numbers 
or 
is equal to 
.
of positive integers that satisfy the equation![\[
x + xy + xyz = 31.
\]](http://latex.artofproblemsolving.com/a/7/8/a78d1e9d66831f11e7de08c109855e7727ec8535.png)
numbers placed in a circular way is either 
or 
.
consecutive numbers.
Find the minimal possible value of this sum.
Find the maximal possible value of this sum.
people who do not know each other. Prove that it is possible to introduce them in such a way that no three of them have the same number of acquaintances.
Y by Adventure10, Mango247
There are 
lamps 
arranged in a circle in that order. At any given time, each lamp is either on or off. Every second, each lamp undergoes a change according to the following rule:
(a) For each lamp
, if 
have the same state in the previous second, then is off right now. (Indices taken mod .)
(b) Otherwise, is on right now.
Initially, all the lamps are off, except for
which is on. Prove that for infinitely many integers all the lamps will be off eventually, after a finite amount of time.
lamps 
arranged in a circle in that order. At any given time, each lamp is either on or off. Every second, each lamp undergoes a change according to the following rule:(a) For each lamp
, if 
have the same state in the previous second, then is off right now. (Indices taken mod .)(b) Otherwise,
is on right now.Initially, all the lamps are off, except for
which is on. Prove that for infinitely many integers all the lamps will be off eventually, after a finite amount of time.This post has been edited 1 time. Last edited by anantmudgal09, Dec 9, 2017, 1:09 PM
works and we get a Sierpienski's triangle 
. Here's my solution: We claim that 
works (for 
)
will remain off, till the lamp just one step closer to 
.
)
.
(where indices are taken modulo 
lamps, and so by the induction hypothesis, all of them will be switched on after a finite number of moves. Until then, the 
lamps 
will remain off. Now, in the next moment, all lamps, except for 
will get closed. Then, one can easily see that, in the next second, all lamps will become on, as desired. Hence, done. 
,
is a positive integer, satisfies the given condition.
,
consecutive lamps are turned on, and the rest are off.
seconds, all lamps will be turned on.
![[asy]
size(200);
int k = 3;
int n = 2^(k+1);
int on = 2^k;
pair P[];
for (int j = 0; j < n; ++j) {
P.push(dir(360*j/n));
}
// Dibujar lámparas
for (int j = 0; j < n; ++j) {
if (j < on) {
fill(circle(P[j], 0.1), black); // Encendidas
} else {
draw(circle(P[j], 0.1)); // Apagadas
}
}
[/asy]](http://latex.artofproblemsolving.com/5/d/f/5dfb115a8833da0ea8b90df37022844719700794.png)
It is evident for 
.
We assume that Lemma P holds for all 
,
.
![[asy]
size(200);
int k = 4; // Example for k=4
int n = 2^(k+1);
pair P[];
for (int j = 0; j < n; ++j) {
P.push(dir(360*j/n));
}
for (int j = 0; j < n; ++j) {
if (j < 2 || j >= n/2 && j < n/2 + 2) {
fill(circle(P[j], 0.1), black); // On lamps
} else {
draw(circle(P[j], 0.1)); // Off lamps
}
}
[/asy]](http://latex.artofproblemsolving.com/9/2/1/921f4afcecaf7550cb653297972eef25e5561956.png)
:
consecutive lamps turned on, as well as their opposite 
adjacent off-lamps on both sides (i.e., 
lamps, each in the form of Lemma P for 
,
,![\[
1 + 2^0 + 2^1 + 2^2 + \dots + 2^{L-2} = 2^{L-1}
\]](http://latex.artofproblemsolving.com/0/4/0/040bb2fbbc672df26bbfe4f97545117266cc841b.png)
which completes the inductio
lamps arranged in a circle 
)
lamps are off.
lamps are off.
