What is the first puzzle in MHB's puzzle contest?

[alyafey22]

Hi guys , this topic is specialized to post some puzzles and challenge members to solve them.

• I will post the first puzzle and anyone solves it should try to post another one .
  
• If he/she didn't have any, in half a day anyone is allowed to post another challenging problem
  
• The scope of this thread is puzzles that require thinking .
  
• We don't want puzzles that require lots of computations .
  
• Also try not to post puzzles that require advanced mathematics or any other field.
  
• Figurative puzzles are good idea also word problems are welcomed.
  
• Each puzzle should stand for one day , if not solved the OP should give some hints or post the complete solution .
  
• Thanking the post by the OP indicates a right answer .
  

We will start later today :)

 

[alyafey22]

First Puzzle

• In a certain bank the positions of cashier, manager, and teller are held by Brown, Jones and Smith, though not necessarily respectively. The teller, who was an only child, earns the least. Smith, who married Brown's sister, earns more than the manager.
  
  what position does each man fill ? illustrate .
  

 

[MarkFL]

First Puzzle

• In a certain bank the positions of cashier, manager, and teller are held by Brown, Jones and Smith, though not necessarily respectively. The teller, who was an only child, earns the least. Smith, who married Brown's sister, earns more than the manager.
  
  what position does each man fill ? illustrate .

Brown cannot be the teller since he has a sister, and thus is not an only child. Also Smith cannot be the teller since he does not earn the least. So the teller must be Jones.

We know Smith is not the manager as he earns more than the manager, so he must be the cashier, leaving Brown to be the manager.

 

[MarkFL]

Here is my puzzle:

• You have a bag that contains two coins. One coin has heads on both sides and the other has heads on one side and tails on the other. You select one coin from the bag at random and observe one face of the coin. If the face is heads, what is the probability that the other side is heads?

 

[anemone]

Here is my puzzle:

• You have a bag that contains two coins. One coin has heads on both sides and the other has heads on one side and tails on the other. You select one coin from the bag at random and observe one face of the coin. If the face is heads, what is the probability that the other side is heads?

\(\displaystyle P(\text{selecting one coin where the other side is heads given the observed face is heads})\)

\(\displaystyle =\frac{P(\text{selecting the coin that has heads on both sides})}{P(\text{selecting a coin with the observed face is heads})}\)

\(\displaystyle =\frac{P(\text{selecting the coin that has heads on both sides})}{P(\scriptsize \text{selecting the coin that has heads on both sides})+P(\text{selecting the coin that has head on one side and tail on the other side AND the observed face is heads})}\)

\(\displaystyle =\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}\cdot \frac {1}{2}}\)

\(\displaystyle =\frac{2}{3}\)

 

[MarkFL]

\(\displaystyle P(\text{selecting one coin where the other side is heads given the observed face is heads})\)

\(\displaystyle =\frac{P(\text{selecting the coin that has heads on both sides})}{P(\text{selecting a coin with the observed face is heads})}\)

\(\displaystyle =\frac{P(\text{selecting the coin that has heads on both sides})}{P(\scriptsize \text{selecting the coin that has heads on both sides})+P(\text{selecting the coin that has head on one side and tail on the other side AND the observed face is heads})}\)

\(\displaystyle =\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{2}\cdot \frac {1}{2}}\)

\(\displaystyle =\frac{2}{3}\)

Correct!

Here's the way I looked at it. Knowing one face of the coin is heads doesn't tell you which coin you have, so you either have the head/tail coin or the head/head coin. There are 3 ways you can be looking at a head, with 2 of those ways having a head on the other side of the coin, so the answer is 2/3.

Your turn to post a puzzle! (Wave)

 

[anemone]

Your turn to post a puzzle! (Wave)

Here's my puzzle::)

Divide the given region in four congruent pieces.
(The figure is obtained by joining three squares.)
View attachment 890

 

[MarkFL]

This is one solution:

View attachment 891

 

[anemone]

This is one solution:

View attachment 891

Your turn to post another riddle, Mark! :p

 

[MarkFL]

Okay, here's one:

• Three antique clocks were put up for auction, the first lost one minute every 24 hours; the second lost one minute every hour; and the third didn't work at all. Which clock would you buy if you chose the one that showed the correct time the most often?

 

[mathworker]

Okay, here's one:

• Three antique clocks were put up for auction, the first lost one minute every 24 hours; the second lost one minute every hour; and the third didn't work at all. Which clock would you buy if you chose the one that showed the correct time the most often?

the one that does not work...:)

 

[MarkFL]

the one that does not work...:)

Can you demonstrate that this is true? (Nod)

 

[mathworker]

Can you demonstrate that this is true? (Nod)

first one will get a minute difference every hour which is same as second and it can't obviously show the correct time every 24 hrs (1 hr per 24 hrs error) which is the time period for third clock to show correct time...;)

 

[MarkFL]

This is how I viewed it:

Since these are antique clocks, they do not distinguish between AM and PM.

