The worlds first riddle!

He's 17.

[size=8]two apples
gopher trap
you'd need a camera

SEQUENCE
O E F T S T F T S T N S S E E F N T F T
Thanks for the hint and good memory.[/size]

12683
02994 +
======
15677

[size=8]each year has a new year's day and a christmas day
old 100 > new 1
silence?[/size]

Of the members of four athletic teams at West Point,
41 are on the football team,
15 are on the basketball team,
21 are on the baseball team,
and 32 are on the wrestling team.

No wrestler is also a basketball player
9 play football and baseball.
5 wrestle and play baseball.
12 wrestle and play football.
6 play basketball and football.
7 play basketball and baseball.
3 play on all the teams except basketball.
4 play on all the teams except wrestling.

How many individuals are involved in these four sports

77 players

breakdown:
18 just wrestling
7 just baseball
6 just basketball
21 just fooball
9 just wrestling and football
2 just wrestling and baseball
3 just baseball and basketball
2 just basketball and football
2 just football and baseball
3 wrestling baseball and football
4 baseball basketball and football

hooray for venn diagrams

Whim:

12683
02994 +
======
15677



Absolutely correct.



Mark:

each year has a new year's day and a Christmas day
old 100 > new 1
silence



Thoh:

77 players

breakdown:
18 just wrestling
7 just baseball
6 just basketball
21 just fooball
9 just wrestling and football
2 just wrestling and baseball
3 just baseball and basketball
2 just basketball and football
2 just football and baseball
3 wrestling baseball and football
4 baseball basketball and football

hooray for venn diagrams



Hooray indeed. To solve this problem, a Venn diagram (or Euler circles) would be useful. Draw three circles which intersect each other (there should be 7 regions formed) and then draw one more circle which intersects only two of those circles (there should now be 11 regions). Each circle represents one of the sports. The 11 regions correspond to the 11 combinations of sports that could be played.

Begin by finding a region where three circles intersect. This could represent the number of athletes who play basketball, football, and baseball. Put the number 4 in that region.

Now find the other region where three circles intersect. This represents the athletes who are on the wrestling, baseball, and football teams. Put a 3 in that region.

Between those two regions is a region representing the players on just the football and baseball teams. Since there are 9 who play football and baseball, you should put 2 in this region (9 minus 4 minus 3) because 2 play only football and baseball.

Continue with this logic and you should get the same result as Thoh.




In an effort to find the shortest, most difficult question that will remain unanswered for 1000 years, I present:


You are in a dark room, find two unequal numbers, each of which is the square of the other

[size=8]MUTUAL SQUARES
-1/2 + i*sqrt(3)/2
-1/2 - i*sqrt(3)/2

Solve (a + i*b)^4 = (a + i*b) for a and b.
a = a^4 - 6(a^2)(b^2) + b^4
b = 4(a^3)b - 4a(b^3)

Cool problem![/size]

Mark:


MUTUAL SQUARES
-1/2 + i*sqrt(3)/2
-1/2 - i*sqrt(3)/2


(-1 + i sqrt(3)) / 2
and
(-1 - i sqrt(3)) / 2
where sqrt(3) represents the square root of 3.

To solve the problem, call the two numbers x and y.
Then y = x^2 and x = y^2.
Substituting, you get x = x^4.
So, x^4 - x = 0
Then x (x^3 - 1) = 0 so
x (x - 1) (x^2 + x + 1) = 0

Using the quadratic formula.
x = 0
x = 1
x = (-1 + i sqrt(3)) / 2
x = (-1 - i sqrt(3)) / 2

You can eliminate the first two solutions because x = y.
The other two solutions x = (-1 + i sqrt(3)) / 2 and
y = (-1 - i sqrt(3)) / 2 satisfy the given conditions.

Needless to say my 1000 year problem lasted 3:59min.


Mark wrote,
"Solve (a + i*b)^4 = (a + i*b) for a and b.
a = a^4 - 6(a^2)(b^2) + b^4
b = 4(a^3)b - 4a(b^3)
Cool problem!"

Is this one of your own? Are you inviting comments?



