I have always loved brain teasers and puzzles. Has anyone come up with some of their own.

Post yours here that you like.

And if you want to answer one, don't cheat and look online for answers. You would only be cheating yourself anyway.



My original that I came up with is this:

The answer is a male's name:

The clue: It will surface in the middle of the year....

Edit to add: Platypus suggested that I should let everyone know this should be a chain type thread like the defending the indefensible thread. So if you like a particular teaser or puzzle post it here.

Here are a couple more sequence puzzles.

What number or letter comes next in these sequences..


Easy one first

1,2,3,5,8,13,21......?

3,3,5,4,4,3.....?

J,M,M,J,S......?

Ack! Does that mean having to solve the previous puzzle before posting one's own?

The answer to your puzzle

Is no of course not, feel free to post away

34. This is part of the famed Fibonacci sequence, where each number is the sum of the two preceding and the ratios between adjacent numbers ever more closely approximate the golden ratio (sqrt(5)-1)/2.

5. This is the number of letters in the English versions of successive numbers starting from one.

N. These are the first letters of odd months January, March, etc.


A new puzzle, picked primarily for succinctness of expression: what is the least number of links you can cut in a chain of 21 links to be able to give someone all possible number of links up to 21?

Good job on the answers Platypus..

If I am reading yours correctly you want to know how many cuts(as few as possible) it would take to separate all the links.

My answer is 11

Let's say that the links are money, and you want to be able to make change for any amount up to 21, so you can have chains of more than one link so long as some combination of the shorter chains you have left will add up to each number. For example, if you need to give someone 7, you could do it as one chain of 7, two chains of 6+1, 5+2, 5+1+1, etc. Does that make it clearer?

56....I think? 10+9+8+7+6+5+4+3+2+1+1

translate this:
NaCl + H2O / ccccccc

Salt Water over the 7 seas?

LOL! Very close! Saline (like sailing) over the seven seas. Hard to think of a good one after my afternoon glass of vino.
Was I right with yours, Plat?

Not quite. The correct answer is 2. What you end up with is sections of 3, 5, and 11 links, plus the two broken ones. To make any number up to 21...

1: 1
2: 1+1
3: 3
4: 3+1
5: 5 or 3+1+1
6: 5+1
7: 5+1+1
8: 5+3
9: 5+3+1
10: 5+3+1+1
11: 11
12: 11+1

etc.


OK, here's another favorite. Some of you might recognize it in another form, but this is the version for polite company. Let's say that you have to shake hands with three people, who might be infected with a horrible disease that's transmitted by skin contact. You have two gloves. How do you use the two gloves to shake hands with all three people without any chance of anyone getting infected through contact with a glove surface that has been contaminated by previous use?

Have two of the people put on the gloves, and shake their hands. Turn each of the gloves inside out. Place one on my hand (uninfected surface facing inward) and one on the third person's hand (unexposed surface facing inward) and shake. Right?

No, you have to shake each person's hand, they can't shake each others'.

Well...If I place one glove on one hand of 2 potentially infected people, and shake each. Then, the gloves are turned inside out....I place one on my hand, and one on the third person's. We shake?

Close enough. The constraints are more obvious in the original, which involves condoms instead of gloves, but I thought that version might be a little inappropriate.

I'm obviously sucking at describing puzzles today, so I'll let someone else pose the next one.

Here's a new attempt: Translate this

(friends) standing/ miss (friends)

Mis-understanding between friends

Yep

Very cool thread.

I like to make up original number puzzles, which can be solved using only basic algebra and logic. Here we go:

1. I have a certain amount of money made up of a certain number of ten-dollar bills and a certain number of one-dollar bills. The number of ten-dollar bills multiplied by the number of one-dollar bills is equal to the total amount of money I have in dollars. If the number of ten-dollar bills is NOT a prime number, how much money do I have?

2. Find the only two-digit number which is equal to the sum of its digits multiplied by the difference between its digits.

