The worlds first riddle!

Mark:

FISH
33


Fantasmagorical.


That is to say N the approximate number of red fish in the pond. One expresses the proportion of marked red fish of green in two different ways:

12/N or 4/11

One obtains a simple equation which amounts calculating fourth proportional:

12/N = 4/11

4?-N = 12?-11

N = 132/4

N = 33
There are roughly 33 fish in the pond of the college. (Translated from French. Notation may differ)



CORN
3 (awfully easy, or I missed the gotcha)

If it looks too easy, than you can be sure it's a trap. The answer as any Australian would know is: Nine nights. Two of the ears each journey belonged to the rabbit.


WORDS
(sat)iate

Days of the week.



What number comes next in this sequence:
8 5 5 3 4 4?

Clue:

Using the same logic, you would add 7 to the start


I have a large money box, 10 inches square and 12 inches tall. I also have a pile of pennies a mile high. Roughly how many pennies can I place in my empty money box


The letters that follow below represent a common phrase containing 18 letters

FRIJUSTENDS



Can you think of a word that changes both number and gender when you add the letter 'S'


A passenger plane is equipped with 100 first class seats and 200 economy class.
It is noted that among the passengers, 10 men out of 21 and 2 women out of 10 choose the first class. For the present flight, the plane is filled with seven eighth in economy class and four fifths in first class.

Could you deduce from this information, crew aside, the number of women and men on the aircraft

No, I'm not Joe King.

[size=7]PHRASE
just between friends

PLANE
105 men, 150 women[/size]

[size=7]NUMBER/GENDER
I was running through all of the names for male/female animals when princes(s) popped into my head.[/size]

I was asked if I knew, what falls but never breaks and breaks but never falls? I said there was no such answer, but within 24 hours I had solved the riddle. Can you do better

The time starts … now!


Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers

This has a 5 star difficulty reting

The time starts … now

Night and Day

[size=7]LOGICIANS
4, 13

S's first statement limits the sum to:
11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53

Each of these provides at least one product that would allow P to determine the numbers. However, only 17 provides one such product. That is required for S to make his second statement. That product is 52.[/size]

Mark:


PHRASE
just between friends

PLANE
105 men, 150 women


NUMBER/GENDER
I was running through all of the names for male/female animals when princes(s) popped into my head.

I wonder what Sigmund Freud would have made of that.

LOGICIANS
4, 13

S's first statement limits the sum to:
11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53

Each of these provides at least one product that would allow P to determine the numbers. However, only 17 provides one such product. That is required for S to make his second statement. That product is 52.

May I point out Marks reasoning is clear, concise and accurate.

I on the other hand…
First of all, trivially, xy cannot be prime. It also cannot be the square of a prime, for that would imply x = y.

We now deduce as much as possible from each of the logicians' statements. We have only public information: the problem statement, the logicians' statements, and the knowledge that the logicians, being perfect, will always make correct and complete deductions. Each logician has, in addition, one piece of private information: sum or product.

P: I cannot determine the two numbers.

P's statement implies that xy cannot have exactly two distinct proper factors whose sum is less than 100. Call such a pair of factors eligible.

For example, xy cannot be the product of two distinct primes, for then P could deduce the numbers. Likewise, xy cannot be the cube of a prime, such as 33 = 27, for then 3×9 would be a unique factorization; or the fourth power of a prime.

Other combinations are ruled out by the fact that the sum of the two factors must be less than 100. For example, xy cannot be 242 = 2×112, since 11×22 is the unique eligible factorization; 2×121 being ineligible. Similarly for xy = 318 = 2×3×53.

S: I knew that.

If S was sure that P could not deduce the numbers, then none of the possible summands of x+y can be such that their product has exactly one pair of eligible factors. For example, x+y could not be 51, since summands 17 and 34 produce xy = 578, which would permit P to deduce the numbers.

We can generate a list of values of x+y that are never the sum of precisely two eligible factors
Eligible sums: 11, 17, 23, 27, 29, 35, 37, 41, 47, 53 .

(We can use Goldbach's Conjecture, which states that every even integer greater than 2 can be expressed as the sum of two primes, to deduce that the above list can contain only odd numbers. Although the conjecture remains unproven, it has been confirmed empirically up to 4×1014.)

P: Now I can determine them.

