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Puzzles-3
Posted Date: 04 Mar 2008 Resource Type: Articles/Knowledge Sharing Category: Entertainments
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Posted By: Deepu Member Level: Diamond
Rating:  Points: 5
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When Socrates was imprisoned for being a disturbing influence, he was held in high esteem by his guards. All four of them hoped that something would occur that would facilitate his escape. One evening, the guard who was on duty intentionally left the cell door open so that Socrates could leave for distant parts.
Socrates did not attempt to escape, as it was his philosophy that if you accept society's rules, you must also accept it's punishments. However, the open door was considered by the authorities to be a serious matter. It was not clear which guard was on that evening. The four guards make the following statements in their defense:
Aaron:
A) I did not leave the door open.
B) Clement was the one who did it.
Bob:
A) I was not the one who was on duty that evening.
B) Aaron was on duty.
Clement:
A) Bob was the one on duty that evening.
B) I hoped Socrates would escape.
David:
A) I did not leave the door open.
B) I was not surprised that Socrates did not attempt to escape.
Considering that, in total, three statements are true, and five statements are false, which guard is guilty?
Answer
David is the guilty.
Note that "All four of them hoped that something would occur that would facilitate his escape". It makes Clement's statement B True and David's statement B False.
Now consider each of them as a guilty, one at a time.
Aaron Bob Clement David True
Stmts
A B A B A B A B
If Aaron is guilty False False True True False True True False 4
If Bob is guilty True False False False True True True False 4
If Clement is guilty True True True False False True True False 5
If David is guilty True False True False False True False False 3
Since in total, three statements are true and five statements are false. It is clear from the above table that David is the guity.
************************************************************
Substitute digits for the letters to make the following subtraction problem true.
S A N T A
- C L A U S
-----------------
X M A S
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.
Answer
One of the simplest brain teaser as there are total 26 possible answers.
It is obvious that S=C+1. Since A-S=S, it is clear that A=2*S or 2*s-10. Also, L and X are interchangeable.
SANTA - CLAUS = XMAS
24034 - 16492 = 7542
24034 - 17492 = 6542
24074 - 15432 = 8642
24074 - 18432 = 5642
24534 - 16492 = 8042
24534 - 18492 = 6042
24794 - 16452 = 8342
24794 - 18452 = 6342
24804 - 15462 = 9342
24804 - 19462 = 5342
24974 - 16432 = 8542
24974 - 18432 = 6542
36806 - 27643 = 9163
36806 - 29643 = 7163
36156 - 27693 = 8463
36156 - 28693 = 7463
62132 - 54206 = 7926
62132 - 57206 = 4926
62172 - 53246 = 8926
62172 - 58246 = 3926
62402 - 53276 = 9126
62402 - 59276 = 3126
62712 - 53286 = 9426
62712 - 59286 = 3426
62932 - 58206 = 4726
62932 - 54206 = 8726
========================
Brain Teaser No : 00844
Scientist decided to do a study on the population growth of rabbits. Inside a controlled environment, 1000 rabbits were placed.
Six months later, there were 1000Z rabbits. At the beginning of the 3rd year, there were roughly 2828Z rabbits, which was 4 times what the scientists placed in there at the beginning of the 1st year.
If Z is a positive variable, how many rabbits would be there at the beginning of the 11th year?
Submitted by : David Johnson
Answer
At the beginning of the 11th year, there would be 1,024,000 rabbits.
At the beginning, there were 1000 rabbits. Also, there were 4000 rabbits at the beginning of third year which is equal to 2828Z. Thus, Z = 4000/2828 i.e. 1.414 (the square root of 2)
Note that 2828Z can be represented as 2000*Z*Z (Z=1.414), which can be further simplified as 1000*Z*Z*Z*Z
Also, it is given that at the end of 6 months, there were 1000Z rabbits.
It is clear that the population growth is 1.414 times every six months i.e. 2 times every year. After N years, the population would be 1000*(Z^(2N)) i.e. 1000*(2^N)
Thus, at the beginning of the 11th year (i.e. after 10 years), there would be 1000*(2^10) i.e. 1,024,000 rabbits.
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A number of 9 digits has the following properties:
* The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.
* Each digit in the number is different i.e. no digits are repeated.
* The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.
Find the number.
Answer
The answer is 381654729
One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.
The other way to solve this problem is by writing a computer program that systematically tries all possibilities.
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Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat the whole of the grass.
How many cows are needed to eat the grass in 96 days?
Answer
20 cows
g - grass at the beginning
r - rate at which grass grows, per day
y - rate at which one cow eats grass, per day
n - no of cows to eat the grass in 96 days
From given data,
g + 24*r = 70 * 24 * y ---------- A
g + 60*r = 30 * 60 * y ---------- B
g + 96*r = n * 96 * y ---------- C
Solving for (B-A),
(60 * r) - (24 * r) = (30 * 60 * y) - (70 * 24 * y)
36 * r = 120 * y ---------- D
Solving for (C-B),
(96 * r) - (60 * r) = (n * 96 * y) - (30 * 60 * y)
36 * r = (n * 96 - 30 * 60) * y
120 * y = (n * 96 - 30 * 60) * y [From D]
120 = (n * 96 - 1800)
n = 20
Hence, 20 cows are needed to eat the grass in 96 days.
