The worlds first riddle!

DIGIT PROBLEM
Coincidentally, the number of non-paren solutions to the 30 problem is 30:

1+2+3+4+5*6+7-8-9
1+2+3+4-5+6*7-8-9
1+2+3+4*5-6-7+8+9
1+2*3+4+5+6+7-8+9
1+2*3-4*5+6*7-8+9
1+2*3*4+5+6-7-8+9
1+2*3*4+5-6+7+8-9
1+2/3-4-5+6*7*8/9
1+2/3/4*5*6+7+8+9
1-2+3+4+5*6-7-8+9
1-2+3*4+5+6+7-8+9
1-2-3+4+5+6*7-8-9
1-2-3+4*5+6+7-8+9
1-2-3-4+5*6+7-8+9
1-2-3-4-5+6*7-8+9
1*2+3+4+5+6-7+8+9
1*2-3-4+5+6+7+8+9
1*2-3-4+5-6*7+8*9
1*2-3-4-5*6+7*8+9
1*2-3-4-5*6-7+8*9
1*2-3*4+5*6-7+8+9
1*2*3+4+5*6+7-8-9
1*2*3+4-5+6*7-8-9
1*2*3+4*5-6-7+8+9
1*2*3*4+5*6-7-8-9
1*2*3*4*5/6-7+8+9
1/2*3*4+5*6-7-8+9
1/2*3*4*5+6-7-8+9
1/2*3*4*5-6+7+8-9
1/2/3*4*5*6-7+8+9

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NEW DIGIT PROBLEM
131 is the smallest positive integer with only one solution that doesn't require parens.

1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 ? 8 ? 9 = 131

Replace the question marks with operations (+, -, *, /).
Precedence rules:
* and / are evaluated first (left to right)
+ and - are evaluated last (left to right)

Other integers less than 200 with unique solutions:
152, 153, 156, 158, 160, 164, 166, 167, 169, 175, 176, 177, 178, 189, 197

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It is surprising how many solutions there are, and that's not allowing digit concatenation I assume.

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markr wrote:

OK guys, I've done a bit of research to find out just what we're dealing with here.

Yugi is not a 14-year old girl from Indiana. He's a character in a game. This is from some site that has a parent's guide to Yu-Gi-Oh.

Check out this comment from the site:
The Japanese versions of Yu-Gi-Oh (at least the early ones) were a bit racy and included some concepts that might not be appropriate for some children. U.S. versions have been sanitized by the licensees to remove any inappropriate content. We do not think many parents would have trouble with the current content, but encourage you to make your own informed decision.

Clearly, we're dealing with an early Japanese version (see language from Yugi in previous posts).

For those who are interested in a bit of Yugi's background, here's the story line:
Yugi's grandfather runs a game store. One day Yugi learns of a very, very old Egyptian card fighting game called "Duel Monsters", In the game, different creatures battle in magical duels. Besides just battling themselves, they also have to contend with a changing battlefield that holds many dangers. When the battle is over, only one beast rises to be called Yu-Gi-Oh! (King of Games).
Along the way Yugi solves the secret of an ancient puzzle (Millennium Puzzle) and releases the spirit of an Egyptian King, Yami Yugi. With help from the King's spirit, Yugi sets out to become King of the Duelists. In the past Yugi led a mild mannered, quiet life. Now his friend, Joey, helps him deal with his much greater responsibilities. With Joey's help and the strong energy he receives from Yami Yugi, he is ready to face the challenges ahead.

One of his opponents, Seto Kaiba is very rich. Kaiba becomes very upset when Yugi beats him. He now wants to beat Yugi however possible. Kaiba and his younger brother Mokuba were raised in an orphanage. Kaiba promised to support and take care of Mokuba and appears to be easily able to do so as head of his own international hi-tech company, Kaiba Corp. But Kaiba focuses all his energy and business skills on Duel Monsters. He challenges Yugi.

Yea... duh... see my name on this site is "yugirules14" simply stating that I am a fan of said cartoon character. Oh... and my nick name in real life is "YUGI" so talk what you knowq bitch not what you think. Dumb ass.

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just out of curiosity markr....whats the smellest number that cant be made from those digits?

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Mark:

It won't run without winding Cool

phone - ph = one Cool

multiple life sentences Cool

Very good, I have:
.....not long at all if it needs winding

.....Take off the letters "st" from "stone" and "one" is left

....."Go to prison for life."

(But don't tell Stu)

Mark:
CRUISE SHIP
You will pass 15 ships in each direction. Cool

Stu wrote, "I'll never tell how many feet I wanted, and there is not information to figure it out from the riddle because you only supplied 1 equation:"

Mark wrote, " But not if 0 <= inches < 12."