P now knows that x+y is one of the values listed above. If this enables P to deduce x and y, then, of the eligible factorizations of xy, there must be precisely one for which the sum of the factors is in the list. The table below, shows some such xy, together with the corresponding x, y, and x+y. The table is sorted by sum and then product.

Note that a product may be absent from the table for one of two reasons. Either none of its eligible factorizations appears in the above list of eligible sums (example: 12 = 2×6 and 3×4; sums 8 and 7), or more than one such factorization appears (example: 30 = 2×15 and 5×6; sums 17 and 11.)

S: So can I.

If S can deduce the numbers from the table below, there must be a sum that appears exactly once in the table. Checking the table, we find just one such sum: 17.

Therefore, we are able to deduce that the numbers are x = 4 and y = 13.

Eligible products and sums
Product x y Sum
18 2 9 11
24 3 8 11
28 4 7 11
52 4 13 17
76 4 19 23
112 7 16 23


Whim:
"what falls but never breaks and breaks but never falls?"

Night and Day

Wow, that was quick


To number all the pages of a large book, starting from page n°1, Whim used twice as many figures than the number of pages of this book.


How many pages in the book



The number 2.525252525252??? can be written as a fraction. What the freek is the sum of the denominator and numerator

Number - 3


Box - None, otherwise the box wouldn't be empty


Book- 108 if the "n°"s aren't included (with modest assumptions)

9 + 2*90 + 3*9 = 216

If they are, the only answer is 0.


Freek- I see triple question marks after the number. If they mean the decimal expansion repeats (maybe it was an ellipsis when you typed it?), the answer is 349*n with nonzero n (n=1 for the reduced fraction).

The mysterious X joins the fray . s/he writes, "Number - 3"


What number comes next in this sequence:
8 5 5 3 4 4 ?

Clue:

Using the same logic, you would add 7 to the start.


I am sorry that is not correct for this sequence. However, the number after the ? is 9


As for the others, the answers will be out tomorrow.

In the meantime:


A car travels downhill at 72 m.p.h. (miles per hour), on the level at 63 m.p.h., and uphill at only 56 m.p.h. The car takes 4 hours to travel from town A to town B. The return trip takes 4 hours and 40 minutes.

Is it possible to find the distance between the two towns

[size=7]FRACTION
100x - x = 250
99x = 250
x = 250/99
sum = 349

CAR
How about 273 miles?[/size]

[size=7]BOX OF PENNIES
Assuming:
diameter = .75 inches
thickness = .06102+ inches
At least 38,445 pennies can be placed in the box.[/size]

"We can generate a list of values of x+y that are never the sum of precisely two eligible factors
Eligible sums: 11, 17, 23, 27, 29, 35, 37, 41, 47, 53 . "

Either I missed something or you did. I have 51 in that list.

Shurp!

Do it this way

The difference is as plain as uphill and downhill.
Let Tl be the time on level ground
and Tu be the time spent in hill country
Then
Time out
Ti+Tu=4:40
and Time back is
Ti-Tu=4:00
so 2Ti=8:40
so
Ti=4:20
and Tu=-0:20

The distance is then the solution to
D=Tl*63mph+Tu(72-65)mph
D=4:20*63-0:20*16=(273-16/3)mi
the towns are 267 2/3 miles apart

Rap

SUN ken
MON asticism
TUE
WED ge
THU mb
FRI volity
-----------------

SAT iate

How can the time spent in the hills be negative?

While puttering around this morning I've had some time to further ponder about this part of the problem.

If level the time spent between the two towns (Tl) is at 63mph. Since Tl=4:20 the distance is 273 miles. To gain 20 minutes the average trip speed is 68 1/4 mph, to lose 20 min the average trip speed is 58 1/2 mph.

So on the trip out (4:00) the car travels 1:40 on level ground and 2:20 going downhill. This checks to 273 miles.

On the way back (4:40) the car travels on level ground for 1:40 and 3:00 going uphill. This also checks to 273 miles.

So the answer, as markr states, is 273 miles.

As I've said before, I'm only a moderately intelligent monkey (with a forehead that is severely atilt)..

Rap

X:

Box - None, otherwise the box wouldn't be empty


Book- 108 if the "n°"s aren't included (with modest assumptions)


Do I know you? You're not my x are you?




Mark:

FRACTION
100x - x = 250
99x = 250
x = 250/99
sum = 349

It looks fine, but I will have to get back to you when I find what I have.

