2450 = 1 * 2 * 5 * 5 * 7 * 7
so list down all the possible sets ( x, y, z) such that xyz = 2450;
some sets are (1,49,50); (2,35,49); (5,10,49); (7,7,50) and so on.....
Now the bartender says that the sum of their ages is equal to your age. Since the person knows his own age but still cannot get the answer, this means that 'x + y + z ' is same for multiple sets say (x1,y1,z1) and (x2,y2,z2).....
only one such combination of multiple sets exists i.e
(5,10,49); (7,7,50) both of which sum up to 64.
Now we have to choose from one of these sets...
Now the bartender says that the oldest lady is older than peterson. Which means that the set must be (5,10,49).
So the ages must be 5,10,49
Sunday, July 17, 2011
Solution to The flippant number
Answer: 17
Solution: We use the notation “|” to mean “divides.”
There is only one flippant 2-digit number, namely 77. Indeed, if 10a + b is flippant
(where a, b are integers 1–9), then 7 | 10a + b and 7 | 10b + a. Thus,
7 | 3(10a + b) − (10b + a) = 29a − 7b = a + 7(4a − b),
so that 7 | a, and similarly 7 | b, so we’d better have a = b = 7
There are 16 flippant 3-digit numbers. First consider the 12 palindromic ones (ones
where the hundreds and units digits are the same): 161, 252, 343, 434, 525, 595, 616,
686, 707, 777, 868, and 959. Now consider the general case: suppose 100a + 10b + c is
flippant, where a, b, c are integers 1–9. Then 7 | 100a + 10b + c and 7 | 100c + 10b + a,
so 7 | (100a + 10b + c) − (100c + 10b + a) = 99(a − c), and so 7 | a − c. In order for this
not to result in a palindromic integer, we must have a − c = ±7 and, moreover, both
100a + 10b + a and 100c + 10b + c must be palindromic flippant integers. Consulting
our list above, we find 4 more flippant integers: 168, 259, 861, and 952.
Solution: We use the notation “|” to mean “divides.”
There is only one flippant 2-digit number, namely 77. Indeed, if 10a + b is flippant
(where a, b are integers 1–9), then 7 | 10a + b and 7 | 10b + a. Thus,
7 | 3(10a + b) − (10b + a) = 29a − 7b = a + 7(4a − b),
so that 7 | a, and similarly 7 | b, so we’d better have a = b = 7
There are 16 flippant 3-digit numbers. First consider the 12 palindromic ones (ones
where the hundreds and units digits are the same): 161, 252, 343, 434, 525, 595, 616,
686, 707, 777, 868, and 959. Now consider the general case: suppose 100a + 10b + c is
flippant, where a, b, c are integers 1–9. Then 7 | 100a + 10b + c and 7 | 100c + 10b + a,
so 7 | (100a + 10b + c) − (100c + 10b + a) = 99(a − c), and so 7 | a − c. In order for this
not to result in a palindromic integer, we must have a − c = ±7 and, moreover, both
100a + 10b + a and 100c + 10b + c must be palindromic flippant integers. Consulting
our list above, we find 4 more flippant integers: 168, 259, 861, and 952.
Solution to funbit
The answer is: a dice. An explanation: "It's always 1 to 6": the numbers on the faces of the dice, "it's always 15 to 20": the sum of the exposed faces when the dice comes to rest after being thrown, "it's always 5": the number of exposed faces when the dice is at rest, "but it's never 21": the sum of the exposed faces is never 21 when the dice is at rest, "unless it's flying": the sum of all exposed faces when the dice is flying is 21 (1 + 2 + 3 + 4 + 5 + 6)
Monday, July 11, 2011
Solution to Monty Hall problem
The solution presented shows the three possible arrangements of one car and two goats behind three doors and the result of switching or staying after initially picking Door 1 in each case:
Door 1 Door 2 Door 3 result if switching result if staying
Car Goat Goat Goat Car
Goat Car Goat Car Goat
Goat Goat Car Car Goat
A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.
