Math Puzzle

Dan

Dan

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Here's a puzzle or more of a math problem to try to figure out.

The situation is that you have to figure out a number somewhere between 1 and 2000. You have to figure it out only by asking 15 yes or no questions. But the catch is that you won't get back any answers until you have asked all 15 questions and then one of the answers might be a lie. What will those 15 questions be?
Here's a hint, it might help to think in terms of computers.
 
If you figure that one out, I have another one from the same source but I haven't figured this one out myself yet.

There are 8 tennis players, you have to rank them in order of skill in the shortest period of time possible but you only have one tennis court and each game will take exactly 1 hour. You can assume that if player 1 beats player 2 and player 2 beats player 3 then player 1 would also beat player 3. You can also assume that these tennis players will never get tired. Now what is the fastest method for ranking them no matter what the outcome of any of the games? You should be able to do it in 17 hours.
 
If a brick weighs a pound and half a brick, how much does a brick-and-a-half weigh?
 
Do I ask all questions in one post or one @ a time?
 
it weights 3 pounds, anyways, don't jack the thread until the first two questions have been answered, I want to see what people get.
 
Dan said:
it weights 3 pounds, anyways, don't jack the thread until the first two questions have been answered, I want to see what people get.
Click to expand...

Half of 1.5 is .75, half a brick weighs .75, 1.5 + .75 = 2.25.
 
read a bit closer

If a brick weighs a pound and half a brick, how much does a brick-and-a-half weigh?
Click to expand...
brick = 1 pound + 1/2 brick
.: 1/2 brick = 1 pound
1.5 brick = 3/2 brick = 3 pound
 
Brian Damage said:
If a brick weighs a pound and half a brick
Click to expand...
This inicates a brick weighs one pound plus the weight of half a brick making both half one pound, so two pounds for thw whole brick. plus another half is three :)

EDIT: Dan beat me :(
EDIT2:w00t 100 posts
 
Do we ask 15 questions each?

Anyway, first question : Is the number even?
 
You're not looking for a specific number, you are looking for a set of questions which will find any number no matter what it is. But that's a good start.
 
Dan said:
You're not looking for a specific number, you are looking for a set of questions which will find any number no matter what it is.
Click to expand...

Ah ok, well is the number even is a pretty good start, eliminates half of the numbers.
 
You could then ask, can I multiply two by an integer value to get this number?

They couldn't lie on both.
 
Dan said:
read a bit closer


brick = 1 pound + 1/2 brick
.: 1/2 brick = 1 pound
1.5 brick = 3/2 brick = 3 pound
Click to expand...

Bah, shit.

I thought he said pound and a half...not and half a brick.
Sorry. :o
 
mortiz said:
You could then ask, can I multiply two by an integer value to get this number?

They couldn't lie on both.
Click to expand...

So your asking the same thing twice. Still doesn't help as you don't know which question was answered truthfully and which might be a lie.
 
but then which would be the lie and which the truth?
 
then you could ask, does this number have a factor of two?

You've asked the same question 3 times, but at least you'd know what to look for.
 
Here, if there is no lying involved, the problem can be solved in 11 questions. And remember the first hint I gave you too. Should simplify things a little bit.
 
So basically you'd have to ask every question 3 times in order to know the true answer to it.

You might as well just ask it the exact same way. No reason to change the wording.
 
nooooooooo!!

I assumed you couldn't ask the same question twice?
 
ok, here;s the start of my proper list.

1. Is the number even?
2. Is this number a power of two?
3. Is the number prime?
4. Is this number less than 1000?
 
mortiz said:
ok, here;s the start of my proper list.

1. Is the number even?
2. Is this number a power of two?
3. Is the number prime?
4. Is this number less than 1000?
Click to expand...