CAR
How about 273 miles?

Yeah, what about it? No wait…You are right, good ?'guess' genius.


Let the total distance travelled downhill, on the level, and uphill, on the outbound journey, be x, y, and z, respectively.
The time taken to travel a distance s at speed v is s/v.

Hence, for the outbound journey

x/72 + y/63 + z/56 = 4

While for the return journey, which we assume to be along the same roads

x/56 + y/63 + z/72 = 14/3

It may at first seem that we have too little information to solve the puzzle. After all, two equations in three unknowns do not have a unique solution. However, we are not asked for the values of x, y, and z, individually; only for the value of x + y + z.

Multiplying both equations by the least common multiple of denominators 56, 63, and 72, we obtain

7x + 8y + 9z = 4 • 7 • 8 • 9
9x + 8y + 7z = (14/3) • 7 • 8 • 9

Now it is clear that we should add the equations, yielding

16(x + y + z) = (26/3) • 7 • 8 • 9

Therefore x + y + z = 273; the distance between the two towns is 273 miles.


A unique solution is possible because the speeds are chosen so that a round trip over a sloping section of road takes the same time as that over a flat section of the same length. Had we chosen to write down an equation for the round trip, the answer would have been immediately apparent.




BOX OF PENNIES
Assuming:
diameter = .75 inches
thickness = .06102+ inches
At least 38,445 pennies can be placed in the box.

Do you know what? When I tried to work out the exact answer, I got to;
38,455 before remembering the, how many times can you take 5 from 25? Same reasoning. :wink:


Mark writes, "Either I missed something or you did. I have 51 in that list."

You are correct in the numbers for the list.
You missed nothing.
I missed nothing. I only listed up to unique sequence to save space and boredom, if you wish to see the full list sit back and enjoy:

Eligible products and sums
Product x y Sum
18 2 9 11
24 3 8 11
28 4 7 11
52 4 13 17
76 4 19 23
112 7 16 23
130 10 13 23
50 2 25 27
92 4 23 27
110 5 22 27
140 7 20 27
152 8 19 27
162 9 18 27
170 10 17 27
176 11 16 27
182 13 14 27
54 2 27 29
100 4 25 29
138 6 23 29
154 7 22 29
168 8 21 29
190 10 19 29
198 11 18 29
204 12 17 29
208 13 16 29
96 3 32 35
124 4 31 35
150 5 30 35
174 6 29 35
196 7 28 35
216 8 27 35
234 9 26 35
250 10 25 35
276 12 23 35
294 14 21 35
304 16 19 35
306 17 18 35
160 5 32 37
186 6 31 37
232 8 29 37
252 9 28 37
270 10 27 37
322 14 23 37
336 16 21 37
340 17 20 37
180 5 36 41
114 3 38 41
148 4 37 41
238 7 34 41
288 9 32 41
310 10 31 41
348 12 29 41
364 13 28 41
378 14 27 41
390 15 26 41
400 16 25 41
408 17 24 41
414 18 23 41
418 19 22 41
132 3 44 47
172 4 43 47
246 6 41 47
280 7 40 47
370 10 37 47
396 11 36 47
442 13 34 47
462 14 33 47
480 15 32 47
496 16 31 47
510 17 30 47
522 18 29 47
532 19 28 47
540 20 27 47
546 21 26 47
550 22 25 47
552 23 24 47

Phew!



BillyFalcon:

SUN ken
MON asticism
TUE
WED ge
THU mb
FRI volity
-----------------

SAT iate

Well done Billy, few could answer that, surprisingly easy when you know. Otherwise, as hard as hell.
Good to see you participating in Riddles.



Raprap:

While puttering around this morning I've had some time to further ponder about this part of the problem.

GOLF! You mean you play golf? I thought you detested golf. Anywho, I always enjoy reading your posts. Remember, it is far better to take part than to win. :wink:

Chin up Rap.





Is the number 2438100000001 prime or composite



Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with the adjacent marble(s). Also, each marble is in contact all around the funnel wall.

The smallest marble has a radius of 8mm. The largest marble has a radius of 18mm.

What is the radius of the middle marble



Two players take turns choosing one number at a time (without replacement) from the set {-4, -3, -2, -1, 0, 1, 2, 3, 4}. The first player to obtain three numbers (out of three, four, or five) which sum to 0 wins.