3. I have a rectangular box. Each of the three dimensions of the box (height, width, and length) is an exact number of inches. The three dimensions are all different. None of the dimensions is a prime number. The volume of the box in cubic inches is the same as the total surface area of all six sides of the box in square inches. What are the dimensions of the box?

For the last three problems, when I say the "reverse" of a number, I mean just writing it backwards. 21 is the reverse of 12, and 321 is the reverse of 123, for example.

4. Find a two-digit number and its reverse such that the sum of the two numbers is a perfect square, and the difference between the two numbers is also a perfect square.

5. Find a two-digit number and its reverse such that the sum of the two numbers is a perfect square, and the difference between the two numbers is a perfect cube.

6. Find a three-digit number and its reverse such that the difference between the two numbers contains the same three digits in yet another order.

I cheat. #2 is zero-zero.

Works for me!

Cute.

This is what mathematicians call a degenerate solution.

#4- Find a two-digit number and its reverse such that the sum of the two numbers is a perfect square, and the difference between the two numbers is also a perfect square.
All seem to work:

00

22

40

88

...or am I reading the question wrong?

Mike

Consider the source.

Hey now... I was on the math team. I even took first place once.

And Wertz, I don't see you answering any them...

Mike

I don't consider "00" to be a "real" two-digit number for the purposes of this question.

22 + 22 = 44; not a square, even if we let 22 - 22 = 0 be a square.

40 + 04 = 44; not a square, even if we let "04" be a two-digit number.

88 + 88 = 176; not a square, even if we let 88 - 88 = 0 be a square.

Maybe you are thinking of 144 (square of 12) and 196 (square of 14)?

Here are the perfect squares: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, and so on. Since 99 + 99 = 198, obviously you don't need to look for any larger squares. Another hint: None of the squares or cubes mentioned in these puzzles are zero.


EDIT: To make it clear: I mean the sum of the two TWO-DIGIT NUMBERS, not the sum of the two DIGITS.

Hehe. I used only the digits, no the two numbers.



Oh, so now you tell me!

Given my mis-read, though, my answers did work.

22 (two and two)
Sum: 4 (perfect square)
Diff: 0 (perfect square)

40 (four and zero)
Sum: 4 (perfect square)
Diff: 4 (perfect square)

88 (eight and eight)
Sum: 16 (perfect square)
Diff: 0 (perfect square)

These are some tough questions, though. #1 had me looking up algebra at 2:30 in the morning.

Mike

I believe the answer is 65.
65+56=121 (11 squared) and 65-56=9 (3 squared)

I can only get this one to work if you accept negative numbers. In which case:
48. 4 + 8 = 12. 4 - 8 = -4. -4 x 12 = -48. But then, -48 isn't exactly equal to 48.


954 - 459 = 495

(For Mike )

mrspigpen and Wertz are quite correct.

(Notice in #2 that I just said "the difference between its digits" -- it doesn't matter which digit is subtracted from the other.)

48 = (4+8)(8-4) = (12)(4)

I suppose I could have been clearer in my wording.

for you both.

OK, here's two, to keep people out of trouble for a while.

The first one's an oldie but a goodie. You have eight coins. One of them is either heavier or lighter than the other seven, but you don't know which, and you don't know how much. The only tool you have is a balance, which can tell you whether loads placed on both sides are the same weight, or which one's heavier. The challenge is to identify the "odd" coin, using no more than three weighings in any case.

The second one's new to me. You have eight tennis players to rank, from top to bottom. How can you devise a complete and accurate ranking of all eight, playing no more than 16 total matches? Matches are always decisive - win or lose, no draws - but there's no limit to how many matches a player can play in succession. I know an answer for a maximum of 17 matches, I know one exists for 16 but I haven't found it yet.

