Adrián Paenza Online Conference
Forzados to be right, we have the mente!
Thursday, 11th June 2020 at 19:00
This was the fourth of the international online sessions taking advantage of the lockdown of the pandemic.
Adrián Paenza is an Argentine mathematician, well known everywhere for his books of puzzles and problems of recreational mathematics, books that he also makes available to everyone freely. In Argentina he is also recognized as a journalist in the sports field. He has received multiple awards as a science communicator.
More information from the speaker
The enigmas posed by Adrián with some of the solutions provided
Suponga que uno tiene 25 authenticities, a track for hacerlos dar vueltas. Only five self-driving cars are allowed. I don't have to time to determine the timing.
Whatever can be done at the end of a career, it is ordered by order of legacy. Can you design a strategy in such a way that you can select the three more quicks usando nada más que siete carreras (in total)?
As it comes, the problem is to design a strategy to select the authenticities for each race. Esa is the part that corresponds to usted.
Yo ya lo "ayudé" cuando le dije que con siete carreras alcanza. Ahora remains usted with the Chance to think.
– Hacemos 5 groups of 5. Every group has a career. Dentro of each group los coches are ordered así: A-B-C-D-E.
– We have a career between the 5 ganadores (los A) and ordenamos los grupos según haya quedado su coche A in this sixth career:
Grupo cuyo A has bequeathed 1º –> Los tagmos A1-B1-C1-D1-E1
Grupo cuyo A has bequeathed 2º –> Los tagmos A2-B2-C2-D2-E2
etc.
– Ya solo hay que tener en cuenta the 3 first "diagonals":
A1 B1 C1
A2 B2
A3
because with the relationships of order stable in the matriz discardmos así all the coches that tie por delante three or más coches más rápidos that ellos.
As we know that A1 is the most fast, solo hay that has to run in the otros cinco the last race. The two that remain first will be the segundo y tercero más rápidos of the total, respectively.
NOTE: in change from what is third in this last race, we can not guarantee that they occupy the absolute positions of 4º, 5º, etc., puesto that we have discarded coches that could also be serlo.
(provided by Carlos Luna)
A río separates two cities. Two boats use it in directions opuestas a velocidad constante, no need for misma, but they maintain misma velocidad a lo largo del trayecto.
More aún: cuando un boat bequeath of the otro lado, da la vuelta immediately sin detenerse y vuelve hacia el lugar de origen. And they reply to the process one y otra vez.
The boats sail to the mismo tiempo. They will be encuent by first vez in the way to 7 kilometers of one of the coasts, and continue his trayecto. Cuando each one bequeath of the otro lado, as Scribí más arrives, da la vuelta immediately. The boats were able to find one in segunda vez, this vez 4 kilometers from the Opuesta coast.
Cuál is the ancho del río?
Sea: x the amplitude of the río, V and W las velocidades, T y T' los tiempos de encuentro.
For the first encuentro tenemos dos ecuaciones:
7=V*T
x-7=W*T
deducimos 7/(x-7) = V/W la razón de las velocidades
For the second encuentro are:
x+4=V*T'
2x-4=W*T'
and por tanto (x+4)/(2x-4) =V/W
igualando y resolviendo x = 17km
(provided by Enric Brasó)
Representing the situation, based on the
semejanza de los triángulos of height 7 and 4 tendremos
If BA1 = m y AB1 = n
(2m – n)/4 = m/7 y por tanto 10m=7n
y yes BA = k
(k-7)/n = 7/m
k – 7 =10
k=17
It is observed that the sprints are proportional to 1/m and 1/n, so much will be in proportion 10/7.
Also tiene interest ver donde se sitúan los puntos de encuentro en 7 viajes del boat rápido (corresponding to 10 of the boat lento).
(provided by Manel Udina)
Cuando los boats will meet for the first time, between the two have toured the ancho del río and the boat lento has traveled 7 km.