Increasing the number of doors
That switching has a probability of 2/3 runs counter to many people's intuition. If there are two doors left, then why isn't each door 1/2? The intuition may be aided by generalizing the problem to have a large number of doors so that the player's initial choice has a small chance of winning.
It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three . In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. His initial probability of winning is 1 out of 1,000,000. The game host then opens 999,998 of the other doors revealing 999,998 goats. (Imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door.) The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. The chance that the player's door is correct hasn't changed. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.
To extend the above, it's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens the first 999,998 of them for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.
This example can also be used to illustrate the opposite situation in which the host does not know where the prize is and opens doors randomly. There is a 999,999/1,000,000 probability that the contestant selects wrong initially, and the prize is behind one of the other doors. If the host goes about randomly opening doors not knowing where the prize is, the probability is likely that the host will reveal the prize before two doors are left (the contestant's choice and one other) to switch between. If the host happens to not reveal the car, then both of the remaining doors have an equal probability of containing a car. This is analogous to the game play on another game show, Deal or No Deal; in that game, the contestant chooses a numbered briefcase and then randomly opens the other cases one at a time.
Door 1 Door 2 Door 3 result if switching result if staying
Car Goat Goat Goat Car
Goat Car Goat Car Goat
Goat Goat Car Car Goat
A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.
Increasing the number of doors
That switching has a probability of 2/3 runs counter to many people's intuition. If there are two doors left, then why isn't each door 1/2? The intuition may be aided by generalizing the problem to have a large number of doors so that the player's initial choice has a small chance of winning.
It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three . In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. His initial probability of winning is 1 out of 1,000,000. The game host then opens 999,998 of the other doors revealing 999,998 goats. (Imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door.) The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. The chance that the player's door is correct hasn't changed. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.
To extend the above, it's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens the first 999,998 of them for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.
This example can also be used to illustrate the opposite situation in which the host does not know where the prize is and opens doors randomly. There is a 999,999/1,000,000 probability that the contestant selects wrong initially, and the prize is behind one of the other doors. If the host goes about randomly opening doors not knowing where the prize is, the probability is likely that the host will reveal the prize before two doors are left (the contestant's choice and one other) to switch between. If the host happens to not reveal the car, then both of the remaining doors have an equal probability of containing a car. This is analogous to the game play on another game show, Deal or No Deal; in that game, the contestant chooses a numbered briefcase and then randomly opens the other cases one at a time.
Monday, July 4, 2011
Solution to CARD GAME
If we approximate the problem by assuming each card has 1/13 chance for each value 1 to 13, we can compute the expected return based on the number of cards remaining, like in the dice problem towr mentioned.
With the last card, the expected return is (1+13)/2 = 7.
With 2 cards remaining, we keep the card if the value is higher than that. The return is (7*7 + 6*(7+13)/2)/13 = 8.6154. (7 chances to continue and get 7, 6 chances to stop and earn (7+13)/2 on average).
With 3 cards remaining, you keep anything above 8.6. The return is (8*8.6154 + 5*(9+13)/2)/13 = 9.5325.
With 4 cards remaining, you keep 10 or more, the return is (9*9.5325 + 4*(10+13)/2)/13 = 10.13791534. That is what you can earn on average.
If you take into account that the cards don't repeat it changes these values. But I don't think it changes them to the point where the strategy should be changed. The average return should be slightly higher due to the fact that when you discard cards, these are likely to be low values, so the return after discarding a card is often higher than computed.
Just for the last card, if you are holding a 7, you can check whether the first 2 cards average to more than 7. If yes, you should better keep your seven. If not, you should draw the last card.
But If I do the same calculations with the rule that you always keep a value 10 or above, I find a result of 10.0068. I think the real value is slightly higher.
With the last card, the expected return is (1+13)/2 = 7.
With 2 cards remaining, we keep the card if the value is higher than that. The return is (7*7 + 6*(7+13)/2)/13 = 8.6154. (7 chances to continue and get 7, 6 chances to stop and earn (7+13)/2 on average).
With 3 cards remaining, you keep anything above 8.6. The return is (8*8.6154 + 5*(9+13)/2)/13 = 9.5325.