I'll try the first one...
Seperate the eight coins into two groups of four. Weigh one of the groups (two and two) on the scale.
1.If it is balanced, we know that there are no odd coins in that group, the odd one is in the other group. If it isn't balanced, we know one group (on the scale) contains the odd coin.
2.Take two of the coins from the odd group and two from the normal group and balance. If it is balanced, we know that the odd group is with the other two. If it isn't balanced, we know that the odd one is one of the two.
3. Take one of the two coins from the odd group and weigh it against one of the normal group. If it balances, it is the other coin. If it doesn't, it is the odd coin.

Your solution would work if we knew whether the odd coin was heavy or light. Let's step through an example:

1. Weigh coins AB against CD. They're equal, indicating that one of EFGH is the odd one.
2. Weigh AB (normal) against EF (unknown). Equal again. Now we know that one of GH is the odd one.
3. Weigh A (normal) against G (unknown). Equal again. Now we know that H is the odd one, but not whether it's light or heavy.

There's a bit of a trick to this. Each weighing can divide the possibilities into at most three groups. The largest group is always the most problematic, and contains at least one third of the original possibilities. Therefore, two weighings can distinguish between no more than nine possibilities, three can distinguish no more than 27, and so on. In the above, we know after step two that we're doomed because we only have one weighing left but four possibilities. In fact, we should start to worry after step one; we've reduced the possibilities from 16 to eight, but can only distinguish nine possibilities with the two weighings we have left. That means the second and third weighings must both do more work than the first in dividing the possibilities.

I also have a comment on the second problem. Anyone who studied computer science should recognize it right away as a sorting problem, and indeed one of the 17-step solutions is a merge sort (there are others as well). However, it's a sorting problem with some special characteristics, that allow a specific algorithm to work with fewer steps than a more general one. I really don't expect anyone to come up with a 16-step solution; very few, even among those who specialize in this particular kind of problem, have cracked that particular nut. I've been gnawing at it in spare moments for the last couple of days, devising at least a dozen solutions that are guaranteed to work in 17 or less and that average less than 16, but every time I think I have one that guarantees 16 or less I find one case that forces me to start over. I've even thought of writing a program to find the algorithm, but that's hard enough by itself that I might suggest it as a task for next year's functional-programming competition.

Doesn't anyone have any good logic (British logic) puzzles we can solve? I don't enjoy math and equations enough for these sorts of puzzles even though I am tempted to spend hours trying to figure them out because I am a puzzle nut. I like word puzzles though

Anyone? Anyone even know what British logic puzzles are?

Geez. I thought I just had to identify the coin. I have to tell if it's heavier or lighter in three moves? This is going to keep me thinking a while...I'll be back

Here's an easy one (meaning it doesn't take too much thought)

You have three boxes each labelled BB, WW and BW.
In each respective box there is two marbels (BB has two black marbles, WW has two white marbels and BW has one of each).
Each label is then switched around so that the boxes are wrongly labelled.

You can remove one marble at a time from any box and must replace it before taking another.
What is the smallest number of drawers you can make before you know what is in all the boxes.

I assume your instructions mean that we know that all the boxes are incorrectly labelled.

1. Take one marble out of the box labelled "BW." If it is black, you know both marbles are black. If it is white, you know both marbles are white. (Since it is labelled "BW," we know it does not have one of each.)

2. If the box labelled "BW" actually has two black marbles, then you know the box labelled "WW" must have one black marble and one white marble. (Since it is labelled incorrectly, it cannot have two white marbles.) Therefore the box labelled "BB" must have two white marbles. If the box labelled "BW" actually has two white marbles, then the box labelled "BB" must have one white marble and one black marble. (Since it is labelled incorrectly, it cannot have two black marbles.) Therefore the box labelled "WW" must have two black marbles.

In either case, you only need to take out one marble to know what is in each box.

SUMMARY:

CASE ONE: One black marble out of the box labelled "BW":

"BW" = two black marbles
"WW" = one of each
"BB" = two white marbles

CASE TWO: One white marble out of the box labelled "BW":

"BW" = two white marbles
"BB" = one of each
"WW" = two black marbles

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Sorry, I don't know what a British logic puzzle is. Can you give us an example, kmsouthern?

