Math riddles :D

[Ryme]

For x != y ? In which case I'll need an order relation.

x =!y

A hint: if you think of the physical situation there obviously is a dimmest point; as you approach a candle it gets brighter and brighter, so you move away to get less light but of course if you move too far you get more light from the second candle. For instance, if the candles were equally bright, the dimmest spot would be halfway between them, but they are not equally bright so the answer depends on x and y. How exactly?

Whilst you ponder that, it made me recall another toughy -

In my school days I remember that when my class was lined up by height, I was right in the middle. Must to my chagrin, the girl that I secretly fancied was taller than me and stood 15th in line, from shortest to tallest, while the boy I knew she secretly fancied was in 26th spot.
I also noticed this was enough information to infer how many of us were in the class. How many were there?

[Traman]

Pinochio says "I am lieing"

(Just a paradox, no math really involved)

[Spitfyre]

In my school days I remember that when my class was lined up by height, I was right in the middle. Must to my chagrin, the girl that I secretly fancied was taller than me and stood 15th in line, from shortest to tallest, while the boy I knew she secretly fancied was in 26th spot.
I also noticed this was enough information to infer how many of us were in the class. How many were there?

theres 27 kids, i'm currently thinking about the second one but having some trouble with it.
to poster above me this is a riddle thread and a paradox by its very nature has no solution, it's amusing but kind of irrelevant.

[Darsey]

Its because the big shape is actually a square and not a triangle

[Spitfyre]

Two candles of luminosity (meaning brightness) x and y are separated by a distance z. Find the position of the dimmest spot between the candles.

Not really a puzzle, but it's stumped my old university's maths undergraduates for years now.

I'm assuming that when you say the position its only on the line at the same vertical hight of the wick of both candles. Secondly in order to correctly define the position I will be defining it on an xy-plane where the wick of candle X (the candle with luminosity x) is at (0,0) and the wick of candle Y (the candle with luminosity y) is at (z,0). I believe that the position of the dimmest point should be at (zx/2y,0).

[v2prwsmb45yhuq3wj23vpjk]

women = time + money (women require time and money)
time = money (but time is money, friend!)
women = money2 (therefore meaning that girls are money + money .. or money squared!)
money = √all evil (but as we all know, money is the root of all evil!!)
women = evil (and if we cancel out the squares and roots we find that... GASP!)

:3 cookie now?

money = 2

Because that's the only way, assuming time = money, that money + time can equal money^2

[Spitfyre]

women = time + money (women require time and money)
time = money (but time is money, friend!)
women = money2 (therefore meaning that girls are money + money .. or money squared!)
money = √all evil (but as we all know, money is the root of all evil!!)
women = evil (and if we cancel out the squares and roots we find that... GASP!)

:3 cookie now?

Correction

time=money (time is money)
women= time+money (women require time and money)
women=2(money) (substitution)
money=sqrroot(evil) (money is the root of all evil)
women=2sqrroot(evil) (second substitution)

[Deleo]

1. As I was going to the fair, I saw a man with golden hair. He had 3 sons each with another one. How many people were going to the fair?

2. There is a pink single story house and everything in it is pink. The doors are pink, the windows are pink and the TV is pink. What color are the stairs?


If you are looking for serious ones:

I- A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:

There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?

II- Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.

III- This one brings back some good memories: You have 100 bags with coins. All coins are equal, but one bag contains coins that are all 1 gram heavier than the coins in all the other bags. If you have a scale (not a balance) that you can use only once, can you find the bag with the bad coins?

[Pitchkart]

Here's one I got from a test my Geometry/Algebra2 teacher gave me back in school.

[BatteredRose]

Nah - a penny and a 1972 dime with a Roosevelt imperfection, today worth exactly 29 cents.

Roffles. The janitor ftw.

[Makamoka]

Pinochio says "I am lieing"

(Just a paradox, no math really involved)

More like, What happens if Pinochio says "Now my nose is going to grow." Because, if the nose dont grow he was lying so the nose will grow but then he was telling the truth.

