Magia con un cuadrado magico

Todos conocemos los cuadrados magicos , y que tienen muchas propiedades de simetria que , enfocandolo de una manera ludica , puede llevarnos a sorprender a personas poco familiarizados con estos cuadrados ; bueno , tambien nos sorprenden a nosotros mismos , por supuesto , aunque sepamos que detras de ello hay un modelo matematico que lo explica perfectamente.

Tenemos una serie de naipes dispuestos segun la imagen de arriba (despues de lo anterior, ya sabemos que es un cuadrado magico , claro).

Pues bien , ahora elige 4 cartas , tal que no esten 2 de ellas en la misma fila ni en la misma columna , es decir , las 4 cartas deben estar en filas y columnas diferentes.

Suma los valores de las cartas elegidas. ( NOTA: AS=1 , J=10, Q=11)

Calcula a que hora corresponde , es decir , restale sucesivamente 12 , hasta que obtengas un numero menor o igual a 12.

Busca en el cuadro un naipe que coincida con esa hora.

Yo puse esa carta en este sobre

pulsa y comprueba si acerté.

Actualizacion: Habia un error en un par de cartas , que gracias a la observacion de un lector , ya hemos corregido.

El acertijo de las 23 cabezas


En el excelente blog El espejo Ludico , proponen este curioso problema :

En una animada tertulia, uno de los amigos del Conde, exclamó sonriente: “Tengo un animal en mi propiedad que tiene 22 cabezas y estoy dispuesto a apostarme cualquier cantidad con ustedes a lo que digo es rigurosamente cierto”.

El Conde Numerales, sin levantar la mirada de su café, replicó: “Querido amigo, o es usted un descuidado con sus animales o está equivocado”.

El interpelado dudó un momento y dijo: “Perdón, es cierto, me refería a 23 cabezas”.

¿Era posible que tuviera un animal semejante?

Hasta aqui el acertijo , como ya he comentado allí , no tengo ni idea y mis propuestas creo que van mal encaminadas.
A ver si alguien da con la solucion.

Esto me ha dado la idea de postear , una vez a la semana, un acertijo que considere interesante (hay muchos , claro , no podré poner todos) de blogs que leo habitualmente y publican acertijos.

Actualizacion: Solucion en comentarios

Acertijo geometrico.


Bueno , este es muy facilito…

1. Divide la figura 1 (morado) en 2 partes iguales

2. Divide la figura 2 (verde) en 3 partes iguales

3. Divide la figura 3 (amarillo) en 4 partes iguales

4. Divide la figura 4 ( rosa) en 5 partes iguales

Actualizacion: Solucion en comentarios por Quatermain

Acertijo optico. otro ejercicio para tus ojos


Hace tiempo puse el post Ejercicio para tus ojos en el que habia que encontrar un caracter distinto entre varios similares. Fue uno de los post mas visitados , y como estamos en vacaciones y bajo el influjo de las reposiciones televisivas , os pongo algunas pruebas mas , distintas , claro está , una cosas es una reposicion y otra copiar el post.

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En los siguientes casos , no te digo la que debes encontrar , tienes que hacerlo por ti mismo 😉
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Y este ya para los que no tengan realmente nada que hacer….

Encuentra la q

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Para los graciosos…
Encuentra el punto final.

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Las hormigas andarinas

Colocamos 1000 hormigas andando sobre un trozo de madera de 1 metro de largo ( a efectos del problema , consideramos solo la longitud como medida importante , y consideramos todo sobre el mismo plano)

Cada hormiga anda a una velocidad constante de 1 m/s

Las hormigas son colocadas todas ala vez y en posiciones al azar en la madera , tambien aleatoriamente , algunas empiezan a andar hacia la derecha y otras hacia la izquierda.

Cuando 2 hormigas se encuentran, cambian su sentido del movimiento:

Cuando una hormiga alcanza el final del palo , caen fuera. Finalmente todas ellas caeran.

¿Cual es el mayor tiempo que aguantará la hormiga que caiga la ultima?

Este acertijo solo requiere imaginacion, ningun conocimiento matematico.Un niño de 7 años podria resolverlo…lo que no quiere decir que sea facil..;)

Actualizacion: Solucion en comentarios , por Lobo