,
:![[asy]
size(250);
int k = 5;
int i = 2;
int n = 2^k - 1;
int on1 = 2^i;
int off1 = 2^(k-1) - 2^i - 1;
int on2 = 2^i;
int off2 = 2^(k-1) - 2^i;
pair P[];
for (int j = 0; j < n; ++j) {
P[j] = rotate(360*j/n) * (1,0);
}
for (int j = 0; j < n; ++j) {
draw(circle(P[j], 0.1));
if (j < on1 || (j >= on1 + off1 && j < on1 + off1 + on2)) {
fill(circle(P[j], 0.1), black);
}
}
[/asy]](http://latex.artofproblemsolving.com/6/e/9/6e97228badb128cf3aeae0eba6e894c100ebce83.png)
,
.
,
steps, all lamps will be turned on.
consecutive lamps are on, and the rest are off.
![[asy]
size(250);
int k = 5; // Puedes cambiar el valor de k
int n = 2^k - 1;
int off = 2^(k-1);
pair P[];
for (int j = 0; j < n; ++j) {
P[j] = rotate(360*j/n) * (1,0);
}
// Dibujar todas las lámparas
for (int j = 0; j < n; ++j) {
draw(circle(P[j], 0.1));
}
// Encender las lámparas fuera de la zona apagada
for (int j = off; j < n; ++j) {
fill(circle(P[j], 0.1), black);
}
[/asy]](http://latex.artofproblemsolving.com/c/5/6/c56c4ab3ca5472506b1353d5ba7e736f74a637a3.png)
.
(it is only necessary to verify the case 
)
,
is an integer.
:![\[
2^{1-1} + 2^{2-1} + \dots + 2^{k-1} = 2^k - 1
\]](http://latex.artofproblemsolving.com/c/1/a/c1aac03a1bd0da82eb2947ff755192a56c38ce13.png)
seconds to reach a state where all lamps are on.
,
:
off-lamps and the 
off-lamps, we obtain a total of 
lamps.
![\[
2^{L+1} - 2^i - 1 \geq 2^{i-1} + 2^{i-1} \quad \text{for all } i \leq L-1
\]](http://latex.artofproblemsolving.com/a/6/4/a6463bc090746176a626a0e7b9ae5affa756daf2.png)
and![\[
2^{L+1} - 2^i \geq 2^{i-1} + 2^{i-1} = 2^i
\]](http://latex.artofproblemsolving.com/a/e/5/ae5ba75be565ddc3df647d04550dec8bc6899b54.png)
since 
.
,
![\[
2^{L+1} - 2^i - 1 = 2^{i-1} + 2^{i-1} - 1
\]](http://latex.artofproblemsolving.com/0/7/2/0720daca00a4a17a600e10a9688f66dd493ccd47.png)
and![\[
2^{L+1} - 2^i = 2^{i-1} + 2^{i-1} = 2^i.
\]](http://latex.artofproblemsolving.com/0/1/f/01f73f128402a666f3785e5398bcb130964dd5d9.png)
This implies that the selected sets of lamps intersect at exactly one lamp.
The statement is trivially true for 
.
Consider 
lamps adjacent to the illuminated lamp, both to the left and right.![[asy]
size(200);
int L = 4; // You may adjust L as needed
int k = L + 1;
int n = 2^k;
pair P[];
for (int j = 0; j < n; ++j) {
P.push(dir(360*j/n));
}
// Draw lamps
for (int j = 0; j < n; ++j) {
draw(circle(P[j], 0.08));
}
// Highlight the initially illuminated central lamp
fill(circle(P[0], 0.08), black);
// Select the 2^{L-1}-1 lamps on both sides
int range = 2^(L-1) - 1;
for (int j = 1; j <= range; ++j) {
draw(circle(P[j], 0.1), red);
draw(circle(P[n-j], 0.1), red);
}
[/asy]](http://latex.artofproblemsolving.com/8/4/2/8428fe66a0fc3225d2cae0dd2283610f7c82592a.png)
,
to the 
selected lamps, we obtain that in 
seconds, 
lamps will be illuminated while the remaining ones stay off.
seconds, all lamps are turned off, as required.