You need to have the first two repeated again. Otherwise it doesn't work. And it never says you can't ask the same question more than once. Also, in order to solve it your probably going to have to think of the whole list all at once not one questin at a time.
 
hmmm.......... I'll probably leave it for the time being,
 
Well this isn't going anywhere so i'll give further hints. Think binary, any number can be represented by a series of 1s and 0s. Yes and No answers are a lot like 1s and 0s. As for having 1 lie, a guy named Hamming worked out some pretty useful parity bits over 50 years ago. And it just so happens that a 15 digit string with a 4 bit parity will be able to correct a single error. I'll still leave it to you to try to figure out how to make simple yes or no questions out of that.
 
i think i may have gotten the 2nd puzzle.

PART I:
start out by splitting the 8 players into 2 groups of 4 players. just call them a,b,c,d.... in each group of two, determine who is the better player by playing a game (call these "Game 1"). it only takes 1 game for each group of 2... 4 groups so that part takes 4 hrs.

PART II:
then take your 4 ranked groups, and put them into 2 groups of 4... as in, put group a and b into, maybe, group "A"; and c and d into group "B". now determine ranking within each of the 2 new groups by having a game (call it "Game 2" just to prevent confusion) between the 2 top players (top player from old "a" vs top player from old "b"... call them a1/a2/b1/b2) then have a game (Game 3) between the loser from Game 2 and the loser in Game 1 (if a1 beats b1, then b1 plays a2)... then if necessary (as in, if b1 wins over a2 -- you don't know where b2 ranks) then play the loser from Game 3 against the other person in the other group (a2--theloserfrombefore-- vs b2)....... so overall the ranking for this part of the game takes 3 hrs for group "A" (3 games maximum to rank between groups "a" and "b") and 3 hrs for group "B" doing the same as with group "A". so this part takes a total of 6 hrs. to rank the players within the two groups of 4 players. 6 hrs.+4 hrs from before gives you 10 hrs; you have 7 hrs. left.

PART III:
now you want to see where the players in "A" rank in relation to the players in "B". to do this you'd use the same systematic games as for the 2nd part. as in, play top 2 players, then take the loser and play him against the next highest ranked person in the other group. kinda hard to explain, but if you write out all the combinations then it's easier to see. yeah... i did write out all the combinations. but the maximum number of games that need to be played in this part is 7 hrs (i can put up combinations if you don't trust that)...... so 7+10 from before gives 17 hrs. maximum.
 
Here's a puzzle or more of a math problem to try to figure out.

The situation is that you have to figure out a number somewhere between 1 and 2000. You have to figure it out only by asking 15 yes or no questions. But the catch is that you won't get back any answers until you have asked all 15 questions and then one of the answers might be a lie. What will those 15 questions be?
Here's a hint, it might help to think in terms of computers.
Click to expand...

Just be be boring...
If you convert the number to binary and subtract 1, is the first digit 1?
If you convert the number to binary and subtract 1, is the first digit 2?
If you convert the number to binary and subtract 1, is the first digit 3?
If you convert the number to binary and subtract 1, is the first digit 4?
If you convert the number to binary and subtract 1, is the first digit 5?
If you convert the number to binary and subtract 1, is the first digit 6?
If you convert the number to binary and subtract 1, is the first digit 7?
If you convert the number to binary and subtract 1, is the first digit 8?
If you convert the number to binary and subtract 1, is the first digit 9?
If you convert the number to binary and subtract 1, is the first digit 10?
If you convert the number to binary and subtract 1, is the first digit 11?

(the subtract 1 isn't nessecary, but I did it to make the number series start at 0.

Then, to find out which one is a lie, you follow a similar process with totals.
Ex: 1000=1111101000
1+1+1+1+1+0+1+0+0+0="6"="0110"
Of corse you have to see if your questions about lies recieve lies...but no one is paying me to write code for a 15 bit 1 to 2000 checksum.

