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River crossing – the harder one

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A dysfunctional family has to cross the river. On one side of the river are a mom and 2 daughters, dad and 2 sons, the maid and the dog. There is a boat only big enough to hold 2 people (counting the dog as 1 person). Only the adults are capable of operating the boat. Everyone has to get to the other side, without anything bad happening.
Now the difficulties are : if the dog is left with anyone and the maid isn’t there to control him, he’ll bite. The dad can’t be left with any of the daughters when the mom isn’t there. Likewise, the mom can’t be trusted alone with either of the sons when the dad isn’t there.
Remember! only an adult can operate the boat, and the boat can’t drive itself.See Solution : River crossing - the harder one

Say
West shore is {W} and East shore is {E}
Lets give the nick names to each family member and dog 😉
Mother – {m}, Father – {F}, Daughters – {d1, d2}, Sons – {s1, s2}
House maid – {h}, Dog – {d}

Initially,
W = {m, d1, d2, f, s1, s2, h, d}
E = {…}

let’s move everyone, over…

housemaid and dog go east, and the housemaid comes back:

W = {m, d1, d2, f, s1, s2, h}
E = {d}

housemaid and s1 go east, h and d come back:

W = {m, d1, d2, f, s2, h, d}

E = {s1}

father and s2 go east, father comes back:

W = {m, d1, d2, f, h, d}

E = {s1, s2}

mother and father go east, mother comes back:

W = {m, d1, d2, h, d}

E = {f, s1, s2}

h and d go east, father comes back:

W = {m, d1, d2, f}

E = {s1, s2, h, d}

father and mother go east, mother comes back:

W = {m, d1, d2}

E = {f, s1, s2, h, d}

mother and d1 go east, housemaid and d come back:

W = {d2, h, d}

E = {m, d1, f, s1, s2}

h and d2 go east, h comes back

W = {h, d}

E = {m, d1, d2, f, s1, s2}

h and d go east

W = {}

E = {m, d1, d2, f, s1, s2, h, d}

And we are done.

Daughter’s ages

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Two MIT math grads bump into each other at Fairway on the upper west side. They haven’t seen each other in over 20 years.

The first grad says to the second: “how have you been?”
Second: “great! i got married and i have three daughters now”
First: “really? how old are they?”
Second: “well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there..”
First: “right, ok.. oh wait.. hmm, i still don’t know”
Second: “oh sorry, the oldest one just started to play the piano”
First: “wonderful! my oldest is the same age!”

How old are the daughters ?

See Solution : Daughter's ages

We know that there are 3 daughters whose ages multiply to 72. Let’s look at the possibilities…

Ages:          Sum of ages:
1 1 72            74
1 2 36            39
1 3 24            28
1 4 18            23
1 6 12            19
1 8 9             18
2 2 18            22
2 3 12            17
2 4 9             15
2 6 6             14
3 3 8             14
3 4 6             13
After looking at the building number the second man still can’t figure out what their ages are, so that means that the sum of the ages (or building number) must be 14, since that is the only sum that has more than one possibility. Finally the man discovers that there is an oldest daughter. That rules out the “2 6 6” possibility since the two oldest would be twins. Therefore, the daughters ages must be “3 3 8”.

Ants on a triangle

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There are three ants on a triangle, one at each corner. At a given moment in time, they all set off for a different corner at random. What is the probability that they don’t collide ?

See Solution : Ants on a triangle

Solution 1:
Let the three ants are a, b, c.

There are two cases when they will not collide, the one is when they all move clockwise and the other is when they all move anticlockwise.
They will collide if any two ants move towards each other, at the same time the third ant can move in clockwise or in anticlockwise. so for each pair there are 2 such cases. And there are 3 pairs possible (a,b), (b,c) and (c,a). So total 3*2 = 6 cases when they will collide.

So probability that they will not collide is 2/(2+6) i.e. 1/4

Solution 2 :

Consider the triangle ABC. We assume that the ants move towards different corners along the edges of the triangle.

Total no. of movements: 8

A->B, B->C, C->A; A->B, B->A, C->A; A->B, B->A, C->B; A->B, B->C, C->B; A->C, B->C, C->A; A->C, B->A, C->A; A->C, B->A, C->B; A->C, B->C, C->B

Non-colliding movements: 2

A->B, B->C, C->A; A->C, B->A, C->B

(i.e. the all ants move either in the clockwise or anti-clockwise direction at the same time)

So probability of not colliding = 2/8 = 1/4

Calendar cube

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A man has two cubes on his desk. Every day he arranges both cubes so that the front faces show the current day of the month. what numbers are on the faces of the cubes to allow this?

See Solution : Calendar cube
We have two cubes, that means 12 faces or 12 numbers one on each face. The all possible dates are 1 to 31, that includes 11 and 22. So 1 and 2 should be there on both cubes. It means we need 12 digits (0-9 and 1, 2). Now if we see the distribution of numbers on each faces, we have 1 and 2 on both cubes. For 30 we need 0 and 3 on different cubes. So lets say, first cube has 1, 2, 3, 4, 5, 6 and other one has 0, 1, 2, 7, 8, 9. It looks fine, but we we notice, the man uses both the cubes for each day, so how do we show 07, 08, 09 ???

So that means we need to have 0 on both the cubes, but that makes it 13, but there are only 12 faces. Thats the trick in this question, we can place cubes upside down too, now which is the number we can use both the ways, yes its 6, it can be used as 9 and then we can have all the possible dates.

so first cube : 0, 1, 2, 3, 4, 5

second cube : 0, 1, 2, 6, 7, 8

Cube Puzzle

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You have a normal six sided cube. I give you six different colors that you can paint each side of the cube with (one color to each side). How many different cubes can you make?

Different means that the cubes can not be rotated so that they look the same. This is important! If you give me two cubes and i can rotate them so that they appear identical in color, they are the same cube.

See Solution : Cube puzzle

Let X be the number of “different” cubes (using the same definition as in the problem). Let Y be the number of ways you can “align” a given cube in space such that one face is pointed north, one is east, one is south, one is west, one is up, and one is down. (We’re on the equator.) Then the total number of possibilities is X * Y. Each of these possibilities “looks” different, because if you could take a cube painted one way, and align it a certain way to make it look the same as a differently painted cube aligned a certain way, then those would not really be different cubes. Also note that if you start with an aligned cube and paint it however you want, you will always arrive at one of those X * Y possibilities.

How many ways can you paint a cube that is already “aligned” (as defined above)? You have six options for the north side, five options for the east side, etc. So the total number is 6! (that’s six factorial, or 6 * 5 * 4 * 3 * 2 * 1). Note that each way you do it makes the cube “look” different (in the same way the word is used above). So 6! = X * Y.

How many ways can you align a given cube? Choose one face, and point it north; you have six options here. Now choose one to point east. There are only four sides that can point east, because the side opposite the one you chose to point north is already pointing south. There are no further options for alignment, so the total number of ways you can align the cube is 6 * 4.

Remember, Y is defined as the number of ways you can align the cube, so Y = 6 * 4. This gives us 6! = X * 6 * 4, so X = 5 * 3 * 2 * 1 = 30.

Reference : http://www.techinterview.org/Solutions/fog0000000128.html

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