With 4 cards remaining, you keep 10 or more, the return is (9*9.5325 + 4*(10+13)/2)/13 = 10.13791534. That is what you can earn on average.
If you take into account that the cards don't repeat it changes these values. But I don't think it changes them to the point where the strategy should be changed. The average return should be slightly higher due to the fact that when you discard cards, these are likely to be low values, so the return after discarding a card is often higher than computed.
Just for the last card, if you are holding a 7, you can check whether the first 2 cards average to more than 7. If yes, you should better keep your seven. If not, you should draw the last card.
But If I do the same calculations with the rule that you always keep a value 10 or above, I find a result of 10.0068. I think the real value is slightly higher.
Sunday, July 3, 2011
Solution to ENVELOPE GAMBLE I
here is the possible outcome in a tabular form
| x | 2x
--------------------------------
dont change | x | 2x
--------------------------------
change | 2x | x
so if i dont change my "profit" obviously is going to be 0.
but if i change my "profit" is going to be 100% in one case and -50% in the other case. that makes the average profit to be 25%.
I am a bit confused about the approach, as intuitively i feel as this is same as suggested by the question so might be the wrong approach.
what do you think,
Update: another solution suggested is why should we consider the gain %age(as intially the amount is not with us, we are given some amount and an option to change it, %age wise though there is a gain if we change). So from the above table as we see, there is equal probabilty of getting x or 2x. Hence in this case if we change or dont change does matter.
Another observation: the answer might change if we say, (a) one amount is 1.5 times the other, or (2) one amount is 3 times the other.
| x | 2x
--------------------------------
dont change | x | 2x
--------------------------------
change | 2x | x
so if i dont change my "profit" obviously is going to be 0.
but if i change my "profit" is going to be 100% in one case and -50% in the other case. that makes the average profit to be 25%.
I am a bit confused about the approach, as intuitively i feel as this is same as suggested by the question so might be the wrong approach.
what do you think,
Update: another solution suggested is why should we consider the gain %age(as intially the amount is not with us, we are given some amount and an option to change it, %age wise though there is a gain if we change). So from the above table as we see, there is equal probabilty of getting x or 2x. Hence in this case if we change or dont change does matter.
Another observation: the answer might change if we say, (a) one amount is 1.5 times the other, or (2) one amount is 3 times the other.
Solution to BUTTON TRAP ROOM
Here is the solution. At the end of any step you may win, otherwise proceed to the next step.
1) push opposite buttons
2) push adjacent. buttons
3) push opposite buttons (if you didn't win by this step then you know you have a 3/1 situation)
4) push any one button (either winning or resulting in a 2/2 situation)
5) push opposites (either winning or resulting in an UUDD situation)
6) push adj. buttons
7) push opposites
1) push opposite buttons
2) push adjacent. buttons
3) push opposite buttons (if you didn't win by this step then you know you have a 3/1 situation)
4) push any one button (either winning or resulting in a 2/2 situation)
5) push opposites (either winning or resulting in an UUDD situation)
6) push adj. buttons
7) push opposites
Solution to MOUSE EATING CHEESE CUBES
Answer - no
Solution - Consider the subcubes to form a three dimensional checkerboard composed of subcubes of swiss and cheddar cheese. Assume the corners and centers of each face are swiss cheese and the rest of the subcubes, including the center, are cheddar cheese. Assume the mouse eats two subcubes an hour. At the beginning of every hour the mouse will be starting on a swiss subcube and will end the hour with a subcube of cheddar cheese. After 13 hours the mouse will have just finished a cheddar subcube. However the center subcube is also cheddar. Since two adjacent cubes must be of different kinds of cheese the mouse can not eat the center cube last.
Solution - Consider the subcubes to form a three dimensional checkerboard composed of subcubes of swiss and cheddar cheese. Assume the corners and centers of each face are swiss cheese and the rest of the subcubes, including the center, are cheddar cheese. Assume the mouse eats two subcubes an hour. At the beginning of every hour the mouse will be starting on a swiss subcube and will end the hour with a subcube of cheddar cheese. After 13 hours the mouse will have just finished a cheddar subcube. However the center subcube is also cheddar. Since two adjacent cubes must be of different kinds of cheese the mouse can not eat the center cube last.