Suppose at noon, you set all three clocks to 12:00.

The first 2 clocks will have to lose 12·60 = 720 minutes to again show the correct time.

For the first clock, this will take 720 days, or almost 2 years.

For the second clock, this will take 720 hours, or 30 days.

The third clock will be correct at the following midnight.


You're on deck! (Cool)

 

[mathworker]

Three boxes are all labeled incorrectly, and you must get the labels right.
The labels on the boxes read as follows:

\(\displaystyle Box 1-nails\)
\(\displaystyle Box 2-screws \)
\(\displaystyle Box 3-nails and screws\)
To gain the information you need to move the labels to the correct boxes, you may remove a single item from one of the boxes. You may not look into the boxes, nor pick them up and shake them, etc.

Can this be done? If so, how? If not, why not?

 

[MarkFL]

Since all boxes are labeled incorrectly, we know box 3 must either be all nails or all screws, so remove one item from box 3 to determine which. Move the appropriate label to box 3, then take the remaining label left on either box 1 or box 2 and move it to the other of those two boxes, then place the nails and screws label on the remaining box without a label. (Whew)

 

[mathworker]

Since all boxes are labeled incorrectly, we know box 3 must either be all nails or all screws, so remove one item from box 3 to determine which. Move the appropriate label to box 3, then take the remaining label left on either box 1 or box 2 and move it to the other of those two boxes, then place the nails and screws label on the remaining box without a label. (Whew)

obsolutely correct...(Lipssealed)

 

[MarkFL]

I will leave this with you, then go to bed:

• You have a 5 gallon jug and 3 gallon jug and more than enough water. You need exactly 4 gallons of water in order to defuse a bomb. Not one drop more than 4 gallons.
  
  How do you get exactly 4 gallons?

(Yawn)(Sleepy)

 

[mathworker]

take water in 5 gln tub and pour it into 3 gln tub and empty 3 gln now pour remaining 2gln to 3gln tub now take water in 5 gln and pour water in 3 gln your will be with four galon

 

[MarkFL]

take water in 5 gln tub and pour it into 3 gln tub and empty 3 gln now pour remaining 2gln to 3gln tub now take water in 5 gln and pour water in 3 gln your will be with four galon

Correct! (Cool)

 

[alyafey22]

Digit's Puzzle

• If the below letters represent digits' multiplications what digit does each letter represent ?

View attachment 892

 

[alyafey22]

Hint : if you match the letters with the digits and order the digits (1 2 ... 9 0) you get a meaningful statement .

EDIT : I made a terrible error , sorry !

 

[alyafey22]

I think this problem requires lots of computations .

Here is the solution

N O S U R E P A T H
1 2 3 4 5 6 7 8 9 0

 

[alyafey22]

Logic Puzzle

• Four men, one of whom was known to have committed a
  certain crime, made the following statements when questioned
  by the police.
  
  Archie : Dave did it.
  Dave : Tony did it.
  Gus : I didn't do it.
  Tony : Dave lied when he said I did it.
  
  If only one of these four statements is true, who was
  the guilty man? On the other hand, if only one of these
  four statements is false, who was the guilty man?
  

 

[mathworker]

gus and tony

 

[MarkFL]

archie and tony

I think it is best to provide the method you used to arrive at these solutions. (Nod)

 

[mathworker]

if we consider the last statement correct then dave didn't do it and others are false so tony didn't do it even and gus said he didn't so he do while in the other case we will get one statement opposing.
and if one statement is false if we consider first one false then then last one doesn't contradict and third one doesn't even so tony did it(second statement)

 

[alyafey22]

if we consider the last statement correct then dave didn't do it and others are false so tony didn't do it even and gus said he didn't so he do while in the other case we will get one statement opposing.
and if one statement is false if we consider first one false then then last one doesn't contradict and third one doesn't even so tony did it(second statement)

The solution to the second question is wrong, the second and the fourth statements contradict each other so one of them has to be false .

 

[mathworker]

sorry i thought it was 'Dave lied when he said "I did it"...i think it should be "Dave said that..." then...then first one too might have been wrong(Crying)

 

[alyafey22]

Squares and Cubes

• Can you find two whole numbers, such that the difference of their squares is a cube and the difference of their cubes is a square? What is the answer in the smallest possible numbers?

 

[mathworker]

\(\displaystyle 1^2-0^2=1^3\)
\(\displaystyle 1^3-0^3=1^2\)

 

[alyafey22]

\(\displaystyle 1^2-0^2=1^3\)
\(\displaystyle 1^3-0^3=1^2\)

In some context 0 is not a whole number ;) .

 

[alyafey22]

• \(\displaystyle 10^2-6^2 = 4^3 \)
  
• \(\displaystyle 10^3-6^3 = 28^2 \)