[Eneth]

You want to build a tent without a bottom with two rectangular sides and two gables in the shape of isosceles triangles with the base towards the ground. Determine the height of the tent if the volume is V and we want to use as little textile as possible.

[kailtas]

More like, What happens if Pinochio says "Now my nose is going to grow." Because, if the nose dont grow he was lying so the nose will grow but then he was telling the truth.

Pinocio does not now know the answer to the question. So the statement cannot be a lie.

[Samthemann]

1. As I was going to the fair, I saw a man with golden hair. He had 3 sons each with another one. How many people were going to the fair?

2. There is a pink single story house and everything in it is pink. The doors are pink, the windows are pink and the TV is pink. What color are the stairs?


If you are looking for serious ones:

I- A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:

There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?

II- Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.

III- This one brings back some good memories: You have 100 bags with coins. All coins are equal, but one bag contains coins that are all 1 gram heavier than the coins in all the other bags. If you have a scale (not a balance) that you can use only once, can you find the bag with the bad coins?

I got the easy ones, but the last 2, phew.

1. Only you're going.
2. There aren't any stairs in the single story house, though I'd imagine if they had any at all they'd be pink.

I. You're a sadist, nty.
II. Wasted my time going backwards through the possibilities as any combination of the 9 would result in a number divisible by 9, but knew it had to start with a 1, and then followed by an even number I built the number up from that premise. I got upto 2 7 digit numbers but the 8th always broke me. Quite possible I didn't do the math correctly since I'm tired.

[crowpala]

You can prove without breaking any math rules that 2 = 5. I wish I had still that paper somewhere.
Basicly you start of with 2=2 and then go from there till you end up with 2=5

This is a really basic question, but what is the mathematic value of: i
No I'm not talking about the i=1 from way back but from higher math you use i in different scenarios

[Verdris]

You can prove without breaking any math rules that 2 = 5. I wish I had still that paper somewhere.
Basicly you start of with 2=2 and then go from there till you end up with 2=5

This is a really basic question, but what is the mathematic value of: i
No I'm not talking about the i=1 from way back but from higher math you use i in different scenarios

i=sqrt(-1). It doesn't have a "value", it has a "definition".

There are certain expressions, however, that produce numerical values:

i^2=-1
e^(i*pi)+1=0
i^i=0.207879576 (principal value for the k=0 branch)
i!=gamma(1+i)=.5-.15i (approx)

etc.

EDIT: you also see i being used to denote the unit vector in the positive-x direction in cartesian space.

---------- Post added 2011-05-30 at 08:04 AM ----------

If you are looking for serious ones:

I- A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:

There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?

II- Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.

III- This one brings back some good memories: You have 100 bags with coins. All coins are equal, but one bag contains coins that are all 1 gram heavier than the coins in all the other bags. If you have a scale (not a balance) that you can use only once, can you find the bag with the bad coins?

1. Only the first locker remains open.

2. 123456789

3. Probably not. You don't tell us if all the bags have an equal number of coins or not. Thus one bag could contain the n+1 gram coins and another bag could contain some amount of n gram coins that is equal in weight to the bag of n+1 gram coins.

Assuming, however, that all bags contain equal amounts of coin, put 50 bags on one side and 50 bags on the other. One side will sag due to extra weight. Then remove bags, one at a time, from each side until the scale is equal - therefore you have just removed the errant bag. I define this as "using the scale once". Compare the heft of the last two bags removed to determine your target bag.

EDIT: A scale, not a balance? Then I assume it has some sort of display. Add bags, one at a time, keeping track of the partial sums of your obviously arithmetic sequence. Again, I define this as "using the scale only once". As soon as you add a term that deviates from your common difference, you've found your bag.

[Deleo]

I- A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony:

There are one thousand lockers and one thousand students in the school. The principal asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?

II- Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.

III- This one brings back some good memories: You have 100 bags with coins. All coins are equal, but one bag contains coins that are all 1 gram heavier than the coins in all the other bags. If you have a scale (not a balance) that you can use only once, can you find the bag with the bad coins?

