muster

wrestle

Humour is correct!

That was much easier than I expected. I guess my hint gave it away.

it was easier, eh? You know how canada got it's name, right? It used to be called Commonwealth Northern Dominion...however, they realized that that was too long of a name, so, they abbreviated it C.N.D...but whenever it was spoken it became C-eh-N-eh-D-eh...

I'm just playing around...

colour: xxx

favour: xxx

You guys are definitely on the right track.

humour?

This Masterword might be a little harder for my friends south of the border. Instructions can be found here.

I'm thinking of a 6-letter word.

word-eh? Just kidding...hmmm...well, i'm not from canada, so, we'll see how it goes:

favour?

I've got 2.28 at least, maybe more:

BLADDER 16 answers/7 letters= 2.28

1) ALE

2) RED

3) LADDER

4) LAD

5) ADD

6) READ

7) LARD

8) BLED

9) BAD

10) DAB

11) DEAD

12) DAD

13) BALD

14) BED

15) LAB

16) BLADE

Well, the rules state that you must use ALL of the letters in the word in your anagrams...so, that example doesn't work.

4 and 13

Correct...It's a great riddle...I'm assuming you figured out how to get the answer and are leaving that up for everyone else to figure out

And yes, that is the only answer to the puzzle...

Sorry, my mistake. I didn't pay due attention to the statement of the problem. I just saw the names P and S, and assumed this problem was the same. (The other puzzle had two two-digit numbers.) So scratch my previous posts.

However, I don't see, why product couldn't be a 4th power of a prime. Say, 81 could be 9*9, or it could be 3*27.

can't be 9*9 since it says they are distinct numbers...so that means it could only be 3 * 27...and therefore he would know the answer just by that

As for this variation the answer could be:

The pair of numbers could be 21 and 10. Their product 210 and the sum 31.

Initially, P has two choices: 15*14 and 21*10. However, if the numbers were 15 and 14, then S would guess after P didn't, since the sum of 29 allows only one such pair with non-unique product.

To find whether this is the only possible solution, you'd still save a lot of time by writing a computer program. I suspect, there are other solutions.

As far as puzzles go, this variation is more fair than the one posted before. Because solving the other one without a computer program was not a feasible task. I would change the question of the puzzle "What could be the numbers?"

After looking at the other problem posted, I'm not sure if this one is actually any easier...in fact I could solve the other one easier than this one (however that could just be the way I think)...

Second, this is not at all what I got for my answer...I'm confused as to how you made the statement that P has two choices and why those were the only 2 choices P had. Unless I have misstated the problem, you over-simplified it...maybe if you walk through how you got the answer...because from what I have found there is only one possible answer to this problem, and it is not the one that you stated...

Initially, P says that he doesn't know the two numbers, which means that his product cannot be the product of 2 distinct factors of the number he was told whose sum is less than 100 (so, they can't be 2 prime numbers...his number cannot be the cube of a prime...or the 4th power of a prime...)

I couldn't find this one on the site...so I figured I'd post it...if it has already been done, I apologize...

There was once a king who wanted to determine who the smartest person in the land was (funny how that seems to happen a lot in these forums). After a bunch of tests, it was down to two men (P and S), who claimed to be "perfect" logicians (meaning there was never any flaw in their logic).

The king then told them their final test:

"I am thinking of two different numbers greater than 1, whose sum is less than 100. I will tell P the product of my two numbers, and I will tell S the sum of the two numbers," the kind stated.

The king whispers in P's ear the product of his two numbers, and whispers to S the sum of the two numbers...The following is the conversation that the logicians had:

P: I don't know the two numbers...

S: I know you don't

P: Ah...now I know them...

S: So do I

Together, both men then stated the two numbers to the king, and he stared in disbelief...

What were the two numbers?

(NOTE: I actually wrote a program to solve this puzzle...however, I know it can be done without the use of any computer...)

Oh wow, lucky guess!

futon: o

laser: o

prime: oo

right: xxxx

sight

For a word of length n, there are at maximum n! possible anagrams. Since our score is (number of anagrams)/(number of letters) then the maximum score is:

n!/n = (n-1)!

that is true...in theory...but I don't think you'll find a 4 letter word with a score of 6, or a 5 letter word with a score of 24. So, I was just saying you could try to find the highest score that you can per word length.

For example, the answer already posted for 7-letter words would be tough to beat (score of 1.29).

And you can just keep coming up with new challenges with this game...like what's the longest word you can come up with with a score over 1 (I think that 7-letter is the highest I've seen so far)...etc..etc...

I'm glad you are enjoying this little puzzle. Like I said, it is really simple, but a good way to use up some time when you are bored (either in class or at work ).

Yes, to clarify, you must use ALL of the letters in the word in your anagrams, and the anagrams MUST be valid words (I would probably just use dictionary.com or something to check for valid words). That's what makes it more challenging...the longer the word you use, the more VALID anagrams you must find to give it a higher score.

EXAMPLE:

if your word is TIN, you can obviously find TIN, TNI, NTI, NIT, INT, and ITN as anagrams, but only TIN and NIT are also words (unless you're a computer programmer, then INT would probably count ). No acronyms allowed. So therefore, there are only 2 VALID anagrams of TIN...and that means the score would be 2/3 or 0.67...hope that makes sense

Besides the 1.75 score that I saw (which I still don't know about ), I have only gotten a score of 1.5 (woon also posted this answer).