Solution to TRUTHS, FALSEHOOD, RANDOMNESS
There are six possible sceanarios:
Truth Lying Random
Guy Guy Guy
----- ----- ------
I A B C
II A C B
III B A C
IV B C A
V C A B
VI C B A
Label the three men A, B, and C and tell each man who is what letter.
Step 1: Ask A, "Is B more likely to tell the truth than C?" If yes go to step 2, if no go to step 5.
Step 2: Ask C, "Are you the random guy?" If yes go to step 3, if no go to step 4.
Step 3: Ask C, "Is A the truth guy?" If yes then scenario IV, if no then scenario II.
Step 4: Ask C, "Is A the lying guy?" If yes then scenario V, if no then scenario VI.
Step 5: Ask B, "Are you the random guy?" If yes then step 6, if no then step 7.
Step 6: Ask B, "Is A the truth guy?" If yes then scenario VI, if no then scenario I.
Step 7: Ask B, "Is A the lying guy?" If yes then scenario III, if no then scenario IV.
Truth Lying Random
Guy Guy Guy
----- ----- ------
I A B C
II A C B
III B A C
IV B C A
V C A B
VI C B A
Label the three men A, B, and C and tell each man who is what letter.
Step 1: Ask A, "Is B more likely to tell the truth than C?" If yes go to step 2, if no go to step 5.
Step 2: Ask C, "Are you the random guy?" If yes go to step 3, if no go to step 4.
Step 3: Ask C, "Is A the truth guy?" If yes then scenario IV, if no then scenario II.
Step 4: Ask C, "Is A the lying guy?" If yes then scenario V, if no then scenario VI.
Step 5: Ask B, "Are you the random guy?" If yes then step 6, if no then step 7.
Step 6: Ask B, "Is A the truth guy?" If yes then scenario VI, if no then scenario I.
Step 7: Ask B, "Is A the lying guy?" If yes then scenario III, if no then scenario IV.
Friday, July 1, 2011
Solution to GLOBE TRAVERSAL
the North Pole. AND, a ring near the South Pole, slightly more than 1 mile North of the South Pole, where after you go 1 mile South, you walk in a 1 mile circle totally around the South Pole, then North back to where you started.
This covers most of the answer, but you could also start out a bit closer to the south pole, such that your walk that circles totally around the south pole is only 1/2 mile in length, and you do it twice. This is another ring of answers, and if you do the circle walk 3 times, this is another rings of answers, and so forth.
This covers most of the answer, but you could also start out a bit closer to the south pole, such that your walk that circles totally around the south pole is only 1/2 mile in length, and you do it twice. This is another ring of answers, and if you do the circle walk 3 times, this is another rings of answers, and so forth.
Solution to WHO AM I?
1) Nothing
2) A human life cycle as it walks on 4 limbs when a infant on two when growing and three when old i.e. 2 legs + a support stick
3) The letter E
4) A river
5) The future
6) Stars
7) A water mellon
8) The letter N
9) puggry(but this in not a popular word but some say it is a word) and answer is "Language"
10) A cofin
11) Darkness
12) A Penny
13) Bookkeeper. An alternate, tricky, answer could be Woollen (where W is a "double u")
14) Charcoal
15) An ear of corn
16) Fire
17) Iron ore
18) Few
19) Stove, fire, smoke
20) A frog. The frog is an amphibian in the order Anura (meaning "tail-less") and usually makes noises at night during its mating season.
