The worlds first riddle!

[size=7]BAD COINS
Number the bags 1-10. From each bag, place N (where N is the bag number) coins on the scale. If W is the weight (in ounces) of the pile of coins, then the bad bag is numbered (55-W)*4.

Note that this can't be done using the usual "find the bad coin" procedure. That's why I asked about the scale and the number of coins in each bag.

THANKSGIVING
7 (517 total guests)[/size]

Rap:

Tom and Rachael

pt+nt+dt+qt=pr+nr+dr+qr+32
pt+5nt+10dt+25qt=pr+5nr+10dr+25qr
qt=qr+4
so
(qt-qr)=100
(pt-pr)+(nt-nr)+(dt-dr)+(qt-qr)=32
(pt-pr)+5(nt-nr)+10(dt-dr)+25(qt-qr)=0
this becomes
(pt-pr)+(nt-nr)+(dt-dr)=36
(pt-pr)+5(nt-nr)+10(dt-dr)=100
doing away with (pt-pr)
4(nt-nr)+9(dt-dr)=64
(nt-nr)<>0
(dt-dr)<>0
and both are nonzero integers
so (nt-nr)=7
so if Rachael has 23 nickels, nr=23, and Tom (nt=nr+7) has 30.




Let's let P, N, and D be the number of pennies, nickles, or dimes that Tom has over and above the numbers that Rachel has. (Then, since Tom has more of each of these kinds of coins, we know that P, N, and D must all be positive integers.) Since Rachel has 4 more quarters, but Tom has 32 more coins total, we must have

P + N + D = 36 .
But also, since they each have the same amount of money, the value of Tom's extra coins must be the same as Rachel's extra 4 quarters. In other words:

P + 5N + 10D = 100 .
A little experimenting shows that the only two solutions to these two equations are

P = 20, N = 16, D = 0
and
P = 25, N = 7, D = 4

The first one, however, is not allowable, since D must be positive. (Tom has more dimes!) So, the second solution must be the right one, so Tom has 7 more nickels. Since Rachel has 23 nickels, Tom must have 30 nickels.



Chain Linking--
Open and close all three links of one chain, use those three to connect the other four

Frog in a well--
So the progression is n=20-4=16.


Last table has 7 * see below


The second guy is lying, so the 3rd is good.


Good guy. The first man can only have said good guy. If he is a good guy, he tells the truth and says he's a good guy. If he is a villain he lies and says good guy. The second man must be lying then, which means the third man told the truth, and is therefore a good guy.



And between 12 and 12 is 10.


10 times. (1:05:27, 2:10:54, 3:16:21, 2:21:48, 5:22:15, 6:32:42, 7:38:09, 8:43:36, 9:49:03, 10:55:30)



Mark:


TOM & RACHEL
Tom has to make up $100 with 36 coins (pennies, nickels, and dimes). This can be done in two ways:
20p 16n
25p 7n 4d

The first way doesn't satisfy the requirement that he has more coins of each type except quarters. Therefore, he has 7 more nickels than she.
Tom has 30 nickels.

CHAIN
$0.75
Open all three links from one chain and use them to connect the remaining four chains.


VILLAINS
The second man is lying because the first would never say he is a villain. Therefore, the third man is a good guy.


FROG
16 days

MINUTE HAND
10


BAD COINS
Number the bags 1-10. From each bag, place N (where N is the bag number) coins on the scale. If W is the weight (in ounces) of the pile of coins, then the bad bag is numbered (55-W)*4.

Note that this can't be done using the usual "find the bad coin" procedure. That's why I asked about the scale and the number of coins in each bag.

Noted

THANKSGIVING
7 (517 total guests) *




*...You both give 7 people at the last table. I have a higher figure. However, as it is two against one I naturally question my result.

After careful consideration, I have confirmed my original prognoses, and ask whether you wish to reconsider your earlier positions?



A witness sees a crime involving a taxi in Carborough. The witness says that the taxi is blue. It is known from previous research that witnesses are correct 80% of the time when making such statements. The police also know that 85% of the taxis in Carborough are blue, the other 15% being green.

a)What is the probability that a blue taxi was involved in the crime

b) How about if it was green



A three-digit number is such that its second digit is the sum of its first and third digits ...







...Easy enough I hear you say, well get this... Now prove that the number must be divisible by 11



During the World Cup, a local supermarket was running a World Cup medallions promotion. Each time you spent a certain amount of money you got a free coin embossed with the face of one of the members of the Canadian squad.