In a an effort to extend on the above time I give you:

Washington State High trains its students to be trustworthy, respectable
Citizens; however, it takes some time for them to get to that point.
In fact, Mr. Rottweiler, who has been at the institution as long as anyone can remember, has noticed the following tendencies among the student body:

Freshmen always lie.
Reflecting their standing as "second-class" citizens, sophomores will always lie unless they are the second ones to speak in a conversation.
Juniors only lie if they are the third ones to speak or if their sentence begins with a J.
Seniors at Washington never lie.

To test the validity of these observations, Mr. R recently brought four randomly chosen students into his computer lab for a talk. Their names were Fred, Sophie, Julius, and Selena. As it happened, no two of them were in the same graduating class. Mr. R asked each student to tell which class another student belonged. They answered as follows:

Fred: Julius is a sophomore.
Sophie: Selena is a senior.
Julius: Sophie is a freshman.
Selena: Fred is a junior.

Mr. R realized that with this information, he could determine the grade of each student. Can you

"Is this one of your own? Are you inviting comments?"

No, I solved the previous problem the hard way. That was my method.

So, was the conversation:
1. Mr. R: Tell which class another student belongs to.
2. Fred: Julius is a sophomore.
3. Sophie: Selena is a senior.
4. Julius: Sophie is a freshman.
5. Selena: Fred is a junior.

or was it:
1. Mr. R: Fred, tell which class another student belongs to.
2. Fred: Julius is a sophomore.
3. Mr. R: Sophie, tell which class another student belongs to.
4. Sophie: Selena is a senior.
5. Mr. R: Julius, tell which class another student belongs to.
6. Julius: Sophie is a freshman.
7. Mr. R: Selena, tell which class another student belongs to.
8. Selena: Fred is a junior.

or was it:
1. Mr. R: Fred, tell which class another student belongs to.
2. Fred: Julius is a sophomore.
1. Mr. R: Sophie, tell which class another student belongs to.
2. Sophie: Selena is a senior.
1. Mr. R: Julius, tell which class another student belongs to.
2. Julius: Sophie is a freshman.
1. Mr. R: Selena, tell which class another student belongs to.
2. Selena: Fred is a junior.

If you can spare a minute to illuminate a non-mathematician, are you saying:
"b = 4(a^3)b - 4a(b^3)" Therefore: b=0?



Washington High:

Freshmen always lie.
Reflecting their standing as "second-class" citizens, sophomores will always lie unless they are the second ones to speak in a conversation.
Juniors only lie if they are the third ones to speak or if their sentence begins with a J.
Seniors at Washington never lie.

Therefore the list cannot be correct as it stands.

To solve this logic problem, try the process of elimination.
Since seniors always tell the truth, assume one of the four Washington students is a senior, and then that student 's statement must also be true. Continue until you find a contradiction or until it works out.

Clue: Selena is the senior.

Full explanation on the morrow

b=0 (a=1) is one solution, but it doesn't satisfy the unequal requirement.

I gave two equations with two unknowns. You have to solve them together, not separately.

------------

I wasn't looking for a clue. I just needed to know how the conversation went since order of speaking is important.

If Selena is a senior, then the conversation went like this (apparently, Mr. R's statement isn't part of the conversation):
1. Fred: Julius is a sophomore. (false)
2. Sophie: Selena is a senior. (true)
3. Julius: Sophie is a freshman. (lie)
4. Selena: Fred is a junior. (true)

Fred: junior
Sophie: sophomore
Julius: freshman
Selena: senior

Mark:
Fred: junior
Sophie: sophomore
Julius: freshman
Selena: senior


Assume Fred is the senior. Then it follows that Julius is a sophomore. Since sophomores lie unless they speak second, it follows that Sophie is not a freshman; therefore, she must be the junior. Since Sophie is the second one to speak and she is a junior, she must be telling the truth which means Selena is the senior and that contradicts our assumption that Fred is the senior.

Similarly, you can reason that Sophie and Julius cannot be seniors because they lead to contradictions.