In the second point of encuentro between the two have toured 3 veces the ancho del río and the boat lento habrá recorrido 3*7=21 km that are 4 más than the ancho del río.
Por tanto el ancho del río is 21-4 = 17 km.
I guess you're going to come into a bar. In the bar hay 25 asientos puestos in a hilera or row (como en cualquier bar). The curiosity is that all the customers who read at the bar are antisocial. What quiero decir con esto? Every evening that alguien viene in the bar, look at the 25 aientos are available and be the rule (not written) but that all cumplen: if all the asientos are vacíos, se sienta en cualquier parte, but if there are some or some employed, it is dejando the maximum possible distance with the otros clients that already occupy some asientos. In particular, sto dice que nadie sienta "al lado de nadie", in the feeling that if alguien enters and advierte that para sentarse tends to have some vecino, you intonate "patch media vuelta" and go.
Algo más que transforms this scene into algo bizarro: the barman, if it were pudiera, would trataria que aun serendo the rule that self-imposed, if it haya haya la mayor cantidad of posible clients. Question: if the barman pudiera elect dónde feel to the first cliente, ¿dónde le convince pedirle que se siente so of alcanzar ese number máximo? It is decided, it is traced that usted elabore a strategy in such a way that empezando con el bar totally vacío, at the moment in which they emp to bequeath clients, the decision of the barman allows to bequeath to the maximum possible fulfilled the rules auto-impuestas.
Razonamiento "de atrás hacia adelante":
The position is 13 persons sentadas in the asientos impares 1,3,5,7,9,11,13,15,17,19,21,23,25
Solo would take care of the asiento 3 if the person occupying the asientos 1 and 5. Solo will take care of the asiento 5 if the person employed 1 and 9. Solo will take on asiento 9 if they are to take care of the asiento 1 and 17.
Hay that place the first person or bien in the asiento 9 or bien in the 17 that are positions simétricas.
If we llamamos "silla de oro" on the silla in which we place the first person, we plant:
Siempre, para cualquier number of sillas, exist the silla de oro?
In lugar de sillas we think of spatios, between 25 sillas hay 24 espacios, the hemos divided into two groups uno of 8 and otro of 16 spatios. The spatios that are powered by two (2,4,8,16,...,...) are those that allow us to divide succinctly by myth and place the people óptimamente
For the golden silla exists the number of sillas must be in the form 2n+1 or bien 2n+2m+1.
3=2+1, 5=2²+1, 7=2²+2+1, 9=2³+1, 11=2+2³+1, 13=2²+23+1, ... , 25=2³+2⁴+1, ...
15 is the smallest integer that is not in this way. There is no silla de oro for a row of 15 sillas. We can check todas las posibilidades.
With a number par of sillas, the colocación óptima permits a single tramo of 2 adjoining sillas libres, stou aumenta the posibilidades of "buenas colocaciones". But the existence of the silla de oro is not guaranteed either. Estudiando caso por caso I have checked that there is no silla de oro for a row of 30 sillas.
(provided by Enric Brasó)
Usted pone in an urn 50 white bolitas (B) and 50 black bolitas (N). The game consists of the serente: usted mete la mano in the ballot box, and if you look, remove two bolitas. If the two that sack are of the mismo color, I receive a black ball. If they are of different color, the white one repeats. And I repite the process (suponga que usted tiene available todas las bolitas que requiera para reponer segun these rules). Obviously, in every step hay a bolita menos dentro of the urn. For the so much, there will be a moment in which there will be a single ball. What colour will be bold?
The siempre white bolitas are reduced in two, so that only their number is reduced if elected 2 white bolitas and replaced by a black one. It is decir, beempre the number of white bolitas will be PAR. Así, puesto that at the beginning tenemos 50 white balls, his number ira diminished to 48, 46, 44, ... 6, 4, 2, 0. The last ball will be black.
It is an ejemplo more of a problem of parity.
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