There were 22 players in the squad, so there were 22 medallions to collect. If we assume that each time you get a medallion it is equally likely to represent any one of the 22 players the question is:

(Leaving aside why would anyone want one)
On average how many will you need to collect before you get a full set



Two bicyclists, Anthony and Beatrice, start 10km apart.
Anthony cycles at 4 m/s towards Beatrice, who cycles at 6 m/s towards him.

Cleopatra the Fly starts on the handlebars of Anthony's bike, and flies at 12 m/s towards Beatrice's. When she gets there she instantaneously reverses direction and flies back to Anthony's, and then repeats the whole process over and over again.

Eventually the two bikes meet. How far has Cleopatra flown

"*...You both give 7 people at the last table. I have a higher figure. However, as it is two against one I naturally question my result.

After careful consideration, I have confirmed my original prognoses, and ask whether you wish to reconsider your earlier positions?"

What do you have?

[size=7]TAXI
a) 95.8%
b) if you mean given that the witness said blue, what is the probability that the taxi is green, then 4.2%

THREE-DIGIT NUMBER
Let N = 100a+10b+c, where b=a+c.
(10a+c)*11 = 100a+10a+10c+c = 100a+10b+c = N
Therefore, 11 divides N.

CLEOPATRA
12km[/size]

[size=7]MEDALLIONS
~81.2 based on several million trials.[/size]

I have 11 guests at the last table.

Suppose that the number of empty packages of rolls is R and the number of empty pie tins is P . Then the number of guests must be
13R + 3 (remember, there were 10 rolls left in the last package, so 3 rolls from that package were eaten). But counting pie slices instead of rolls, the number of guests must also have been 9P + 5

(there were 4 slices left in the last pie tin). So what we need is a number between 500 and 600 that is of both of the forms 13R + 3 and 9P + 5.

A little experimenting shows that the only possibility is 536 -- there were 536 guests! The number of people seated at the last table will now be the remainder when 536 is divided by 15. Since 525 is a multiple of 15, this remainder is 11. So, there are 11 people seated at the last table.

Or were there?

There were. I screwed up and did:
13R+10, 9P+4

OK! Double or quits:

Mr. and Mrs. Field are proud grandparents of new baby triplets. Three new grandchildren at once -- and it had been six years since their last grandchild was born.

Mr. Field likes doing arithmetic in his head, and he happened to notice that right now the sum of his age and Mrs. Field's age is exactly equal to the sum of the ages of their grandchildren. However, six years ago, when the last grandchild was born, the sum of the two grandparent's ages was twice the sum of the grandchildren's ages.

Mrs. Field, who also enjoys a good arithmetic challenge, points out that (assuming no new grandchildren come along) in 30 years the grandchildren's ages will sum to three times the sum of the grandparents' ages.

How many grandchildren do the Fields have

[size=7]FIELDS
10 who average 11.4
1 six year old
3 newborns
---
14[/size]

murder or suicide
:wink:
Alfred Hitchcock walks into a crime scene. No one has touched a single thing It appears like a suicide. He walks over to the body and looks at it. Gun shot wound to the head. Yes the gun is in his mouth. Then he looks and see a table beside the body with a recorder on it. He presses play. The recorder plays, "I can't go on living anymore, life has treated me like ****", Then there is a gunshot and the tape stops. How immediately after listening to this recording does the detective know it is a murder not a suicide.

Had it been a suicide, the tape would have kept recording.

Mark:

TAXI
a) 95.8%
b) if you mean given that the witness said blue, what is the probability that the taxi is green, then 4.2%
(My poor wording)


We know that 85% of the taxis in Carborough are blue, the other 15% of taxis being green.
Suppose that there's no bias for any particular blue or green taxi to be involved in such incidents. Then consider 100 possible cases (contingencies), in which taxis are involved, in proportion to their numbers. Each case is equally probable. We expect 85 cases to involve blue taxis, 15 to involve green. If a blue taxi were really involved, the witness might report blue (with 80% probability) or green (with 20% probability). If a green taxi were really involved, the witness might report blue (with 20% probability) or green (with 80% probability). The number of times that these outcomes are expected to occur in each of these four cases are shown below



Reality v. Witness reports……….BLUE Taxi………………………….GREEN taxi

Reality No. of cases……………………85……………………………………..15


Witness reports:

Blue taxi………………………………………68=80% of 85…………………..3=20% of 15
Green taxi……………………………………17=20% of 85……………………12=80% of 15


So, given that the witness reported seeing a blue taxi, we must use the row in the table corresponding to the reported blue taxi. We have 68+3 = 71 equally likely cases and the probability that a blue taxi really was involved is 68/71 = 96%.

A more interesting and surprising result is obtained in the case when the witness reports a green taxi. Many people on being asked this problem will say that if the witness reports a green taxi, then there is an 80% probability that a green taxi was involved. This however is incorrect, as the following argument shows.

