The worlds first riddle!

SYSTEMS
a) y=-1/(2x) + 1/2 (I'm guessing there was a missing operator between x and y in the second equation)
b) x=3, y=6, z=-4
c) w=-1, x=2, y=3, z=2

ah, markr was quicker and I was so proud with myself for solving b)

DINNER
If the smaller table has an even number of people, the answer is always 3.
Smaller table: 2P people (divide into two groups: p1, p2)
Larger table: 2P+Q people (divide into three groups: p3, p4, q)

Note: p1=p2=p3=p4

Meal 1: p1/p2 - p3/p4/q
Meal 2: p1/p3 - p2/p4/q
Meal 3: p2/p3 - p1/p4/q

I'm headed out for an all day activity. I'll attempt to follow up later with a 2P+1, 2P+Q solution.

Enjoy your day out, while I catch up with some answers.

DINNER
If the smaller table has an odd number of people, and the larger table has twice as many or more people, then the answer is 3.

Smaller table: 2P+1 people (one group: p1)
Larger table: 4P+2+Q people (divide into three groups: p2, p3, q)

Note: p1=p2=p3

Meal 1: p1 - p2/p3/q
Meal 2: p2 - p1/p3/q
Meal 3: p3 - p1/p2/q

Otherwise, the answer can be at least as large as 4 (I don't think it is ever greater than 4), but I don't know what conditions force that.

SYSTEMS
If the second equation of (a) was supposed to be
2x+y+3z=0
then the solution is
x=1, y=-2, z=0

If the second equation of (a) was supposed to be
2x-y+3z=0
then the solution is
x=-2, y=-1, z=1

Shock - horror, I thought the questions would last until the middle of next week at last! I was wrong again. Even without Paula's help you sure fixed my wagon.

Ps. Is it true that Paula is ?'walking out' with Francis? Oh sachet blue.



Mark:
BOAT
3:36:50.07pm

the boat will be closest when the line from the radar station to the boat is perpendicular to the path of the boat. That boat will reach that point when it travels 12.5 * cos(57.3) = 6.75 mi about 6.75 mi since the boat is travelling 11 mph the boat travels that distance in;
6.75/11 = 0.613636363636 hours-about 36.8 minutes.

Mark has a better stopwatch.




SYSTEMS
I don't claim to be right, but where we agree, we must be right.

Mark:
a) y=-1/(2x) + 1/2 (I'm guessing there was a missing operator between x and y in the second equation)

Well, yes, and no.
x + y + 2z = -1
2x + y + 3z = 0
-y + z = 2
z = 2 + y
x + 2y + 2(2 + y) = -1
2x + y + 3(2 + y) = 0
x + 2y + 4 + 2y = -1
2x + y + 6 + 3y = 0
x + 4y = -5
2x + 4y = -6
x = -1
y = -1
z = 1

Mark:
b) x=3, y=6, z=-4

MYO:
"markr was quicker and I was so proud with myself for solving b"

So you should be, they were far from easy.


x + y + 2z = 1
3x - y + z = -1
-x + 3y + 4z = -1
x + y + 2z = 1
-x + 3y + 4z = -1
4y + 6z = 0
y = (-3/2)z
x - (3/2)z + 2z = 1
3x + (3/2)z + z = -1
x + (1/2)z = 1
3x + (5/2)z = -1
3x + (5/2)z = -1
3x + (3/2)z = 3
z = -4
x = 3
y = 6

c) w=-1, x=2, y=3, z=2



z - 2w = 3
-x + y + z + w = 2
-x + 2y + 2z - w = 9
x - y + 2z + w = 2
z = 2w + 3
-x + y + 2w + 3 + w = 2
-x + 2y + 2(2w+3) - w = 9
x - y + 2(2w+3) + w = 2
-x + y + 3w = -1
x - y + 5w = -4
8w = -5
w=-5/8
z = 7/4

-x + y - 15/8 = -1
-x + 2y - 15/8 = 3
y = 4
x = 3 1/8






Mark:
NECKLACE
If I kept up with the problem correctly, there were 128*1161=148608.

