The worlds first riddle!

Error

Box:
The volume of the box is reduced by 87.5%

Calculator:
He will have $9.76 after his purchase.

ACHELEONS
The answer is no for the given populations.

Consider the three populations modulo 3 (remainder after division by 3 - 0, 1, or 2). A solution is only possible if the populations are not 0, 1, and 2 mod 3.

Here are all ten of the possibilities:
0 0 0
0 0 1
0 0 2
1 1 0
0 1 2
2 2 0
1 1 1
1 1 2
2 2 1
2 2 2

I've bolded the interesting cases. These are interesting because they represent the possible desired end states (0 = 0 mod 3).

There are three cycles of length three (experiment and see):
0 0 0 -> 2 2 2 -> 1 1 1 -> 0 0 0
0 0 1 -> 2 2 0 -> 1 1 2 -> 0 0 1
0 0 2 -> 2 2 1 -> 1 1 0 -> 0 0 2

Note that there is a desired end state within each cycle. That means that if you start with any of these nine states, you can get to a desired end state. Also note that where two elements are equal, they stay equal throughout the cycle.

That leaves 0 1 2 as the only state that does not have a solution.

Now I need to show that two of the populations can actually be driven to 0 (real 0).

Each member of each cycle has (at least) two elements that are equivalent mod 3 (call their corresponding real populations A and B, the other one will be C). That means their difference is a multiple of 3. Take that difference and divide it by 3 (call it N). Do N operations which combine the larger of A and B with C. The larger of A and B will decrease by N/3. The smaller of A and B will increase by 2N/3. A and B are now equal (C doesn't matter). Do A operations which combine A and B. A and B will both be driven to 0.

Finally, due to the algorithm described in the previous paragraph and the underlined statement above, the surviving type will be the one with the initial population that is not equivalent mod 3 to the other two populations. When all three populations are equivalent mod 3, then any type can be chosen to survive.

Error

Is this a riddle?

FRANK'S CANDIES
I started to fill out a table on this. Looks too complicated for me to figure out a formula. Looks too time consuming to calculate the entry in the fifth row and tenth column. I give up.

Twelve soldiers had to get to a place twenty-five miles distant with the quickest possible dispatch, and all had to arrive at exactly the same time. They requisitioned the services of a man with a small car.
"I can do ten miles an hour," he said, "but I cannot carry more than four men at a time. At what rate can you walk?"
"All of us can do a steady three miles an hour," they replied.
"Very well," exclaimed the driver, "then I will go ahead with four men, drop them somewhere on the road to walk, then return and pick up four more (who will be somewhere on the road), drop them also, and return for the last four. So all you have to do is keep walking while you are on your feet, and I will do the rest."
They started at noon. What was the exact time that they all arrived together?

Merlin works his magic:

Box:
The volume of the box is reduced by 87.5%

Calculator:
He will have $9.76 after his purchase.
What he meant to say was: $9.67

CHESS clarification. Corners five moves each way. Middle Three each.

Mark:
ACHELEONS
"The answer is no for the given populations."
He is a genius.

Another way to look at the problem:
Let a, b, c be the number of azure, blue and cream respectively, then a necessary and sufficient condition that we may reduce to a single color is that some two of a,b,c be congruent mod 3. From this, it follows that the desired reduction is impossible.

Necessity is obvious, since 2 = -1 (mod 3), thus the given operation (of touching and color change) maps each of a,b,c to an element equal to the (previous value)-1 (mod 3), thus two of a,b,c are equal mod 3 after the operation if and only if they were before. Since the desired final state consists of two 0 values, it follows that at least two of the three must have been equal mod 3 initially.

This is enough to solve our problem, but let's go further and prove sufficiency. Sufficiency follows from induction. The truth is obvious if max(a,b,c)=1. Assume that the statement is true for max(a,b,c)=n, and note that we may assume that no two of a,b,c are equal, since if they are, we may easily reduce to the desired final state. Assume without loss of generality that a>b>c and that a=n+1. We then have that a=n+1, b ≤ n, and c ≤ n-2; note that c ≤ n-2 since a>b>c and we can't have a=n+1, b=n, c=n-1, since this does not have two of a,b,c congruent mod 3. Since a,b>0 we may pair one azure and one blue to form two cream, giving the state a-1, b-1, c+2, but now max(a-1,b-1,c+2) = n since a-1=n, b-1 ≤ n-1 and c+2 ≤ n. By the above observation two of a-1,b-1,c+2 are congruent mod 3 if and only if two of a,b,c are congruent mod 3, and since the statement is true for max(a,b,c)=n, the truth for n+1 now follows.