************************************************************
Substitute digits for the letters to make the following relation true.
W O R L D
+ T R A D E
-------------
C E N T E R
Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter W, no other letter can be 3 and all other W in the puzzle must be 3.
Answer
A tough one.
It is obvious that C=1. Also, the maximum possible value of E is 7. Now, start putting possible values of D, E and R as they occure frequently and use trial-n-error.
W O R L D 5 3 6 8 4
+ T R A D E + 7 6 0 4 2
------------ ------------
C E N T E R 1 2 9 7 2 6
************************************************************
In the following multiplication, certain digits have been replaced with asterisks (*). Replace all the asterisks such that the problem holds the result.
* * 7
X 3 * *
----------
* 0 * 3
* 1 *
* 5 *
-------------
* 7 * * 3
Answer
A simple one.
* If both are open, close both of them. Now, all holes are close.
* If both are close, open both of them. Now, all holes are open.
* If one is open and one is close, invert them i.e. close the open hole and open the close hole. Now, the diagonal holes are in the same status i.e. two holes in one diagonal are open and in other are close.
5. Check any two diagonal holes.
* If both are open, close both of them. Now, all holes are close.
* If both are close, open both of them. Now, all holes are open.
************************************************************
How many possible combinations are there in a 3x3x3 rubics cube?
In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)?
How many for a 4x4x4 cube?
1 1 7
X 3 1 9
----------
1 0 5 3
1 1 7
3 5 1
-------------
3 7 3 2 3
************************************************************
A blindfolded man is asked to sit in the front of a carrom board. The holes of the board are shut with lids in random order, i.e. any number of all the four holes can be shut or open.
Now the man is supposed to touch any two holes at a time and can do the following.
* Open the closed hole.
* Close the open hole.
* Let the hole be as it is.
After he has done it, the carrom board is rotated and again brought to some position. The man is again not aware of what are the holes which are open or closed.
How many minimum number of turns does the blindfolded man require to either open all the holes or close all the holes?
Note that whenever all the holes are either open or close, there will be an alarm so that the blindfolded man will know that he has won.
Answer
The blindfolded man requires 5 turns.
1. Open two adjacent holes.
2. Open two diagonal holes. Now atleast 3 holes are open. If 4th hole is also open, then you are done. If not, the 4th hole is close.
3. Check two diagonal holes.
* If one is close, open it and all the holes are open.
* If both are close, open any one hole. Now, two holes are open and two are close. The diagonal holes are in the opposite status i.e. in both the diagonals, one hole is open and one is close.
4. Check any two adjacent holes.
Answer
There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.
Let's consider 3x3x3 Rubics first.
There are 8 corner cubes, which can be arranged in 8! ways.
Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)
Similarly, 12 edge cubes can be arranged in 12! ways.
Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)
Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.
Total different possible combinations are = [(8!) * (3^7)] * [(12!) * (2^11)] / 2
= (8!) * (3^7) * (12!) * (2^10)
= 4.3252 * 10^19
Similarly, for 4x4x4 Rubics total different possible combinations are
= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24
= 7.4011968 * 10^45
Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.
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There is a number that is 5 times the sum of its digits. What is this number? Answer is not 0.
Answer
The number is 45, simply because
45 = 5 * (4 + 5)
How does one find this number?
Let T be the digit in the tens place and U be the digit in the units place. Then, the number is 10*T + U, and the sum of its digits is T + U.
The following equation can be readily written:
10*T + U = 5*(T + U) or
10*T + U = 5*T + 5*U or
5*T = 4*U
Thus, T / U = 4 / 5
Since T and U are digits, T must be 4 and U must be 5.
************************************************************
Eleven boys and girls wait to take their seats in the same row in a movie theater. There are exactly 11 seats in the row.
They decided that after the first person sits down, the next person has to sit next to the first. The third sits next to one of the first two and so on until all eleven are seated. In other words, no person can take a seat that separates him/her from at least one other person.
How many different ways can this be accomplished? Note that the first person can choose any of the 11 seats.
Answer
There are 1024 different ways.
This is the type of Brain Teaser that can be solved using the method of induction.
If there is just a one person and one seat, that person has only one option.
If there are two persons and two seats, it can be accomplished in 2 different ways.
If there are three persons and three seats, it can be accomplished in 4 different ways. Remember that no person can take a seat that separates him/her from at least one other person.
Similarly, four persons and four seats produce 8 different ways. And five persons with five seats produce 16 different ways.
It can be seen that with each additional person and seat, the different ways increase by the power of two. For six persons with six seats, there are 32 different ways.
For any number N, the different possible ways are 2(N-1)
Thus, for 11 persons and 11 seats, total different ways are 210 i.e. 1024
************************************************************
Find the next in the set:
AZFR, LARU, AMAS, SBNS, KICI, ????
Answer
The next word in the series is AAEA.
If you collect the corresponding letter from each word, they are names of the places.
Consider the following five names :
A L A S K A
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