You sure did.

let x = number of pearls on the necklace
(1/3) x collected by the maid servant
(1/6) x on the bed
(1/3)x + (1/6) x = (1/2) x
(1/4) x + (1/8) x + (1/16) x + (1/32) x + (1/64) x + (1/128) x : the six scatterings
x = (1/2) x + (1/4) x + (1/8) x + (1/16) x + (1/32) x + (1/64) x + (1/128) x + 1161
x = (127/128) x + 1161
(1/128)x = 1161
x = 128 * 1161 = 148608 pearls originally on the necklace


Mark:
DINNER
If the smaller table has an even number of people, the answer is always 3.
Smaller table: 2P people (divide into two groups: p1, p2)
Larger table: 2P+Q people (divide into three groups: p3, p4, q)

Note: p1=p2=p3=p4

Meal 1: p1/p2 - p3/p4/q
Meal 2: p1/p3 - p2/p4/q
Meal 3: p2/p3 - p1/p4/q

If the smaller table has an odd number of people, and the larger table has twice as many or more people, then the answer is 3.

Smaller table: 2P+1 people (one group: p1)
Larger table: 4P+2+Q people (divide into three groups: p2, p3, q)

Note: p1=p2=p3

Meal 1: p1 - p2/p3/q
Meal 2: p2 - p1/p3/q
Meal 3: p3 - p1/p2/q

Otherwise, the answer can be at least as large as 4 (I don't think it is ever greater than 4), but I don't know what conditions force that.

Mark, you continue to astound.




Assume throughout that N >= M. Let's dispose of some strange cases first. If M=0 and N is either 0 or 1, then no sessions are required, since there are no pairs of people to worry about. If M=0 and N >= 2 then one session is required. If M=N=1 the problem is impossible: the two people can never sit together. So we will assume from here on that N >= M >= 1 and (N,M) not equal to (1,1) . If 2M>N >= M >= 1 and both M and N are odd, you need exactly four sessions.

In all other cases you need exactly three sessions. Having disallowed the strange cases, you always need at least three sessions. That's because (in an optimal setup, using the least number of sessions) the arrangement on the first session is necessarily different from that of the second session, so there is one diner x who was at the N table the first night and the M table the second night. There is another diner y who sat at the M table the first night and the N table the second night. You need a third session for x and y to sit together. (You may need more.) CASE 1: N >= 2M Groups (A,B,C,D) of size (N-2M,M,M,M) .

• 1 session: (ABC)(D) at N table and M table respectively
• 2 session: (ABD)(C)
• 3 session: (ACD)(B)

CASE 2: N<2M ; M even Groups (A,B,C,D) of size (N-M/2,M/2,M/2,M/2)

• 1 session: (AB)(CD)
• 2 session: (AC)(BD)
• 3 session: (AD)(BC)

CASE 3: N<2M ; N even Groups (A,B,C,D) of size (M-N/2,N/2,N/2,N/2)

• 1 session: (BC)(AD)
• 2 session: (BD)(AC)
• 3 session: (CD)(AB)

CASE 4: N<2M ; both M and N odd This is the only case that requires four sessions. First let's show that three sessions do not suffice for this case. Without loss of generality the first two sessions use groups (A,B,C,D) of size (X,X,N-X,M-X) for some integer X with 0 <= X <= M .

• 1 session: (AC)(BD)
• 2 session: (BC)(AD)

If all four groups are nonempty, and if only three sessions are to be used, then the third session must seat (AB) together and must also seat (CD) together. (AB) is of size 2X , which is even, so can't fully occupy one table. Neither can (CD). So the third session cannot satisfy everyone. On the other hand if one of the groups is empty, then either X=0 or X=M . If X=0 then the first two sessions are identical, impossible. If X=M then we really had groups (A,B,C) of size (M,M,N-M) , and

• 1 session: (AC)(B)
• 2 session: (BC)(A)

and the third session would have to put (AB) at one table, but the size of (AB) is 2M , which is larger than N . Therefore you need at least four tables for this case. We cannot have M=1 , since (N,M)=(1,1) is ruled out and we have assumed N<2M . So M is at least 3. Create seven groups (A,B,C,D,E,F,G) of sizes |A|=|B|=|C|=1 , |D|=(M-3)/2 , |E|=|F|=(M-1)/2 , |G|=N-(M+1)/2 .