Another way of saying this is that if a,b,c are not all different mod 3, either two of a,b,c are equal, or we may reduce max(a,b,c). This is why induction works.

As for the final color, in the case where not all of a,b,c are equal mod 3 the final color must clearly be the one which was initially unique mod 3. If a,b,c are initially equal mod 3, then it can easily be proven by induction that the final color may be any unless two of a,b,c are initially 0.


FRANK'S CANDIES
I said they were just about impossible. Food for thought though.

See what you think of this:

Let D(n) be the number of different flavors that can be made with n candies, and let E(n) be the number of different flavors that can be made with n candies, but not with any smaller number. The number we are seeking is F = E(1) + E(2) + ... + E(10). Note that D(n) = Sum E(d), where the sum is over all d such that d divides n. This gives us that D(6) = E(1) + E(2) + E(3) + E(6), D(7) = E(1) + E(7), D(8) = E(1) + E(2) + E(4) + E(8), D(9) = E(1) + E(3) + E(9), D(10) = E(1) + E(2) + E(5) + E(10). Thus, we can express F as D(6) + D(7) + D(8) + D(9) + D(10) - 4*E(1) - 2*E(2) - E(3) = D(6) + D(7) + D(8) + D(9) + D(10) - D(3) - 2*D(2) - D(1).

To find what D(n) is, imagine we have 4 sticks and n flavorless candies. Arrange the sticks and candies in a straight line; this can be done in (n+4) choose 4 = (n+4)!/(n!4!) ways. Note that each such arrangement corresponds to a different flavor by the simple mapping of letting all the candies to the left of the first stick be apple, the candies between the first and second sticks be banana, etc. This is a standard discrete math technique.

It follows that F = 14C4 + 13C4 + 12C4 + 11C4 + 10C4 - 7C4 - 2*6C4 - 5C4 = 2681.


Error
"Is this a riddle?"

It is a ?'Riddle' to me why I don't make more. As it happens, I wanted to edit the previous entry and pressed ?'Quote' in error. Then the system would not let me delete it, as you cannot post a blank quote. Oh well, you did ask.


"Twelve soldiers"
The Quartermaster is on the case.




Jane says, "I'm the 6th child in my family, and I have at least as many brothers as sisters." Her brother Jim adds, "I have at least twice as many sisters as brothers."

How many boys and girls are in their family


Suppose you have 5 weights, weighing 1, 3, 9, 27, and 81 grams.
How can you balance a 100 gram object using some or all of the other weights


My first three and last three are the same, and many people stay with me when they die.
Who am I

Hint: You're looking for an 11-letter word.

JANE
3 boys, 4 girls

WEIGHTS
81, 27, 1 on one side; 9 on the other side

Mark:
JANE
3 boys, 4 girls

WEIGHTS
81, 27, 1 on one side; 9 on the other side


Twelve soldiers:
The nearest I got (before I lost the plot) was 5 hours 37 minutes.

I would suggest you let this one run, to see if any other answers are forthcoming. It's good.


Inside a hat are cards numbered from 1 to 12. Al, Bo, Cal, Dan, and Eli each pick two cards without replacement and add the numbers on their cards. Two cards remain in the hat. Al's total was 11, Bo's total was 4, Cal's total was 16, Dan's total was 7, and Eli's total was 19.

What are the numbers picked by each participant


The pattern AABBBCCCCAABBBCCCC continuously repeats.
What is the 2003rd letter in the pattern

HATS
A: 4,7
B: 1,3
C: 6,10
D: 2,5
E: 8,11

PATTERN
The third B

Try:
Twelve soldiers:
The nearest I got (before I lost the plot) was 5 hours 37 minutes.

Sorry, try again.

I will follow your suggestion and leave this for a few days.

Mark:
HATS
A: 4,7
B: 1,3
C: 6,10
D: 2,5
E: 8,11

PATTERN
The third B (Because 9 and 12 remained in the bag).

"Sorry, try again."

Your sympathies are well received. The quest for a simple formula is continuing, although I hold deep suspicion that you are seeking revenge for the racing car fiasco. You know I have difficulty with objects travelling in the same direction. Therefore, to have them coming and going at the same time may well set back my recovery by some time. :wink:


A raffle was held and 1,200 tickets were sold for $2.50 each.
There were 17 winners. The first prize winner received $1,000.
The four second prize winners each received $250.
The remaining winners each received $50.