Another thing you can do is figure out the highest score for each length of word.

IE. we know that for 2 letters, the highest score can ONLY be 1.0...

what's the highest score you can find for 3, 4 5, 6, etc letters?

In case you haven't already seen, my 1.5 score and the 1.75 score were done with 4 letter words...

I didn't come up with these, just stumble upon it.

anestri, asterin, eranist, nastier, ratines, resiant, restain, retains, retinas, retsina, sainter, stainer, starnie, stearin.

7 letters, 14 words

So that would 2.0

Using dictionary.com as my guide, I don't see these words (and I sure don't know them...but maybe they're just not on dictionary.com):

sainter

anestri

asterin

eranist

starnie

So, without those, that word is only 9 words, 7 letters for a score of 1.29

The plural of the greek letter Eta would be Etas, no?

That gives:

seat, sate, teas, eats, east, seta, etas = 7/4 = 1.75

Yeah, that's a good one. I had looked at that one, but

I'll be honest, I didn't think about seta or etas, but after looking them up, I guess they work...etas is still on the fence for me, but, I won't argue with it :c)

I found one 1.5

Spot

Pots

Stop

Opts

Post

Tops

6 / 4 = 1.5

That was my 1.5 as well...I'm just wondering if anyone can come up with any better ones. Nice job woon

So, when I was in school, I got bored in class a lot...so I had to occupy my time somehow, and brain teasers/puzzles were a great way to do so...So here's a word game that I came up with (hopefully I can describe it well):

Take a word of any length, and see how many anagrams you can make out of it that are also words. Then, to score that word, you take the number of valid anagrams and divide it by the number of letters in the word

EXAMPLE (2 letter):

WORD: on ANAGRAMS: on, no SCORE: 2 anagrams / 2 letters = 1.0

EXAMPLE (3 letter):

WORD: tin ANAGRAMS: tin, nit SCORE: 2 anagrams / 3 letters = 0.67

EXAMPLE (4 letter):

WORD: nuts ANAGRAMS: nuts, stun, tuns SCORE: 3 anagrams / 4 letters = 0.75

EXAMPLE (5 letter):

WORD: alter ANAGRAMS: alter, later, alert SCORE: 3 anagrams / 5 letters = 0.6

I know a word that has a score of 1.5 (not claiming it to be the highest, but the highest I found so far, and without the help of any anagram finders...this was just me sitting in class...so there probably are better scores)...what's the highest you can find? Obviously you can use as many letters as you want, but the more letters you use, the more VALID anagrams you must come up with to get a higher score.

Again, #3 has something in common with numbers 1 and 2...and the formula for this sequence can be expressed in a simple fraction:

X₁/X₂

Fibonacci Primes:

2, 3, 5, 13, 89, 233, 1597, 28657...

You are correct dwilly on the first sequence, and snowman is correct on the second sequence (assuming you know WHY it is the answer you gave)...

So, yes, the first sequence is fibonacci primes, the second is the index of those primes in the fibonacci sequence.

The third sequence isn't quite as tightly coupled to the other two...but still involves fibonacci...more math involved here...

I cannot tell a lie, although you may think I am with these sequences:

1) 2, 3, 5, 13, 89...

2) 3, 4, 5, 7, 11...

3) 1, 2, 3, 8, 24, 90...

What are the next numbers in these sequences?

All of them have something in common...really...I'm not lying. It may help to do them in the order that I have written them

I don't really expect you to go very much further with the sequence without the use of a computer of some sort...

That is right for the first one, but...

How did you get 37.36?

And the second sequence uses a different formula, where x= {0, and 2, 3, etc.}

I got 37.36 by the following:

f(5) = 26.42 (given)

f(6) = 6² + 6²/f(5) = 36 + 36/26.42 = 36 + 1.362604... = 37.36

And it seems like the formula for the second sequence is the same formula, just a different set of possible X values...I mean everything else after that second term uses the same formula as the first...so, that's why I said that...

And I got the same answer for the second sequence because:

f(5) = 26.44 (given)

f(6) = 6² + 6²/f(5) = 36 + 36/26.44 = 36 + 1.361573... = 37.36

The other two are:

f(n) = n² + n²/f(n-1)

However, technically speaking, the second sequence doesn't work, since

f(1) = 0

f(2) = 2² + 2²/f(1) and f(1) = 0 in this case...so, you're dividing by 0...however, the rest of the numbers work out

So the next two numbers in BOTH sequences is 37.36 (and from then on, the two sequences will have the same numbers)

north

1. All wavelengths+salty secretion+Scottish diamonds

Spectrum salty jewels?

2. Distilled spirits+women's headdress+stinging plant

burbonnettle?

Ahh...thank you, I had the last part of #2, but I couldn't get the first part...the fact that it was Distilled spirits (plural) was throwing me off...I was trying to get a plural word there.. Nice job

This one was pretty tough...

I think the first one is:

whiteargyles

All Wavelengths = WHITE

Salty Secretion = TEAR

Scottish Diamonds = ARGYLES