There are 8 tennis players, you have to rank them in order of skill in the shortest period of time possible but you only have one tennis court and each game will take exactly 1 hour. You can assume that if player 1 beats player 2 and player 2 beats player 3 then player 1 would also beat player 3. You can also assume that these tennis players will never get tired. Now what is the fastest method for ranking them no matter what the outcome of any of the games? You should be able to do it in 17 hours.
Click to expand...
(assuming I read it correcty...I'm lazy)
Simple....just alittle notation first so I don't have to type so much.
p1=player1
r1=winner of round 1
l1=looser of round 1

Round 1= p1 vs p2
Round 2= p3 vs p4
Round 3= p5 vs p6
Round 4= p7 vs p8
Round 5= r1 vs r2
Round 6= r3 vs r4
Round 7= r5 vs r6 (1st & 2nd place established)
Roung 8= l5 vs l6 (3rd & 4th place established)
Round 9= l1 vs l2
Round 10= l3 vs l4
Round 11= r9 vs r10 (5th & 6th place established)
Round 12= l9 vs l10 (7th & 8th place established)

If a brick weighs a pound and half a brick, how much does a brick-and-a-half weigh?
Click to expand...
1x=1+.5x
1x-.5x=1=.5x
x=2
Therfore
1.5x=3

And no I didn't check any of my figures....although I should probably use my stats training :)


Edit: I guess dan beat me to it.
 
EDIT2: I didn't check dfc05's figures closely, but he's more correct than I am on the second question.
 
I wouldn't call that a maths puzzel, it's more like a mathematical process of elimination using binary.

I thought you had to use questions relating to the characteristics of the number you're trying to find in denary form.
 
well another problem is that you can't just say "convert the number to binary" You have to give them the instructions to do so. I'll give you the answer that I came up with:

1. Is the number odd?
2. Is the integer value of the number divided by 2 odd?
3. Is the integer value of the number divided by 4 odd?
4. Is the integer value of the number divided by 8 odd?
5. Is the integer value of the number divided by 16 odd?
6. Is the integer value of the number divided by 32 odd?
7. Is the integer value of the number divided by 64 odd?
8. Is the integer value of the number divided by 128 odd?
9. Is the integer value of the number divided by 256 odd?
10. Is the integer value of the number divided by 512 odd?
11. Is the integer value of the number divided by 1024 odd?
12. Is the sum of the integer values of the number divided by 1, 4, 16, 64, 128, 256, and 1024 odd?
13. Is the sum of the integer values of the number divided by 1, 8, 32, 64, 128, 512, and 1024 odd?
14. Is the sum of the integer values of the number divided by 2, 8, 16, 64, 256, 512, and 1024 odd?
15. Is the sum of the integer values of the number divided by 2, 4, 32, 128, 256, 512, and 1024 odd?

1-11 carry binary information written from right to left (yes is 1, no is 0).
The integer part basically means ignore anything to the right of the decimal place.

12-15 are parity checks
They sum up the indicated bitvalues and then will make the result even by being either 1 or 0 (yes or no). Now when we go back and check those parity bits against their corresponding data bits you will notice a discrepancy when you get an error. Because each parity bit detects an error in a block of data bits, you can isolate the specific error by seeing which parity bits the error shows up in.

Error Location/ Parity Bit
1/ 12, 13
2/ 14, 15
3/ 12, 15
4/ 13, 14
5/ 12, 14
6/ 13, 15
7/ 12, 13, 14
8/ 12, 13, 15
9/ 12, 14, 15
10/ 13, 14, 15
11/ 12, 13, 14, 15
12/ 12
13/ 13
14/ 14
15/ 15
 
Oh...just noticed Dan was the one who started the thread.

:)

Not many people here will have math degrees...and most of the math graduates I graduated with would probably not be able to figure that out. Now I'm wondering if he asked the question as an attempt to prove forum members were incompitent, to display his infinite wisdom, or just be weird.
 
Well I guess it was more of a show off. The problem came from a series of math puzzles by some math prof. Our teacher assigned them to us as some sort of final project.
 
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