( The various inequalities imply that all these sizes are nonnegative integers.)

• 1 session: (ACDG) (BEF)
• 2 session: (AEG) (BCDF)
• 3 session: (BFG) (ACDE)
• 4 session: (CEG) (ABDF)

So this case can be done with four sessions. A related problem from information theory is how to arrange that each pair avoids each other at least one meal, and there the answer is related to \log2(M+N) .






Five people go to a French restaurant. Not being familiar with such food, they do not recognize any of the names for the dishes. Each orders one dish, not necessarily distinct. The waiter brings the dishes and places them in the middle, without saying which is which. At this point, they may be able to deduce some.

For example, if two people ordered the same item, and everyone else ordered different dishes, then the item of which two copies arrive must be the one of which two were ordered.

They return to the restaurant two more times, following the same drill, though with different orders. After three meals, they have eaten all nine items on the menu, and can tell which is which.

What pattern of ordering fulfils this

What if the number of items on the menu were ten instead of nine




A "directed graph" consists of a collection of "vertices" {A,B,C,....} and "edges" {(B,C),...}.

An edge goes from one vertex to another: (B,C) goes from the source B to the destination C. There may or may not be another edge from C to B. In the present problem no edge will go from a vertex to itself (loops), nor will two edges have the same source and same destination (multiple edges).

A "path of length two" from B to C is a pair of edges (B,D) and (D,C) where the destination of the first edge is the source of the second.
Our particular directed graph has between 30 and 40 vertices, inclusive. For any two nodes B and C, there is either an edge (B,C) or exactly one path of length two from B to C, but not both.

This is true even if C=B: there is exactly one path of length two from B to B.


The question: How many vertices are in the graph



Phew! Time to relax;


Of no use to one
Yet absolute bliss to two.
The small boy gets it for nothing.
The young man has to lie for it.
The old man has to buy it.
The baby's right,
The lover's privilege,
The hypocrite's mask.
To the young girl, faith;
To the married woman, hope;
To the old maid, charity.

What am I



Who makes it, has no need of it.
Who buys it, has no use for it.
Who uses it can neither see nor feel it.

What an I



What book was once owned by only the wealthy, but now everyone can have it? You can't buy it in a bookstore or take it from a library?

What am I

Tryagain wrote:

"Ps. Is it true that Paula is ?'walking out' with Francis? Oh sachet blue."


Paula writes:---> .......<scratches head>

Tryagain, if you find out if that is true or not...would you tell me? I'm always the last to know.
-------------------------------------------------------------------------------------

Tryagain wrote:

Of no use to one
Yet absolute bliss to two.
The small boy gets it for nothing.
The young man has to lie for it.
The old man has to buy it.
The baby's right,
The lover's privilege,
The hypocrite's mask.
To the young girl, faith;
To the married woman, hope;
To the old maid, charity.

What am I?

Love?

FRENCH FOOD
AADEG
BBDFH
CCEFI

GRAPH
36 (guess)

BOOK
telephone book?

Paula (taking a leaf from Enid Blighton) wrote, "if you find out if that is true or not...would you tell me?"

My dear Paula, I have been reliably informed that Francis offered you his ?'arm' and you readily accepted.

Well honey, I gotta tell you… that aint his arm.


Of no use to one
Yet absolute bliss to two.
The small boy gets it for nothing.
The young man has to lie for it.
The old man has to buy it.
The baby's right,
The lover's privilege,
The hypocrite's mask.
To the young girl, faith;
To the married woman, hope;
To the old maid, charity.

What am I?


Love?
You are on the right track. Sorry for the derailment. :wink:




What book was once owned by only the wealthy, but now everyone can have it? You can't buy it in a bookstore or take it from a library?

Mark:
BOOK
telephone book


FRENCH FOOD
AADEG
BBDFH
CCEFI



At the first meal, order dishes A,B,B,C,D.
At the second meal, order dishes D,E,E,F,G.
At the third meal, order dishes G,H,H,I,A.

