The worlds first riddle!

WHOLE NUMBERS
10256
1016949

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oh.... I thought, cause of the three dots... I'm not in smart mode this week. Maybe the flue is killing my brain?

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PLANETS:

.....4593
...20163
.695163
.358691 +
======
1078610

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NEPTUNE
1078610

oops - Whim beat me to it.

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Re: The worlds first riddle!

Tryagain wrote:

Hi all.

I know newbie's must seem dumb to you, and I do not seek to change your mind.
However, I offer for your consideration the following:

Whilst digging in a 4,500-year-old burial mound at Xiaguan in Yunnan province in central China, a parchment was found sealed inside a clay vessel for its owner to take with him on his journey to his ancestors. The parchment talks of a traveller, getting lost on a mountain, and whilst seeking safety sees two old men sitting by a fire next to the entrance of a cave. As he approached, he overheard the last part of their conversation, and he later recorded it thus.

Tryagain, I just read this for the first time. Is that story true? No kidding around mister, straight answer please. Very Happy

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You have a number of consecutive numbers starting from 1, 2, 3, ... upto x. Can you place them in a row in such a way that every adjacent numbers sum up to a square.

a+b = square
b+c = square
c+d = square
etc.

What is the longest row you can find?

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The Question: What number does neptune represent?

Whim:

PLANETS:

.....4593
...20163
.695163
.358691 +
======
1078610 Cool

Mark:
1078610 Cool

Liessa:
!"£$%^ Laughing

Pa ulaj writes, "Is that story true? No kidding around mister, straight answer please." Twisted Evil

I just love it when you are forceful. My lawyer says, "go to page two, where all will be explained" Failing that, your wish is my command. The answer is; No! Now how much extra cheese did you want on your pizza? Laughing

A man is going to an Antique Car auction. All purchases must be paid for in cash. He goes to the bank and draws out $25,000.

Since the man does not want to be seen carrying that much money, he places it in 15 envelopes numbered 1 through 15. Each envelope contains the least number of bills possible of any available US currency (for example, no two tens instead of a twenty).

At the auction he makes a successful bid of $8322 for a car. He hands the auctioneer envelopes 2, 8, and 14. After opening the envelopes the auctioneer finds exactly the right amount.

The Question: How many ones did the auctioneer find in the envelopes

In a distant, dark forest, lives a population of 400 highly intelligent dwarfs. The dwarfs all look exactly alike, but only differ in the fact that they are wearing either a red or a blue hat.

There are 250 dwarfs with a red hat and 150 dwarfs with a blue hat. Striking however, is that the dwarfs don't know these numbers themselves and that none of them knows what the colour of his hat is (there are for example no mirrors in this forest). But the dwarfs do know that there is at least one dwarf with a red hat.

During a certain period of their year, there is a big party in this village, to which initially all dwarfs will go. However, this party is only intended for dwarfs wearing a blue hat. Dwarfs with a red hat are supposed never to return to the party again, as soon as they know that they are wearing a red hat.

The Question: How many days does it take before there are no more dwarfs with a red hat left at the party

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ah, hats, my favourite (except for rivers crossing)
And I should regain some respect after disaster on whimsical pandigit thread...

So, I would say...249 or 250 days, depending of what is consider "after 1 day". Let's imagine that there was only one dwarf with red hat - knowing that there must be one, he would left immidiately, first day. If there are two they will see one hat on another dwarf, but if the same dwarf shows next day, they will both know they are the only two. So, they will leave on second day - or after one day.
And so on...
250 days is needed for them to find out - or after 249 days they will figure it out if first day is 0.

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$25,000
Except for the last one, envelope N contains 2^(N-1) dollars. Therefore, 14 contains $8,192, 8 contains $128, and 2 contains $2.
14 contains no $1 bills (it does contain a $2 bill)
8 contains one $1 bill
2 contains no $1 bills (it also contains a $2 bill)
So, there was one $1 bill.

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WHIM SQUARES

This doesn't quite fit your description. Is yours longer? If so, I'll keep looking.

16 9 7 2 14 11 5 4 12 13 3 6 10 15 1 8 17

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MOU:
"ah, hats, my favourite (except for rivers crossing)"

"250 days is needed for them to find out - or after 249 days they will figure it out if first day is 0." Cool

You are good, very good. Laughing

Mark:
So, there was one $1 bill. Cool

Is also very good. Laughing

Whim, is this what you are after?

2, 23, 13, 12, 4, 21, 15, 10, 6, 19, 17, 8, 1, 3, 22, 14, 11, 5, 20, 16, 9, 7, 18

2 Know or not 2 Know

Two whole numbers, m and n, have been chosen. Both are unequal to 1 and the sum of both is less than 100. The product, m × n, is given to mathematician X. The sum, m+n, is given to mathematician Y. Then both mathematicians have the following conversation:

X: I have no idea what your sum is, Y.
Y: That's no news to me, X. I already knew you didn't know that.
X: Ahah! Now I know what your sum must be, Y!
Y: And now I also know what your product is, X!

The Question: What are the numbers m and n

Here are three answers:

1. Answer A
2. Answer A or B
3. Answer B or C

The Question: There is only one correct answer to this question. Which answer is this

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Tryagain:

Seeing your sequence, I wondered what the longest row could be without restrictions on which number comes first or last?