Does either player have a forced win

[size=7]PRIME OR COMPOSITE
composite: 73 divides it

MARBLES
[(18/8)^(1/4)]^2 * 8 = 12

NUMBER GAME
I'll have to think about this.[/size]

I think this is the correct answer to the fraction question.


The number 2.525252525252??? can be written as a fraction. What is the sum of the denominator and numerator?

Answer
2.5252… = a/b

100 x 2.5252… = 252.5252… = 100a/b

100a/b - a/b = 252.5252… -2.5252… = 250

99a/b = 250

a/b = 250/99

a + b = 250 + 99 = 349.




Mark:

PRIME OR COMPOSITE
composite: 73 divides it


Why did I not think of that?

Firstly, notice that 2438100000001 = x5 + x4 + 1, with x = 300.

Then f(x) = x2 + x + 1 is a factor of g(x) = x5 + x4 + 1, since f(w) = g(w) = 0, where w is a (complex) primitive cube root of unity.
(More formally, since f(w) = g(w) = f(w2) = g(w2) = 0, by the Polynomial Factor Theorem, (x - w) and (x - w2) are factors of f and g. Hence (x - w)(x - w2) = x2 + x + 1 is a factor of f and g.)

Therefore 2438100000001 is composite, with factor f(300) = 90301.


MARBLES
[(18/8)^(1/4)]^2 * 8 = 12


Ok, that's the long way. However: :wink:



Consider two adjacent marbles, of radii a < b. We will show that b/a is a constant, whose value is dependent only upon the slope of the funnel wall.

The marbles are in contact with each other, and therefore the vertical distance between their centers is b + a.

The marbles are also in contact with the funnel wall. Since the slope of the funnel wall (in cross section) is a constant, hence the horizontal distance from the center of each marble to the funnel wall is bc and ac, respectively, where c = sec(x) is a constant dependent upon the slope of the funnel wall. (x is the angle the funnel wall makes with the vertical.)

Let the slope of the funnel wall be m.
Then m = (b + a) / (b - a)c.
Rearranging, b/a = (mc + 1)/(mc - 1).

Hence the ratio of the radii of adjacent marbles is a constant, dependent only upon the slope of the funnel wall. Let this constant be k.

In this case, we have 18 = 8k4.
So k2 = 3/2.

Therefore the radius of the middle marble is 8 • (3/2) = 12mm.






Two ladders are placed cross-wise in an alley to form a lopsided X-shape. The walls of the alley are not quite vertical, but are parallel to each other. The ground is flat and horizontal. The bottom of each ladder is placed against the opposite wall. The top of the longer ladder touches the alley wall 5 feet vertically higher than the top of the shorter ladder touches the opposite wall, which in turn is 4 feet vertically higher than the intersection of the two ladders.

How high vertically above the ground is that intersection



You are given n > 0 of each of the standard denomination US coins: 1¢, 5¢, 10¢, 25¢, 50¢, $1. What is the smallest n such that it is impossible to select n coins that make exactly a dollar



Following complaints that you freekin math geeks with your talk of a+b= n-x gobblywaffle dum dee dum, cant riddle your way out of a paper bag. Check this out; The questions no one dares to ask:


1. One train leaves from New York City heading towards Los Angeles at 100 mph. Three hours later, a train leaves from LA heading towards NYC at 200 mph. Assume there is exactly 2000 miles between LA and NYC. When they meet, which train is closer to NYC


2. What number is two-thirds of one-half of one-fourth of 240


3. If today is Monday, what is the day after the day before the day before tomorrow


4. Mary promised Kay today that she would tell Kay a big secret on the day before four days from the day after tomorrow. If today is Saturday the 12th, on what day and date will Mary tell Kay her big secret


5. If some coffee is 97% caffeine-free, how many cups of it would one have to drink to get the amount of caffeine in a regular cup of coffee

[size=7]LADDERS
6 (similar triangles: x/4 = (9+x)/(4+x))

COINS
77

GEEKS
1. It depends on the definition of "meet." The train that left NYC will be closer to NYC until the front of the other train passes its rear.

2. 20

3. today (Monday)

4. Thursday the 17th

5. Does 97% caffeine-free mean that the coffee is 3% caffeine or has 3% of the regular amount of caffeine? If the former, the amount of caffeine in a regular cup would need to be known. If the latter, 33 1/3 cups.[/size]