Dish "A" is the only dish that appears at both the first and third meals; dish "B" is the only one that appears twice at the first meal; dish "C" is the only one that appears once at the first meal and never again; and so on.
To show that ten dishes is impossible, assign each dish a triple of non-negative integers (x,y,z) if it is ordered x times at the first meal, y times at the second, and z times at the third. (x,y,z) cannot be (0,0,0) because of the stipulation that the diners have eaten all the items on the menu. The triples (x,y,z) corresponding to two different dishes must be different, or else the diners would never learn to distinguish those dishes. There are only three triples whose sum x+y+z is 1, namely (1,0,0), (0,1,0), and (0,0,1). There are only six triples whose sum x+y+z is 2, namely (2,0,0), (0,2,0), (0,0,2), (1,1,0), (1,0,1), and (0,1,1).

On the nine-dish menu, these nine patterns can be used, giving a total of 1+1+1+2+2+2+2+2+2=15=5*3 dishes ordered by the five diners at three meals.
(In this solution they correspond to dishes C,F,I,B,E,H,D,A,G, respectively.)
But on a ten-dish menu, an additional pattern must be used, so that the total number of dishes ordered must be at least 1+1+1+2+2+2+2+2+2+3=18, too much food for five diners in three meals.



The question: How many vertices are in the graph?

GRAPH
36 (guess)

That was a close ?'guess', more like an educated estimate.

From each vertex B to each vertex C there is either a one-step path (a single edge) or a unique two-step path. But we'll analyze the problem by looking at three-step paths as well. Fix two vertices B and C. Suppose there are k different vertices H such that (H,C) is an edge. Call these vertices the "predecessors" of C. Suppose also that there are m different vertices G such that (B,G) is an edge; call these vertices the "followers" of B. How many paths, of length either two or three (but not one) are there from B to C? First answer: there are m such paths. Each two- or three-step path from B to C consists of an edge from B to some vertex G, followed by the unique one- or two-step path from that vertex G to the destination C. Since there are m edges leading out of B, and each can be completed to exactly one path of length 2 or 3 to C, there are m such paths. Second answer: k. Each two- or three-step path from B to C is a one- or two-step path from B to one of C's predecessors H, followed by a one-step path from that vertex to C. Each of the k predecessors of J accounts for exactly one such path. So we must have k=m. By the same reasoning, each vertex D has exactly k two- or three-step paths to C, and so has exactly k successors. Similarly we can see that each vertex has exactly k predecessors. (In graph terminology, each vertex has in-degree k and out-degree k.) Now consider one- or two-step paths starting at B. There are exactly k vertices H reachable by a one-step path. Since each of these vertices also has exactly k followers, there are exactly (k*k) vertices reachable from B via two-step paths. This accounts for all the vertices, so the total number of vertices is k+(k*k). We know there are thirty to forty vertices, so we must have k=5 and the number of vertices is thirty.

It turns out (though this is harder to prove) that the vertices can be given consistent labels of the form Vij, where i and j are two different indices from the set {1,2,...,k+1}, and where an edge goes from Vij to Vkh if and only if j=k. Illustrating in a smaller case k=2, we would have 6=2+4 vertices V12, V13, V21, V23, V31, V32, and twelve edges (two coming out of each vertex): (V12,V21), (V12,V23), (V13,V31), (V13,V32), (V21,V12), (V21,V13), (V23,V31), (V23,V32), (V31,V12), (V31,V13), (V32,V21), (V32,V23).





Consider the integer N=2^1999 (2 raised to the 1999 power).
Is there a positive integer multiple of N whose decimal representation does not contain the digit 0



We are playing baseball on an airless planet. A fly ball is hit directly towards an outfielder. The fielder can detect the changing angle that the ball makes with the horizon, but cannot directly judge changes in distance (apparent size of the ball), and there is no apparent lateral motion.

How does he decide whether to move forward or backward to catch the ball



Suppose C is a positive real number such that for all positive integers N is it true that (N raised to the C power) is an integer. Does C have to be an integer?

Why or why not

Me----> Are,...you saying...what I think ...your saying?

Francis is a gentleman,...isn't he?

2^1999
Since you're asking, I'll guess yes.

N^C
I think C has to be an integer. If it is not, then I don't think 2^C is an integer.


BASEBALL
Can he determine the rate of change of the angle that the ball makes with the horizon?

