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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”.

Red and blue marbles

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You have two jars, 50 red marbles and 50 blue marbles. You need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. When picking, you’ll first randomly pick a jar, and then randomly pick a marble out of that jar. You can arrange the marbles however you like, but each marble must be in a jar.

See Solution : Red and blue marbles

Lets say, we put all the red marbles in jar A and all blue marbles in jar B. Then the probability of getting a red marble is :

jar A : (1/2)*1 = 1/2 (selecting the jar A = 1/2, red marble from jar A = 50/50)
jar B : (1/2)*0 = 0 (selecting the jar B = 1/2, red marble from jar B = o/50)
So probability of getting red marble is 1/2 . Now as we need to maximize the P(getting a red marble), we have to increase the prob of getting a red marble in jar B. If we select jar A, then getting a red marble is guaranteed, but it will also be guaranteed if there is only one red marble in that jar, then also the probability of getting a red marble from jar A is 1/1=1. So now we can place remaining 49 red marbles in jar B, so it increases the prob of getting red marbles in jar B.

So the maximum probability will be :
jar A : (1/2)*1 = 1/2 (selecting the jar A = 1/2, red marble from jar A = 1/1)
jar B : (1/2)*(49/99) = 0 (selecting the jar B = 1/2, red marble from jar B = 49/99)
Total probability = 74/99 (~3/4)

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

Balls in a bag

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You have 20 blue balls and 14 red balls in a bag. You put your hand in and remove 2 at a time. If they’re of the same color, you add a blue ball to the bag. If they’re of different colors, you add a red ball to the bag. (assume you have a big supply of blue & red balls for this purpose. note: when you take the two balls out, you don’t put them back in, so the number of balls in the bag keeps decreasing). What will be the color of the last ball left in the bag?

Once you tackle that, what if there are 20 blue balls and 13 red balls to start with?

See Solution : Balls in a bag

There are 3 possible cases of removing the two balls…

a) If we take off 1 RED and 1 BLUE, in fact we will take off 1 BLUE
b)If we take off 2 RED, in fact we will take off 2 RED (and add 1 BLUE)
c) If we take off 2 BLUE, in fact we will take off 1 BLUE
So In case of (a) or (c), we are only removing one blue ball, but we always take off red balls two by two.

1) 20 Blue, 14 Red balls

If there are 14 (even) number of red balls, we can not have one single red ball left in the bag, so the last ball will be blue.

2) 20 Blue, 13 Red balls

Now as the no. of red balls is odd, there will be one single red ball in the bag with other blue balls, and whenever we remove 1 red and 1 blue ball, we end up taking off only the blue ball. So the red ball will be the last ball in the bag.

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

Zeroes in 100 factorial

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How many trailing zeroes are there in 100! (100 factorial) ?

See Solution : Zeroes in 100!
For every factor of 10, there will be one trailing zero similarly for every factor of 5 there will be one trailing zero (as 5*2 =10, and there are enough number of 2’s). But we have to take care of repetitions.

10, 20,…., 90 = 9 zeros

100 = 2 zeros

5, 15, 25……95 = 10 zeros

and 1 extra 5 in each of 25, 50 and 75 = 3 zeros

so total 9+2+10+3 = 24 zeros.

Average salary

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Three coworkers would like to know their average salary. How can they do it, without disclosing their own salaries to other two?

See Solution : Average salary

Let the three persons are A, B and C. Now first A tells the sum of his salary and a random number to B , lets say (SA + A(random no. of A)). Now B adds the sum of his salary and a random number to the number given by A i. e. B makes the total = (SA +A) + (SB+B) and he passes this number to C. (at each step, they don’t show the no. to the third person). Now C does the same, adds his salary and a random number to the amount told by B. Now C passes this total equals to (SA+SB+SC+A+ B+C) to A. Now A subtracts his random no. (A), passes to B then B subtracts his random no. (B) and passes to C. Finally C subtracts his random no. (C) and tells everybody. They divide it by 3. Now they have the average salary.

100 doors

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There are 100 doors, all closed. In a nearby cage are 100 monkeys. The first monkey is let out, and runs along the doors opening every one. The second monkey is then let out, and runs along the doors closing the 2nd, 4th, 6th,… all the even-numbered doors. The third monkey is let out. He attends only to the 3rd, 6th, 9th,… doors (every third door, in other words), closing any that is open and opening any that is closed, and so on. After all 100 monkeys have done their work in this way, what state are the doors in after the last pass, which doors are left open and which are closed ?

See Solution : 100 doors

Consider door number 56, monkeys will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1 1st monkey will open the door, pass 2 2nd one will close it, pass 4 open, pass 7 close, pass 8 open, pass 14 close, pass 28 open, pass 56 close. For every pair of divisors the door will just end up back in its initial state. But there are cases in which the pair of divisor has same number for example door number 16. 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is repeated because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will be open at the end.

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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