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Here is a word puzzle, as requested. Find a common word in the English language that starts with the letters "he" and ends with the letters "he." (The word "he" is not an acceptable answer; this is not a trick puzzle.) HINT: I hope you do not have one of these things.

I like to play old japanese puzzle games, their is a sliding bock one on the internet that takes 1024 correct slides to finish, i have never won Well at anyrate here is a link to some cool puzzles if you get bored, i lost the link to the original but these i am sure are worth a few minutes.

puzzle page

You have no idea how happy I am no one beat me to this....

Seperate the eight coins into three groups ABC, DEF, and GH

Balance ABC against DEF on th scale. If they are equal, G or H is odd, and can be determined easily (within two steps) by measuring F versus G and then F versus H.

If they are not equal,arrange the scale so that the heavy side is on the right. We will assume that the heavy side is DEF for this purpose. It might be a light ABC or heavy DEF.
Remove AF and rotate so that BE is on one (left) side, and CD on the right.

1 EB(up) versus CD (down)
If the tips the same way (right heavy) it can't be E or C. It must be a light B or heavy D
Balance E against B. If it is equal, D is heavy. If it isn't B must be light.

2 EB(down) versus CD (up)
If left (B,E)is now heavy, it must be either a heavy E or light C
Balance E against B. If it is equal, C is light. If it isn't, E must be heavy.

3.Balanced EB versus CD
If If the scales balace, either A was light or F was heavy. Compare either against one of the normal coins.

A serpant swam in a silver urn.
A golden bird in it's mouth abide
the serpent drank the water, this in turn
Killed the serpent. Then the gold bird died.

What is it?












An oil lamp.

headache and heartache are the two that immediately came to mind



Sure thing.

Here is an example of one:

There are 5 houses in 5 different colours. In each house lives a person of a different nationality. The 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet. Using the clues below can you determine who owns the fish?

The Brit lives in a red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
The green house is on the immediate left of the white house.
The green house owner drinks coffee.
The person who smokes Pall Mall rears birds.
The owner of the yellow house smokes Dunhill.
The man living in the house right in the middle drinks milk.
The Norwegian lives in the first house.
The man who smokes Blend lives next door to the one who keeps cats.
The man who keeps horses lives next door to the man who smokes Dunhill.
The owner who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blend has a neighbour who drinks water.

Your solution is not quite the same as the one with which I'm familiar, but I do think it works. Well done! In case you're interested, here's the one I already knew.

Same first step: ABC vs. DEF. If they're equal, you can do either A vs. G and A vs. H (your solution) or A vs. G and G vs. H (or similar). If they're unequal - let's say DEF is heavier, because it doesn't really matter - you can distinguish the remaining possibilities a second way. Try DE vs. AG and you get the following:

• DE is heavier. Therefore, either D/E is heavy, or A is light. Weigh D vs. E.
• The two are equal. Therefore, either F is heavy or B/C is light. Weigh B vs. C.
• AG is heavier. This is actually impossible, because neither D nor E could be light and neither A nor G could be heavy.

Neat, huh? Both solutions have some rather elegant kinds of symmetry.

Kmsouthern - it's the green, prince smoking, coffee swigging German. Am I right?

Nationality HColour Drink Smoke Pet

1. Norwegian - yellow - water - dunhill - cats
2. Dame - blue - tea - blend - horses
3. Brit - red - milk - pall mall - birds
4. German - green - coffee - prince - ??????
5. Swede - white - beer - blue master - dogs

(House numbers using 1. as the left most house)
Only missing bit of info is the germans pet so he must have the fish.


Here's another teaser:
You are stuck in a room with no doors, and no windows (feel free to ask other questions about the room) and there is nothing in the room except you but you are able to escape. How???
(Edited to add: the walls are several feet of concrete and the ceiling is 20 feet high, so you can't break out through the walls or climb out (there is a ceiling but it has nothing to do with getting out). This is more of a word puzzle than a mechanical one. You'll kick yourelf if I have to tell you. I added more info here so I didn't have to add a new reply)

Really thin walls?