21) The past. (Longfellow)
22) Wine
23) Wind
24) Your heart
25) Time
26) Gold
27) A fart
28) Iceberg
29) Shadow
30) A candle
31) Blue
32) Icicle
33) An echo
34) Water
35) Music
2) A human life cycle as it walks on 4 limbs when a infant on two when growing and three when old i.e. 2 legs + a support stick
3) The letter E
4) A river
5) The future
6) Stars
7) A water mellon
8) The letter N
9) puggry(but this in not a popular word but some say it is a word) and answer is "Language"
10) A cofin
11) Darkness
12) A Penny
13) Bookkeeper. An alternate, tricky, answer could be Woollen (where W is a "double u")
14) Charcoal
15) An ear of corn
16) Fire
17) Iron ore
18) Few
19) Stove, fire, smoke
20) A frog. The frog is an amphibian in the order Anura (meaning "tail-less") and usually makes noises at night during its mating season.
21) The past. (Longfellow)
22) Wine
23) Wind
24) Your heart
25) Time
26) Gold
27) A fart
28) Iceberg
29) Shadow
30) A candle
31) Blue
32) Icicle
33) An echo
34) Water
35) Music
Solution to COIN UNBIASING
Solution 1 - 1) Flip the coin twice.
2) If you get HT, guy #1 wins. If you get TH, guy #2 wins. These are equally probably, regardless of p.
3) If you get HH or TT, go to step 1)
Solution 2 - You hide the coin in one of your hands, and ask your rival to guess which hand has it.
Solution 3 - we know that heads is more likely. So both parties take turns flipping until they get tails. Whoever gets tails in fewer flips wins. If you both get tails on the first try, restart.
Solution 4 - A very practical solution: Why can't the girl just choose which bloke she likes better? :P
2) If you get HT, guy #1 wins. If you get TH, guy #2 wins. These are equally probably, regardless of p.
3) If you get HH or TT, go to step 1)
Solution 2 - You hide the coin in one of your hands, and ask your rival to guess which hand has it.
Solution 3 - we know that heads is more likely. So both parties take turns flipping until they get tails. Whoever gets tails in fewer flips wins. If you both get tails on the first try, restart.
Solution 4 - A very practical solution: Why can't the girl just choose which bloke she likes better? :P
Solution to KNIGHT VS. DRAGON
since both dragon and knight are enemies , they will try their best bet.
So dragon bring a glass from 7th well to which their is no antidote.
but now knight has trick .
before coming to the duel he drinks a glass from the 1st well...
hence the glass from the 7th well given to him by dragon acts an antidote :-)
now how does the dragon die ....
dragon knows that knight can bring water from only 1-6 numbered wells
so it can drink a glass from 7th well after the duel to neutralize the effect ....
but knight is smart enough , he anticipates this and bring a glass of water from the SEA/OCEAN (after all they live on a island ;-) ) which is poison-less....
dragon unaware of this fact still drinks a glass from 7th well and dies....
So dragon bring a glass from 7th well to which their is no antidote.
but now knight has trick .
before coming to the duel he drinks a glass from the 1st well...
hence the glass from the 7th well given to him by dragon acts an antidote :-)
now how does the dragon die ....
dragon knows that knight can bring water from only 1-6 numbered wells
so it can drink a glass from 7th well after the duel to neutralize the effect ....
but knight is smart enough , he anticipates this and bring a glass of water from the SEA/OCEAN (after all they live on a island ;-) ) which is poison-less....
dragon unaware of this fact still drinks a glass from 7th well and dies....
Solution to MYSTERIOUS TRIANGLE AREA
When you put the pieces that form the hypothenuse, you do not get the real value what it's supposed to be (square root of 13^2 plus 5^2). This is because the angle of the blue piece is smaller than the angle of the red piece (with angle i mean the left lower corner of each piece). When we rearrange them, the red and blue switch places and now the angle of the red is bigger than the angle of the blue. The first time the triangle bent in a bit if you would zoom in. Now the triangle bends out. You cannot see this because the line is thick. The area added when the triangle changed from bent in to bent out (really thin parallelogram) is now subtracted by the empty square. That way matter is conserved. Otherwise (it may seem weird) matter would be lost!!!