Does he know how far he is from home plate?

Is the batter on steroids? Was he created, or did he evolve? Is the president trying to keep his wife on life support?

Please let me know if any of the above questions are irrelevant, as that will keep me from heading in a wrong direction.

Paula, I am sure Francis is as Parisian as French Baguette. However, have you seen the size of those baguettes?




Consider the integer N=2^1999 (2 raised to the 1999 power).
Is there a positive integer multiple of N whose decimal representation does not contain the digit 0?

Mark:
2^1999
Since you're asking, I'll guess yes.

Your guesses are as good as the answers. (Well, almost)


Yes. Work from right to left. Start with N, and notice that its rightmost digit is not zero. Find the rightmost zero; say it's in the Kth position. Then add (10^K)*N (that is, N shifted over K positions) to change that digit without affecting any to the right of it. Now find the next zero, and again add (10^J)*N for some appropriate J. With each iteration, the rightmost zero moves further to the left. Continue until the rightmost zero is at least 1999 places to the left (or there are none). Now our integer is (10^1999)*A + B, where B has no zeros. Subtracting off (10^1999)*A = (2^1999)*(5^1999)*A, we are left with B, which is a multiple of (2^1999) with no zeros in its decimal representation.


We are playing baseball on an airless planet. A fly ball is hit directly towards an outfielder. The fielder can detect the changing angle that the ball makes with the horizon, but cannot directly judge changes in distance (apparent size of the ball), and there is no apparent lateral motion.
How does he decide whether to move forward or backward to catch the ball?

BASEBALL
Can he determine the rate of change of the angle that the ball makes with the horizon?

Does he know how far he is from home plate?

Is the batter on steroids? Was he created, or did he evolve? Is the president trying to keep his wife on life support?

Please let me know if any of the above questions are irrelevant, as that will keep me from heading in a wrong direction.

Good questions, and vital to formulate the answer.

If he is correctly positioned, the derivative of (tangent theta) with respect to time is constant. Put another way, there are constants c and d such that tan(theta)=c*t+d where t=time. This is because the ball's height (above the fielder) is a quadratic function of time, with value 0 when t=0 (i.e. when it reaches the fielder), and its horizontal distance from the fielder is a linear function of time, also achieving the value 0 when t=0. Dividing, we see that the slope is of the form c*t+d.



Suppose C is a positive real number such that for all positive integers N it is true that (N raised to the C power) is an integer. Does C have to be an integer?
Why or why not?

Mark:
N^C
I think C has to be an integer. If it is not, then I don't think 2^C is an integer.


The answer is that C must be an integer. I will assume that C is a non-integer that lies between 3 and 4. The proof generalizes but the notation gets harder. Pick an arbitrary large integer N. Consider the function f(x)=x**C, by which we denote "x raised to the C power". We will take the Taylor expansion of f(x) around the point N, and evaluate f(x) at the various points N+i as i=0,1,...,4. By hypothesis, since N+i is a positive integer, so is f(N+i). We will take a certain integer linear combination of the values f(N+i), concocted to make most of the terms of the Taylor expansion cancel out, and we will end up with a quantity which is supposed to be an integer, but instead lies strictly between 0 and 1.

The Taylor expansion tells us f(N+i) = f(N) + f'(N)i + f''(N)(i**2)/2 + f'''(N)(i**3)/6 + f''''(N)(i**4)/24 + f'''''(xi)(i**5)/120, where xi is some number between N and N+i. Now multiply the integers f(N+i) by (1, -4, 6, -4, 1), respectively, and sum. So we have: 1*f(N+0) - 4*f(N+1) + 6*f(N+2) - 4*f(N+3) + 1*f(N+4). On the one hand, we are multiplying integers by integers, so our resulting sum must be an integer. On the other hand, the sum comes out as 1*f(N) - 4*(f(N) + f'(N) + f''(N)/2 + f'''(N)/6 + f''''(N)/24 + f'''''(x1)/120) + 6*(f(N) + 2 f'(N) + 4 f''(N)/2 + 8 f'''(N)/6 + 16 f''''(N)/24 + 32 f'''''(x2)/120) - 4*(f(N) + 3 f'(N) + 9 f''(N)/2 + 27 f'''(N)/6 + 81 f''''(N)/24 + 243 f'''''(x3)/120) + 1*(f(N) + 4 f'(N) + 16 f''(N)/2 + 64 f'''(N)/6 + 256 f''''(N)/24 + 1024 f'''''(x4)/120). Now, many of the terms cancel. The coefficient of f(N) is (1-4+6-4+1)=0. The coefficient of f'(N) is (0-4*1+6*2-4*3+1*4)=(-4+12-12+4)=0. Similarly the coefficients of f''(N) and f'''(N) both vanish. The f''''(N) term has a coefficient of 1: (0 - 4*1 + 6*16 - 4*81 + 1*256)/24 = 24/24 = 1.