The contingency is the same as above but now we must use the row in the table corresponding to the reported green taxi. We have 17+12 = 29 equally likely cases and the probability that a green taxi really was involved is 12/29 = 41%. The cases of correct identification of green are swamped by the false identifications of blue, and the witness's evidence is of no practical value.



THREE-DIGIT NUMBER
Let N = 100a+10b+c, where b=a+c.
(10a+c)*11 = 100a+10a+10c+c = 100a+10b+c = N
Therefore, 11 divides N.


If our number is n and the digits are represented by x, y and z (reading from left to right) then we can write:
n = 100x + 10y + z
We are told that
y = x + z
Therefore
n = 100x + 10(x + z) + z
n = 110x + 11z
n = 11(10x + z)
therefore n must be divisible by 11!

CLEOPATRA
12km


MEDALLIONS
~81.2 based on several million trials.

I only had time for one trial with a difference of .0022

The answer to the question is : 81 medallions on average (rounded number).

Explanation: the equivalent problem is to fill 22 empty boxes sequentially. Each time one fills one box, he moves to fill the next one by drawing in average a number of medallions before finding the one different from the ones in the boxes behind.
Let N be the number of boxes to fill (N=22).
Let n be the position of the nth box to fill after the n-1 before filled.
Let M_n be the average of medallions one needs to fill the nth box with the right medallion.
Let P_n the probability to find the right medaillon for the nth box.
Then: P_n = 1/( M_n ) = (N-n+1)/N
So, for the nth box, the average number of medallions one needs is: M_n = N/(N-n+1)
And to get N=22 different pictures one needs in average to buy :
M = M_1 + M_2 +....+ M_22 = 22*(1/22 + 1/21 + 1/20+...+1/2 + 1)
Result: M = 81.1978 rounded to 81.


How many grandchildren do the Fields have?

Mark:


FIELDS
10 who average 11.4
1 six year old
3 newborns
---
14


WOW! That's great - you win.

It would appear news of your decline was grossly exaggerated.



Hi ACHARAJ, good to see you posting.

You wrote…"Then there is a gunshot and the tape stops"…


Mark replied, "Had it been a suicide, the tape would have kept recording."


Who re-wound the tape? Otherwise, it would start on the blank after the conversation.






Let's pretend for the sake of argument that a piece of candy corn is a perfect cone (with a circular base and an altitude perpendicular to that base).

Suppose that the cone is 1 inch high and has a base radius of 1/4 inch. Finally, suppose that each of the three colors occupies 1/3 of an inch in the height of the candy cone. (As is traditional, we'll have yellow on the bottom, orange in the middle, and white on top of the cone.)

Now suppose we do surgery on this piece of candy. We carefully separate the three colors, melt them down, and reassemble it into another cone of the same dimensions, but with the three colors reversed (white on the bottom and yellow on top).

Your Halloween challenge is to calculate the new height of the yellow part of the candy piece

Or, if it is easier: The % of the total height of the new cone



Here's a simple little riddle about nothing more than good ol' numbers.

Suppose I have two two-digit numbers, and I write them one after another (with the larger two-digit number first) to give a four-digit number.

From this four-digit number, I subtract the difference between the two two-digit numbers.

The result is 5689. (Yes!!! I give you the result )

What was the smaller of the two-digit numbers

The tape stoppeth.

Of course he could be wrong is the last recording was at the last of the tape.

Rap

[size=7]GOOD OL' NUMBERS
23 (larger was 57)[/size]

[size=7]CANDY CORN
[19^(1/3)]/3 = .889"[/size]

Mark:

CANDY CORN
[19^(1/3)]/3 = .889 .

You rock!

The height of the yellow part in the newly rearranged candy corn would be cuberoot(19)/3 inches, or about 0.889 inches. (That's nearly 90% of the total height of the cone!)

The easiest way to do this one is to get a formula for the volume of a cone with these proportions in terms of its height h. A cone with height h and base radius r will have volume (1/3)(pi)r2h (where r is the radius of the circular base of the cone). Our cones have radius equal to one-fourth of their height, so we get
V(h) = (1/3)(pi)(h/4)2h = [(pi)/48]*h3.

Now:
Yellow volume = (total volume) - (combined white and orange volume)
= (cone of height 1) - (cone of height 2/3)
= V(1) - V(2/3)
= (pi)/48 - [(pi)/48]*(2/3)3
= [(pi)/48]*(1 - 8/27)
= [(pi)/48]*(19/27).