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If answer 1 is correct, then also answer 2 would be correct.
If answer 2 is correct, then either answer 1 or 3 would be correct as well.

Therefore - Answer 3 is only possible correct answer with given conditions (and "answer C" is correct answer on given question).

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WHIM

Here's 53:

1 3 6 10 15 21 43 38 26 23 13 12 4 5 11 14 50 31 33 48 52 29 35 46 18 7 9 16 20 44 37 27 22 42 39 25 24 40 41 8 17 32 49 51 30 19 45 36 28 53 47 2 34

I'm going to stop looking for solutions with the 1-8 constraint and start looking for longer general solutions. I wouldn't be surprised if there is no upper bound.

Here's 71:

1 3 6 10 15 21 4 5 11 14 2 7 9 16 48 33 67 54 27 22 42 39 25 24 40 60 61 20 29 52 69 12 13 36 45 19 30 70 51 49 32 68 53 28 8 41 59 62 38 43 57 64 17 47 34 66 55 26 23 58 63 37 44 56 65 35 46 18 31 50 71

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I'm going to stop looking for solutions with the 1-8 constraint.

ok.

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Tryagain wrote:

Pa ulaj writes, "Is that story true? No kidding around mister, straight answer please."

I just love it when you are forceful. My lawyer says, "go to page two, where all will be explained" Failing that, your wish is my command. The answer is; No! Now how much extra cheese did you want on your pizza?

Bats eye lashes...right there------> Smile

I would like a generous amount of extra cheese. Are you feeling...generous?

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WHIM
77:
1 3 6 10 15 21 4 5 11 14 2 7 9 16 20 29 35 46 18 31 50 71 73 8 17 19 30 51 70 74 26 55 66 34 47 53 28 72 49 32 68 13 12 24 76 45 36 64 57 43 38 62 59 22 42 58 63 37 27 54 67 33 48 52 69 75 25 39 61 60 40 41 23 77 44 56 65

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2 KNOW OR NOT 2 KNOW
For Y to know that X can't know the sum, the sum must not be expressable as the sum of two prime numbers. Therefore, the sum must be in this set:
(11, 17, 20, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97)

For X to know the sum after Y's statement, the product must be expressable in exactly one way as the product of two numbers with their sum in the list above.

Starting with 11 (fewest ways to express sum) and working up through the list, I found a solution when the sum is 17 (product is 52).

Y knows these are the possible sums:
2+15 (prod would be 30) (possible sums from X's view are 2+15*, 3+10, 5+6*)
3+14 (prod would be 42) (possible sums from X's view are 2+21*, 3+14*, 6+7)
4+13 (prod would be 52) (possible sums from X's view are 2+26, 4+13*)
5+12 (prod would be 60) (possible sums from X's view are 2+30, 3+20*, 4+15, 5+12*, 6+10)
6+11 (prod would be 66) (possible sums from X's view are 2+33*, 3+22, 6+11*)
7+10 (prod would be 70) (possible sums from X's view are 2+35*, 5+14, 7+10*)
8+9 (prod would be 72) (possible sums from X's view are 2+36, 3+24*, 4+18, 6+12, 8+9*)

* indicates the sum is in the list of valid sums above

X knows these are the possible products:
2*26 (sum would be 28)
4*13 (sum would be 17)

Before Y makes his first statement, X knows the sum is 17 or 28.
After Y makes his first statement, X knows the sum can't be 28 because the product could have been 5*23, and Y couldn't have made the statement.

After X makes his second statement, Y knows the product must be 52 because it is the only product with exactly one possible sum from the list of valid sums above.

Therefore, the numbers must be 4 and 13.

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markr wrote:

2 KNOW OR NOT 2 KNOW

For Y to know that X can't know the sum, the sum must not be expressable as the sum of two prime numbers. Therefore, the sum must be in this set:
(11, 17, 20, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97)

For X to know the sum after Y's statement, the product must be expressable in exactly one way as the product of two numbers with their sum in the list above.

Starting with 11 (fewest ways to express sum) and working up through the list, I found a solution when the sum is 17 (product is 52).

Y knows these are the possible sums:
2+15 (prod would be 30) (possible sums from X's view are 2+15*, 3+10, 5+6*)
3+14 (prod would be 42) (possible sums from X's view are 2+21*, 3+14*, 6+7)
4+13 (prod would be 52) (possible sums from X's view are 2+26, 4+13*)
5+12 (prod would be 60) (possible sums from X's view are 2+30, 3+20*, 4+15, 5+12*, 6+10)
6+11 (prod would be 66) (possible sums from X's view are 2+33*, 3+22, 6+11*)
7+10 (prod would be 70) (possible sums from X's view are 2+35*, 5+14, 7+10*)
8+9 (prod would be 72) (possible sums from X's view are 2+36, 3+24*, 4+18, 6+12, 8+9*)

* indicates the sum is in the list of valid sums above

X knows these are the possible products:
2*26 (sum would be 28)
4*13 (sum would be 17)

Before Y makes his first statement, X knows the sum is 17 or 28.
After Y makes his first statement, X knows the sum can't be 28 because the product could have been 5*23, and Y couldn't have made the statement.

After X makes his second statement, Y knows the product must be 52 because it is the only product with exactly one possible sum from the list of valid sums above.

Therefore, the numbers must be 4 and 13.

Mark, this is a most beautifully crafted answer. Superlatives elude me.

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The worlds first riddle!

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