What percent of the total ticket sales was profit


The length of a rectangle is 3 times longer than the width.
If the perimeter of the rectangle is 96 cm, what is the area of the
rectangle


What is the largest two-digit prime number whose digits are both prime numbers


Molly is playing the videogame Space Squeakers, and she finally loses her last turn after several hours (and after killing numerous invading space mice). She was rather surprised to discover that she had scored the maximum number of points she could have while averaging exactly 9975 points per turn.

If in Space Squeakers you start with 3 turns and you earn an extra turn with every 10000 points you score (e.g. you earn an extra turn at 10000, another at 20000, another at 30000 etc.), what was Molly's final score?

Optional: For a gold star, describe the set of scores Molly could have achieved while still averaging 9975 points per turn.

RAFFLE
13.3%

RECTANGLE
432 sq cm

PRIME
73

hello there I have also question tha i myself cannot answer,

so if anyone can convince me i'll be pleased

-you have a bag with a white ball in it, when blinded you put either a red ball OR another with ball in it.
so you have 2 balls, now you grap blinded a ball out of the bag this one is white.

what is the chance that the second ball is also white?????

White balls:
2/3

Ball 1 Ball2
White Black
or
White White

Case1: Draw Ball1 (White), Ball2 (Black) in bag
Case2: Draw Ball1 (White), Ball2 (White) in bag
Case3: Draw Ball2 (Black), Ball1 (White) in bag
Case4: Draw Ball2 (White), Ball1 (White) in bag

You have a 1/4 chance of pulling a black ball. We eliminate that case. Out of the three remaining cases two result in a white ball remaining in the bag.

MOLLY
11,970,000
She could have scored 9975*N, 801 <= N <= 1200

that looks like a fine answer to me

Magnum, good to see you still remember us. As you live the nearest to ?'Santa Land' perhaps you could put in a good word with the man in the red suit.

Merlin, your answer was a model of clarity, to which we should all aspire.

Mark:
RAFFLE
13.3%

RECTANGLE
432 sq cm

PRIME
73

MOLLY
11,970,000 , ,
She could have scored 9975*N, 801 <= N <= 1200

Let T be the total number of turns Molly had (including the initial 3), and let S be her final score. Then by the rules of the game we have that S = 9975*T and T = 3 + floor[S/10000], where floor[x] is the greatest integer less than or equal to x. This gives that T = 3 + floor [9975*T/10000] = 3 + floor[(10000*T-25*T)/10000] = 3 + T + floor[-25*T/10000] = 3 + T - ceil[T/400], where ceil[x] is the smallest integer greater than or equal to x. We therefore get that ceil [T/400] = 3, and it follows that we must have 801 ≤ T ≤ 1200. Since Molly scored the maximum number of points, T = 1200, and so her final score was 9975*1200 = 11,970,000.


Soldiers:

I have it at last; 5 hours 46 minutes +/- 2 hours. (That should cover any correct answer).

For those that missed Mark's challenging problem:

Twelve soldiers had to get to a place twenty-five miles distant with the quickest possible dispatch, and all had to arrive at exactly the same time. They requisitioned the services of a man with a small car.
"I can do ten miles an hour," he said, "but I cannot carry more than four men at a time. At what rate can you walk?"
"All of us can do a steady three miles an hour," they replied.
"Very well," exclaimed the driver, "then I will go ahead with four men, drop them somewhere on the road to walk, then return and pick up four more (who will be somewhere on the road), drop them also, and return for the last four. So all you have to do is keep walking while you are on your feet, and I will do the rest."

They started at noon. What was the exact time that they all arrived together



The mean high temperature in Honolulu for the first twenty-one days of this month was 78 degrees. If the mean high temperature for the first 24 days of the month is 80 degrees,

What must the mean high temperature be between 22nd to 24th


The ratio of boys to girls in Bill's class is 3 to 4.
If there are 28 students in the class, how many of them are girls


What is the surface area of a rectangular prism with dimensions 6 ft by 8 ft by 15 ft


Spike is taking a series of exams, and it turns out that he'll have to score a 97 on the last one in order to average 90 for the entire series. But even if he scores as low as a 73, he'll still average an 87.

How many exams were in the series

mean temp - 94
girls - 16

oh, and poor Spike had to do eight exams.

However, as a pacifist, I refuse to solve military problems (and my religion bans prisms and I forgot formulas too)