ROPE
wanted: 110" Cool

got: 33"

These questions always appear to be missing a vital piece of information, that is why they are in the riddles forum and are far more mind stretching than straight math.

"DIGIT PROBLEM
Coincidentally, the number of non-paren solutions to the 30 problem is 30:"

Now that is amazing!!!

To play the lottery in A2K, 6 numbers are selected out of 36. In how many ways can 6 numbers be chosen from 36 Question

(Order is NOT important, meaning 1 2 3 4 5 6 is the SAME selection as 2 3 4 5 6 1, etc.)

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Quote:

These questions always appear to be missing a vital piece of information, that is why they are in the riddles forum and are far more mind stretching than straight math.

Ok, fair enough.

Quote:

To play the lottery in A2K, 6 numbers are selected out of 36. In how many ways can 6 numbers be chosen from 36 Question

(Order is NOT important, meaning 1 2 3 4 5 6 is the SAME selection as 2 3 4 5 6 1, etc.)

How is this not just permutation?

36 P 6... = 36!/(36-6)!

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[size=8]LOTTERY
It's combinations, not permutations.
C(36,6) = 36! / (30! * 6!) = 1,947,792[/size]

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Here's a very simple problem (unless I'm wrong). Same rules as the last problem I posted (digits 1-9 in order, 8 operators from the set +, -, *, /).

What's the largest number you obtain?

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stuh505: Yes, concatenation was excluded.

thoh13: Below 200, these are the numbers without solutions: 154, 155, 165, 168, 171, 182.

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markr wrote:

[size=8]LOTTERY
It's combinations, not permutations.
C(36,6) = 36! / (30! * 6!) = 1,947,792[/size]

Read the problem wrong Razz

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[size=7]largest #:

if u have to insert an operation between every digit:

1*2*3*4*5*6*7*8*9 = 9! = 362880

if u dont have to insert an operation between every digit:

12345678*9 = 111111102

if u dont have to insert an operation at all:

123456789[/size]

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thoh13: One operation between each pair of digits is required. Your answer is not correct, but it's the one I expected (which is why I posed the problem).

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ok got it

[size=7]1+2*3*4*5*6*7*8*9 = 9! = 362881
[/size]

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yep!

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Mark:

LOTTERY
It's combinations, not permutations.
C(36,6) = 36! / (30! * 6!) = 1,947,792 Cool

!

It sure is.
The formula for determining how many ways 'k' things can be selected from 'n' things is:
n! / (k! * (n-k)!), where the ! symbol means factorial. A factorial is determined by getting an integer, then multiplying it by that integer minus 1, and so on, until 1 is reached. For example 5! = 5 * 4 * 3 * 2 * 1, which equals 120.

Returning to our problem, n = 36 and k = 6. Putting these into the formula we have: 36!/(6! * (36 - 6)!) which equals 36!/(6! * 30!)
which simplifies to: 36 * 35 * 34 * 33 * 32 * 31 / 6! which equals: 1,947,792
So there are 1,947,792 ways to pick 6 numbers from 36. If in doubt ask Mark.

Returning to our lottery; when 3 of 6 numbers match the winning number, you win a free ticket. Very Happy

What is the probability of getting 3 out of the 6 winning numbers Question

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[size=8]In general, the probability of selecting exactly K of R items chosen from a set of N items is:

C(R,K) * C(N-R,R-K) / C(N,R)

Note that when K=R, the numerator is 1, and we have the probability for the original problem.

In this case, the answer is:
C(6,3) * C(30,3) / C(36,6) = 0.0417[/size]

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Did anyone come up with an answer for this one, it may have got lost from a few pages back?

Mrs Stormy grumbles about bad time-keeping trains like everybody else. On one particular morning she was justified, though. The train left on time for the one hour journey and it arrived 5 minutes late. However, Mrs Stormy's watch showed it to be 3 minutes early, so she adjusted her watch by putting it forward 3 minutes. Her watch kept time during the day, and on the return journey in the evening the train started on time, according to her watch, and arrived on time, according to the station clock. If the train traveled 25 percent faster on the return journey than it did on the morning journey.

So I ask; was the station clock fast or slow, and by how much Question

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To be honest, I'm confused by it. Aren't there two stations and clocks involved (one at each end)? Left on time relative to what (real time, station time, watch time)? Same for arrived five minutes late.

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im confuzzled...

too many questions...
1. r the clocks at the end stations synchronized and running at same pace?
2. the amount of time between arriving and leaving doesn't matter?
3. so is the running slow/fast a rate? x time/hr?
4. is it implied the return trip also should be 1hr long despite the faster speed?
5. or is it 80% the time of the 1st leg?

blah

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The worlds first riddle!

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