P.S. Anyone who still doesn't get it e-mail me at a_red_sani@yahoo.co.in
P.S. Anyone who still doesn't get it e-mail me at a_red_sani@yahoo.co.in
Solution to LOGICAL SIGNS I
Answer: SILVER CHEST
Suppose the inscription on the gold chest is true.This means,that the silver chest inscription is false.So it would mean that the silver chest does not contain the python as it says but it contains the treasure.And if the inscription on the gold chest is false,then it would mean that both the inscriptions are false.That would too imply that the silver chest contains the treasure and not the dragon.So either way we know that the silver chest contains the treasure and it would be safe to open it
Suppose the inscription on the gold chest is true.This means,that the silver chest inscription is false.So it would mean that the silver chest does not contain the python as it says but it contains the treasure.And if the inscription on the gold chest is false,then it would mean that both the inscriptions are false.That would too imply that the silver chest contains the treasure and not the dragon.So either way we know that the silver chest contains the treasure and it would be safe to open it
Solution to HOURGLASSES
1. Start both the 7 minute hour glass & 11 minute hour glass.
2. Wait till the 7 minute hour glass times out. Time is 7 minute!
3. Restart the 7 minute hour glass. At this time 11 minute hour glass will have 4 minutes left to time out.
4. As soon as 11 minute glass times out invert the 7 minute hour glass. Total time now is 11 minutes.
5. After inverting 7 minute hour glass, it will now have 4 minutes left for time out.
6. After these 4 minutes times out, the total time is 15 minutes
2. Wait till the 7 minute hour glass times out. Time is 7 minute!
3. Restart the 7 minute hour glass. At this time 11 minute hour glass will have 4 minutes left to time out.
4. As soon as 11 minute glass times out invert the 7 minute hour glass. Total time now is 11 minutes.
5. After inverting 7 minute hour glass, it will now have 4 minutes left for time out.
6. After these 4 minutes times out, the total time is 15 minutes
Solution to ARAB SHEIKH CAMELS
The old man said change places and race with the other brothers horse
Solution to MARBLE JARS
Move 49 of the white marbles to the black marble jar. This leaves a near 50/50 chance of survival if given one jar, and 100% chance of survival if given the other jar. Therefore your chances have increased from the initial 50/50 chance you had to about 75%
Solution to Poisened Drink Puzzle
Solution 1- the poison was in the ice and not in the scotch. When Jeremiah first consumes the scotch "on the rocks", the ice is still - ice. After some drinks the ice has started to melt. When Jeremiah leaves the last drink the ice has fully melted. So when Wallace takes the drink, he goes cold.
Solution 2- Poison density will be higher than scotch so it will set to bottom...initially when a glass was filled by scotch poison did not came to glass so he did not die...when last glass is gulped by his ancestors poison came to glass and he dies
Solution 2- Poison density will be higher than scotch so it will set to bottom...initially when a glass was filled by scotch poison did not came to glass so he did not die...when last glass is gulped by his ancestors poison came to glass and he dies
Solution to Prime Number Upside down
Here are few:
3718103 = edibile
5355307 = loesses
5734051 = isohels
53517001 = idolises
55076051 = isogloss
55373603 = edgeless
77185181 = ibisbill
378163771 = illegible
537816173 = eligibles
3216070301 = ideologize
03216070301 = ideologized
3718103 = edibile
5355307 = loesses
5734051 = isohels
53517001 = idolises
55076051 = isogloss
55373603 = edgeless
77185181 = ibisbill
378163771 = illegible
537816173 = eligibles
3216070301 = ideologize
03216070301 = ideologized
Solution to Lateral Thinking - The Elevator Puzzle
The man is a dwarf. He can't reach the upper elevator buttons, but he can ask people to push them for him. He can also push them with his umbrella.
Solution to Prisoner's Hat Puzzle
For the sake of explanation let's label the prisoners in line order A B and C. Thus B can see A (and A's hat colour) and C can see A and B.
The prisoners know that there are only two hats of each colour. So if C observes that A and B have hats of the same colour, C would deduce that his own hat is the opposite colour. However, if A and B have hats of different colours, then C can say nothing. The key is that prisoner B, after allowing an appropriate interval, and knowing what C would do, can deduce that if C says nothing the hats on A and B must be different. Being able to see A's hat he can deduce his own hat colour. (The fourth prisoner is irrelevant to the puzzle: his only purpose is to wear the fourth hat).