No ceiling and the walls are short enough to climb?

Answer:

There are openings for doors and windows - just no doors and windows.

Yep - Just walk through the door frame. Cyan had you heard this before or did you work it out.
Platypus or Amlord did you get this before it was posted.

You said Cyan, but I think you were addressing me.

I worked it out.

Sorry, I did mean you Beladonna. Didn't mean to wrongly attribute your prowess.

Here's one more.

A lorry driver has to get accross a desert and can carry 1 unit of fuel but the desert is 1.5 units of distance. What is the smallest number of trips you can make in order that you can cross the desert. You can leave fuel in the desert to return to it later and you can ignore evaporation of fuel that you have left for this puzzle. how many trips must you make. (You can only carry a maximum of 1 unit of fuel).
Also you can only refuel from where you started your journey and must return to the start to refill your tank but you can top up your tank with the fuel you have left anywhere on your journey.
(1 unit of fuel will allow a lorry to travel 1 unit of distance)

(if you solve it for a distance of 1.5, try it with a distance of 1.9 (or more accurately 1.912) and 2.000 units distance(decimal places are important for this one and for 2.000 it is a lot of trips)
Can you work out the formula.

This method uses the fact that, from the refueling point, the vehicle can transport 1/3 unit of fuel, 1/3 unit of distance into the desert and still drive 1/3 unit distance back to the refueling point and arrive with an empty tank.

Step 1 - drive 4/3 units of fuel out to a point 1/3 distance away from the starting point, this will take 7 trips.

Step 2 - return to the start and fill up with a full tank, and drive out to where you left the fuel - you should have 2/3 units left in your tank (2 trips)

Step 3 - top up the tank and drive another 1/3 unit distance into the desert, leave behind 1/3 unit of fuel. (Note that at this point the vehicle is less than 1 unit distance away from the finish - ie with a full tank you could drive straight there.) (1 trip)

Step 4 - return to the point where you left the fuel before and fill up. (1 trip)

Step 5 - drive out to the point where you left the 1/3 of fuel and top up your tank - you can now drive directly to the finish with fuel to spare (2 trips)

1.5 units in 13 trips (whew! )

A distance of up to 5/3 units can be covered in this way.
for up to 2 units, you will need an additional 27 trips (I think) taking the total to 41.

---

Here's mine:

A fish has a head 4 1/2 inches long. The tail is the same length as the head plus half the length of the body and the body is the length of the tail added to that of the head.

How long is the fish?

I think your calculations are right but I haven't explanied the puzzle very well and I didn't say how a trip was counted either.
but here is some more info, I hope it clarifies what I meant.

I want to extend the puzzle (this is what it's meant to be but I didn't explain it very wellthe first time):
In this version you can only turn around once and must return to the beginning after every turn, so;
1.5 takes four trips
1.9 takes sixteen trips
And how many does 2 take.
The puzzle with more info:
You must travel a distance of 2 without leaving any fuel at a location and returning to it so that you can go out again, what I mean is if you leave 1/3 of your fuel at a distance of 1/3 you must pick up this fuel next time you get there and leave no fuel there, if there is more fuel there than you can carry then just say that this extra fuel evaporates. To travel a distance of 2 using this technique is a huge number of trips (a lot of thousands) but can anyone say how many.
hint: 1/3rd's are important and 2/3 has it's place.
In this extension you must leave and travel straight out picking up the fuel you need and then return to where you started, you cannot turn round more than once. Each trip must have you setting out once and then all the way back to the start. (e.g. going north then south. NOT north, south, north, south, north, south and then return to the begining)

(This sounds a bit confusing even to me and I wrote it, I can't find the book with the puzzle in it so I can't remember how it phrased it but have a go)

I'm solving the fish thing at the same time, I'll add it on Edit in a little while.
Nevermind, Amlord Beat me to it.