Since the yellow part of the rearranged candy piece will form the top of the cone, the question becomes "what height of cone gives us this volume?":

V(h) = [(pi)/48]*(19/27)
[(pi)/48}*h3 = [(pi)/48]*(19/27)
h3 = 19/27
h = cuberoot(19/27) = cuberoot(19)/3.


GOOD OL' NUMBERS
23 (larger was 57)

(with the arithmetic in question being 5723 - 34 = 5689).


Here is an interesting trick: a number is a multiple of nine if and only if the sum of its digits is a multiple of nine. For example, the number
2,489,220,423 is a multiple of nine (in fact, it's 9 times 276,580,047).
I could tell this without a calculator and without doing any long division just by adding 2+4+8+9+2+2+0+4+2+3 = 36 which is a multiple of nine.

Your challenge is this: find the greatest possible eight-digit number which is a multiple of nine and which has all digits distinct from each other (no two digits can be the same)

Happy hunting!



Probably everyone has played the "guess my number" game. I choose a target number in some pre-determined range, you try to guess it, and I tell you if you're too high, too low, or if you happened to be lucky and guess it exactly.

Now suppose that the target number is chosen randomly from the numbers 1 to 127. (This simply means that each of the 127 possibilities has the same chance of being the target number.)

What is the average number of guesses needed to find the target



To unravel me
You need a simple key,
No key that was made
By locksmith's hand,
But a key that only I
Will understand.

I am





As a whole, I am both safe and secure.
Behead me, and I become a place of meeting.
Behead me again, and I am the partner of ready.
Restore me, and I become the domain of beasts.

What am I




When young, I am sweet in the sun.
When middle-aged, I make you mellow.
When old, I am valued more than ever.

I am therefore

[size=7]87654321

7 guesses[/size]

Rap

[size=7]8-DIGIT NUMBER
98763210
From 0-9, we must remove two digits that sum to nine while minimizing the larger of the two.

GUESSES
log2(127) rounded up to an integer is 7

RIDDLE 2
stable

RIDDLE 3
grape/wine[/size]

Rap:

87654321 (Close)

7 guesses




Mark:

8-DIGIT NUMBER
98763210
From 0-9, we must remove two digits that sum to nine while minimizing the larger of the two.

We need to build an eight-digit number with every digit distinct. That means we need to throw out two of the numbers 0,1,2,3,4,5,6,7,8, or 9. Now it just so happens that
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
which is a multiple of 9. To have our eight-digit number be a multiple of nine, the sum of the digits we leave behind must still be a multiple of nine. So (and this is the key), the two digits we throw out had better add up to nine. So, we have the choice between throwing out 0 and 9, 1 and 8, 2 and 7, 3 and 6, or 4 and 5.

People often think the best solution would be to throw out the 0 and the 9. This, however, isn't the best choice. If I throw out the 9, the largest leading digit I can use is 8. If fact, it's best to throw out the 5 and the 4. The largest number is then obtained by putting the digits in decreasing order: 98,763,210 .


GUESSES
log2(127) rounded up to an integer is 7

As most people know, the best strategy for this game is to guess the middle number in the range. If you're right, the game is over in a hurry. But if not, at least you've narrowed the field down to half the original size. You can then make your next guess the middle number of the remaining set of choices.

The correct answer is that an average of 769/127 (or about 6.055) guesses will be needed to find the target number.

[1 + 2*2 + 4*3 + 8*4 + 16*5 + 32*6 + 64*7]/127 = 769/127.

There's really a very straightforward solution to this one, though it might help to see it if we look at a simpler example. Suppose the range was 1 to 15 instead of 1 to 127. Our first guess would be the middle number 8. Depending on if that guess is high or low, our next guess (if one is needed) would be either 4 or 12. The third guess (if one is needed) could be any of 2, 6, 10, or 14. Finally, if those first three guesses don't pin it down, the fourth guess would be an odd number and would be guaranteed to be correct.



RIDDLE 2
stable

RIDDLE 3
grape/wine



A woman with no driver's license goes the wrong way done a one-way street and turns left at a corner with a no left turn sign. A cop sees her but does nothing. Why


A square moat surrounds a castle. The moat is ten feet wide. You have two 9 1/2 foot planks. How can you get across the moat


How many nines are there counting from 1 to 100


Which one of the following does not belong in this group
Uncle, cousin, mother, sister, father, cousin, aunt.


A hill is a mile up and down. If you drive up the hill at 30 mph, how fast must you come down to average 60 mph for the whole hill

woman with no drivers licence was walking

Moat... Use the drawbridge?

9s..... 11 (9, 18, 27,) or 20 (9, 19, 29.)

Family tree......... sister (she has red hair, all the others are blonde)

average speed 120 mph
When are you guys going to drag yourselves into line with the rest of the world and go metric (Kilometres)