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King and wine bottles

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A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine. Fortunately (or say unfortunately) the bad king’s guards catch the servant after he has only poisoned one bottle. Alas, the guards don’t know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect. The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time. Explain what is in mind of the king, how will he be able to do so ? (of course he has less then 1000 prisoners in his prisons)

See Solution : King and wine bottles
Hint : Think in terms of binary numbers. (now don’t read the solution without giving a try…)

Number the bottles 1 to 1000 and write the number in binary format.

bottle 1 = 0000000001 (10 digit binary)
bottle 2 = 0000000010
bottle 500 = 0111110100
bottle 1000 = 1111101000

Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.

prisoner = 10 9 8 7 6 5 4 3 2 1
bottle 924 = 1 1 1 0 0 1 1 1 0 0
For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die.
After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned.
1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.

Wires on fire

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A guy has two wires of varying thickness, which each burns in 60 minutes. He actually wants to measure 45 mins. How can he measure 45 mins using only these two wires. (he can’t cut the one wire in half because the wires are non-homogeneous and he can’t be sure how long it will burn)

See Solution : Wires on fire

He will burn the first wire at both the ends and the second wire at one end. After half an hour, the first one burns completely and at this point of time, he will burn the other end of the second wire so now it will take 15 mins more to completely burn.. so total time is 30+15 i.e. 45mins.

Jelly beans

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You have three jars that are all mislabeled. One contains peanut butter jelly beans, another grape jelly jelly beans and the third has a mix of both (not necessarily half-half mix). How many jelly beans would you have to pull out and out of which jars, to find out how to fix the labels on the jars?

Labels on jars are as follows
Jar 1 : Peanut butter
Jar 2 : Grape
Jar 3 : P.b. / Grape

See Solution : Jelly beans

Only one jelly bean from the p.b./grape jar will do the trick.

The trick here is to realize that every jar is mislabeled. Therefore you know that the peanut butter jelly bean jar is not the peanut butter jelly bean jar and the same goes for the rest.
You also need to realize that it is the jar labeled p.b./grape, labelled as the mix jar, that is your best hope. If you choose a jelly bean out of there, then you will know whether that jar is peanut butter or grape jelly jelly beans. It can’t be the mix jar because i already said that every jar is mislabeled.
Once you know that jar 3 is either peanut butter, or grape jelly, then you know the other jars also. If it is peanut butter, then jar 2 must be mixed because it can’t be grape (as its labeled) and it can’t be peanut butter (that’s jar 3). Hence jar 1 is grape.
If jar 3 is grape, then you know jar 1 must be the mix because it can’t be p.b. (as its labeled) and it can’t be grape (that’s jar 3). Hence jar 2 is peanut butter.

Bridge

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Four friends A, B, C, D
Need to cross a bridge..
A maximum of 2 people can cross at a time..
It is night and they have just 1 lamp.
People that cross the bridge must carry the lamp to see the way..
A pair must walk together at the speed of slower person…

Speeds of
A: 1 minute to cross bridge
B: 2 minutes to cross bridge
C: 5 minutes to cross bridge
D: 10 minutes to cross bridge

Now question is… “what is the total minimum time required by all 4 friends to cross the bridge….?

See Solution : Bridge

Person A : 1 minute
Person B : 2 minutes
Person C : 5 minutes
Person D : 10 minutes

Person C and D are the slowest guys, if they don’t walk together that it self will make it 15 minutes, In this case (the best way to save time is) :
Initially A,B,C,D  ========
First A& B will cross the bridge and A will come back  => time taken 2 + 1 => 3
A,C,D ======== B
Now C & D should cross the bridge together and B will come back => time taken 10 + 2 => 12
Now A & B will cross the bridge together => time taken 2
total time 3 + 12 + 2 => 17

Supersonic bee

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Two trains enter a tunnel 200 miles long, traveling at 100 mph at the same time from opposite directions. As soon as they enter the tunnel a supersonic bee flying at 1000 mph starts from one train and heads toward the other one. As soon as it reaches the other one it turns around and heads back toward the first, going back and forth between the trains until the trains collide in a fiery explosion in the middle of the tunnel. How far did the bee travel?

See Solution : Supersonic bee
This puzzle is a little tricky one. One’s thinking about solving this problem goes like this “ok, so i just need to sum up the distances that the bee travels…” but then you quickly realize that its a difficult (not impossible) summation.
The tunnel is 200 miles long. The trains meet in the middle traveling at 100 mph, so it takes them an hour to reach the middle. The bee is traveling 1000 mph for an hour (since its flying the whole time the trains are racing toward one another) – so basically the bee goes 1000 miles.

Palindrome dates

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This year on October 2, 2001, the date in MMDDYYYY format will be a palindrome (same forwards as backwards).
10/02/2001
When was the last date that this occurred on?See Solution : Palindrome dates

One year can have only one palindrome as the year fixes the month and date too, so the year has to be less than 2001 since we already have the palindrome for 10/02. It can’t be any year in 1900 because that would result in a day of 91, same for 1800 down to 1400. it could be a year in 1300 because that would be the 31st day. So whats the latest year in 1300 that would make a month? When i first solved it, 12th month came to my mind as we have to find the latest date, so i thought it would be 1321. But we have to keep in mind that we want the maximum year in 1300 century with a valid date, so lets think about 1390 that will give the date as 09/31, is this a valid date… ? No, because September has on 30 days, so last will be the 31st August. Which means the correct date would be 08/31/1380.

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.

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