In common with many puzzles of this type, the solution relies on the assumption that all participants are totally rational and are intelligent enough to make the appropriate deductions.
After solving this puzzle, some insight into the nature of communication can be gained by pondering whether the meaningful silence of prisoner C violates the "No communication" rule (given that communication is usually defined as the "transfer of information").
Solution for the difficult form -:
In the case where every player has to make a guess, but they are free to choose when to guess, there is a strategy that allows every player to guess correctly. Each player should act as follows:
Count the numbers b of black hats and w of white hats.
Wait min(b,w) seconds.
If nobody has yet spoken, guess that your hat is black if you can see fewer black hats than white hats, or white if you can see fewer white hats than black hats.
If you have not yet spoken, guess that your hat is of the opposite colour to that of one of the first people to speak.
Suppose that in total there are B black hats and W white hats. There are three cases.
If B = W then those players wearing black hats see B−1 black hats and B white hats, so wait B−1 seconds then correctly guess that they are wearing a black hat. Similarly, those players wearing a white hat will wait W−1 seconds before guessing correctly that they are wearing a white hat. So all players make a correct guess at the same time.
If B < W then those wearing a black hat will see B−1 black hats and W white hats, whilst those wearing a white hat will see B black hats and W−1 white hats. Since B−1 < B ≤ W−1, those players wearing a black hat will be the first to speak, guessing correctly that their hat is black. The other players then guess correctly that their hat is white.
The case where W < B is similar.
Solution to Fork in the Road
A logician vacationing in the Bahamas finds himself on an island inhabited by two proverbial tribes of liars and truth-tellers. Members of one tribe always speak the truth, while other always lie. He stands at the fork of a road (a Tee-junction) and has to ask a bystander which leg he needs to follow to reach the village. He knows not whether the native is a truth-teller or a liar. The logician thinks for a moment and then asks only one question. What does he ask?
Another solution- Pick a road and ask, “Will the other guy tell me that this is the right road to take to get to town?” If it’s the right road and you ask the truth teller, he’ll know that the liar will say no, and thus say no. If you ask the liar, he’ll know that the truth teller would says yes, so he’ll lie and say no. If the road was the wrong road to take, the truth teller will know that the liar would says yes, and thus say yes. If you ask the liar, he’ll know that the truth teller would say no, so he’ll lie and say yes.
Another solution- Pick a road and ask, “Will the other guy tell me that this is the right road to take to get to town?” If it’s the right road and you ask the truth teller, he’ll know that the liar will say no, and thus say no. If you ask the liar, he’ll know that the truth teller would says yes, so he’ll lie and say no. If the road was the wrong road to take, the truth teller will know that the liar would says yes, and thus say yes. If you ask the liar, he’ll know that the truth teller would say no, so he’ll lie and say yes.
Solution to Licence To Kill
1)The first bit is easy, as the first letter of each plate spells WHODUNNIT IX
(A challenge to our Inspector.)
2)The second bit is a little trickier, If you read the last three letters in each plate from the bottom up and right to left you get
ADVANCE EACH LETTER BY THE NUMBER SHOWN, so advance W by 5 to get B,
H by 13 to get U and so on until you spell BUTLER DID IT
(A challenge to our Inspector.)
2)The second bit is a little trickier, If you read the last three letters in each plate from the bottom up and right to left you get
ADVANCE EACH LETTER BY THE NUMBER SHOWN, so advance W by 5 to get B,
H by 13 to get U and so on until you spell BUTLER DID IT
Solution to Match and Lemon-aid Puzzle
First, the waiter stuck the match into the lemon wedge, so that it would stand straight. Then he lit the match, and put it in the middle of the plate with the lemon. Then, he placed the glass upside-down over the match. As the flame used up the oxygen in the glass, it created a small vacuum, which sucked in the water through the space between the glass and the plate. Thus, the waiter got the water into the glass without touching or moving the plate.
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