1- numbers are: 1 - 4 - 9 - 16 - 25 - ... - 31^2 (Square numbers that are smaller than 1000 get changed odd number of times the rest get changed even number of times)

2- 381654729 (no freaking way i explain how you get to it. way too difficult with my bad english =). But simply assume your number is "abcdefghi" then come up with relations between those using the given conditions. example: b is even, e is either 0 or 5, since we don't have 0 so it is 5)

3- pick one coin from bag 1, two from bag 2, three from bag 3, .... , and one hundred coins from bag 100. when weighting, the extra weight you get is the number of the bag containing those heavy coins.

[mascaron]

II- Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.

381654729

Took a little guessing on my part.. didn't have an easy way to find divisibility by 7.

How I went about it (for anyone curious):
I didn't do this, but to explain positioning a little better.. we'll label the digits ABCDEFGHI
1. When dividing by 5, end number has to be 0 or 5. No zero, so.. E = 5.
2. When dividing by even numbers, the end number must be even. So, digits B D F H had to be either 2, 4, 6, or 8. This left digits A C G I with 1, 3, 7, or 9.
3. When dividing by 8, the last three digits (FGH) must be divisible by 8. Since we know the 100's digit is even (step 2), we can simply look at the last 2 digits (all even 100's are divisible by 8). Quick check through multiples of 8 beginning with an odd number leaves only 2 and 6 as possible options for digit H.
4. When dividing by 4, the last 2 digits (CD) must be divisible by 4. Quick check through multiples of 4 beginning with an odd number leaves only 2 and 6 as possible options for D. Since 2 and 6 are the only options for D and H... that means if 2 is in D, 6 is in H or vice versa. We can thereby eliminate those 2 as options for anywhere else. Combine this with knowledge from step 2, and we find B and F can only be 4 or 8.
5. To be divisible by 3, the digits must add up to a number divisible by 3 (i.e. 1+2+9 = 12. 12 is divisible by 3, so 129 is divisible by 3). Combined with knowledge from step 4, this leaves a handful of options left for this combo (I just went in order 1+4+3, not a multiple. 1+4+7, check. 1+4+9, not a multiple. 3+4+1, not a multiple. etc). We're left with 147, 183, 189, 387 and their reversals.
6. To be divisible by 6, the digits must add up to a number divisible by 3 and 2. Digit F is already even, so we don't need to worry about 2. Just something to keep in mind.
7. To be divisible by 9, the digits must add up to a number divisible by 9. Since we use all the digits here, any number can go in digit I. Also, all are divisible by 1, so any number can go in digit A.

What it boils down to:
A = 1,3,7,9
B = 4, 8
C = 1,3,7,9
D = 2,6
E = 5
F = 4,8
G = 1,3,7,9
H = 2,6
I = 1,3,7,9

Annnd now I started doing a little guessing with the rules already laid out. I just picked 2 numbers for A and I out of the list, followed divisibility rules laid out above, and I came up with a number that I knew was followed the rules perfectly except for dividing by 7.. so I checked ABCDEFG divided by 7. If it didn't work, I picked another 2 numbers for A and I. So.. for example.. the solution: Randomly picked 3 for A, 9 for I. Step 5 shows only 8 as a possibility for B, which then makes F=4. I then checked step 6. Add up all the digits I knew (A+B+E+F; 3+8+5+4) = 20. So, now I need 2 numbers from CD (1 or 7; 2 or 6) that turn 20 into a number divisible by 3. Only C=1 and D=6 make the sum divisible by 3. This forces G = 7, H = 2. And there is the number :P

Sorry for the long explanation. Harder to explain really than to actually do it.

[Pisholina]

Correction

time=money (time is money)
women= time+money (women require time and money)
women=2(money) (substitution)
money=sqrroot(evil) (money is the root of all evil)
women=2sqrroot(evil) (second substitution)

It's actually Women = time * money because you need time AND money, not time OR money. If you have time, but not money, then you are proper fucked IRL.

[Spitfyre]

It's actually Women = time * money because you need time AND money, not time OR money. If you have time, but not money, then you are proper fucked IRL.

When a question says: "you have 3 apples and 5 oranges, how many fruits do you have?"
the math you would perform is 3+5=8 not 3x8=24.
"and" is a clear marker of addition.