Assume Selena is the senior. Then Fred must be the junior. Since Fred's sentence starts with a 'J', and since he is the junior, he must be lying, so Julius is NOT a sophomore. Therefore, Julius is the freshman and Sophie must be the Sophomore. If Julius is the freshman, he must be lying, which fits since Sophie is not the freshman. Since Sophie is the sophomore, she is telling the truth because she speaks second, and this fits since it was our assumption -- Selena is a senior


Cool problem: Message received, but don't hold your breath.




I could use a little help here; I have lost the phone number of a chick I met last night. All I can remember is;

it is a ten digit telephone number (including area code) in which each digit, 0 through 9, appears once.

The first digit on the left is divisible by 1,
the first two digits form a number divisible by 2,
the first three digits form a number divisible by 3,
the first four digits form a number divisible by 4,
and so on, until we get to the whole 10-digit number which is divisible by 10.

What is the phone number

(381) 654-7290
hooray for divisibility rules

ill try explaining....

the 10th digit has to be 0, so the 5th digit can only be 5

4th digit can be 2 or 6 (because 3rd and 4th digits must form a number divisible by 4), and 6th can be 8 or 4

6th 7th & 8th digit must form a number divisible by 8 so the number has a form of 4-- or 8-- and the middle number cant be even. the only possibility for both ends in 2 so 8th digit is 2, so 4th must be 6 and 6th must be 4, and 2nd must be 8 so we have -8-654-2-0. only numbers left are 1, 3, 7 and 9

4-2 must be divisible by 8 so the 7h number can only be 3 or 7

1st 3 dgits must be divisible by 3 and since the 2nd digit is 8, the other two can only be 1&9, 1&3, or 7&9. here i used trial and error to arrive at the 7 digit number divisible by 7. after that, the only digit left was 9 to go in the 9th place

I gave her a call and she told me to get lost, because she was going out with a guy called Thoh who had called earlier.




As I was eating breakfast this morning, I thought: Canadian Wild Strawberries (CWS) are tiny but tasty.

A and B each have a jar containing 400 CWS:

They decide to have a CWS eating race; A wins, swallowing his last CWS when B still has 23 left. It took A 13.2 seconds; burp!

Next, B takes on C, each with a jar containing 261 CWS; B wins, C left with 117 CWS (C has a bad toothache).

Jack: well, John, A took on C next
John: ya ya I'm sure he did
Jack: each had a jar containing N CWS's
John: oh boy
Jack: wanna try figure out what N is?
John: not really
Jack: here's a hint: in the 200 to 500 range, and they both swallowed at same speed as in their 1st race
John: oh ya? (comes back with a printout)
Jack: A beat C by an integral amount
John: ya; I figured as much; need another clue
Jack: the sum of digits of the number of CWS that C had left when A finished is equal to this number here ...and John knew.

What is the value of N

250 CWS

rate of A from race 1 is 1000/33 CWS/sec and rate of C from race 2 is 520/33 CWS/sec

the time it takes A to eat all the CWS is 33N/1000
in that time C eats (33N/1000)(520/33) CWS, which must be an integer [this simplifies to .52N]

so the number of CWS he has left is 1-.52N or .48 CWS, which is also an integer.

this happens in increments of 25 between 200 and 500 (200, 225, 250 etc.)

the sum of the answers varies between 3 and 15, but 3 appears only once (when N=250, .48N=120 so sum is 3) while the others appear several times so knowing the sum won't give a unique answer unless the sum is 3

Thoh:

(381) 654-7290
hooray for divisibility rules

Good clear explanation, shorter than the one I have.


250 CWS


Nice!




Three pirates -- Abe, Bob, and Cal -- went to a cave one night. Each carried a bag of stolen gold coins.

During the night, Abe woke up, stole half of Bob's coins and half of Cal's, added them to his own bag, and went back to sleep. At this point,Abe had 300 coins.

A little later, Bob woke up and did the same thing to Abe and Cal; and an hour after that, Cal woke up and did the same thing to Abe and Bob.

When each thief checked his bag in the morning, he discovered that he had the same number of coins he had brought to the cave.

How many did each pirate bring

Eureka! At last, after months in the wilderness, I may have the answer to the rain in the jars that Mark posted months and months ago. I think the answer is: 132 minutes

Can you remind us of the question?