The error terms are all constant multiples of f'''''(xi). Now we know that f''''(N) = C*(C-1)*(C-2)*(C-3)*N**(C-4), and since C is between 3 and 4, we have that N**(C-4) grows smaller as N grows larger. For N large enough, f''''(N) is strictly between 0 and 1/2. (The coefficient C*(C-1)*(C-2)*(C-3) doesn't vanish because C is not an integer; this coefficient is some constant E independent of N.) The error term, involving the fifth derivative of f at various places, is bounded by some constant times N**(C-5). So the total sum is a nonzero constant times N**(C-4), plus an error term bounded by another constant times N**(C-5). For N large enough, this sum is bounded strictly away from 0, but smaller than 1 in absolute value. So we have an integer lying strictly between 0 and 1 (or between 0 and -1). By this contradiction we see that such a C cannot exist.





Let's borrow notation from LaTeX and let "^" denote exponentiation, so that x^2 means "x squared".

Consider the numbers 1233 = 12^2 + 33^2, and 990100 = 990^2 + 100^2.

Can you find an eight-digit number N with the same property, namely that if you break N into two four-digit numbers B and C, and add their squares, you recover N

(N really has to have eight digits; its leading digit is not zero. B is the number formed by the first our digits of N, and C is the number formed by the last four digits of N.)





I have decided to revamp the tax code in a major way. All residents will line up, left to right, with Sam standing at the far left at position 0 and the rest of us in positions 1, 2, 3, etc. On January 1 of each year, each resident will observe the net worth of the person on his left. On April 15, this resident pays that much in tax out of his own pocket. (So if Jack is at position 20 and Amanda at position 19, on April 15 Jack will pay the amount that Amanda had on January 1.)

"But what if I don't have that much money?"
Just run a deficit. Does that bother you?

"But what if my neighbor is running a deficit on January 1?"
If your neighbor was running a ten dollar deficit on January 1, then on April 15 you give the government a ten dollar deficit. That's the same as if the government pays you ten dollars.

Sam: "I don't have a neighbor, since I'm at position 0."

That's okay, Sam, you alone don't have to pay taxes. Just pretend your non-existent neighbor has zero net worth.

"What if people change places, drop out, come into the system...?"

None of that is allowed. Furthermore, no money ever changes hands except for payment of taxes. Okay, no further questions.

"Wait, wait. Why are we doing a US-centric April 15 tax day, if IBM is an international company?"

I said no further questions.

Sam starts out with one dollar, and the rest of you start with varying amounts. Some of you will start with no money, in particular the four people on the far right-hand end of the line will be broke. Some of you might even start with deficits.

Four years pass. At the end of those four years, Sam still has his one dollar, and the only other people with nonzero net worth are at positions q, r, s and t, with q < r < s < t. Here q,r,s,t are unknown integers, not assumed to be consecutive, and the answer to the problem will involve only the values q,r,s,t.

How much did the government collect in the first year (net)
(That is, collections minus payouts.) Guesses are most acceptable.

8-DIGIT NUMBER
94122353

Need some time on the other.

Tryagain Wrote:
"Paula, I am sure Francis is as Parisian as French Baguette. However, have you seen the size of those baguettes?"

No I have not. Are you saying,...Francis is like a baby in a topless bar, and I accidently started dancing, with said baby?

I'm confused, again, Try.

Since you've been such a bloomin' gem at helping people with your resplendent knowledge, I thought I would bring you a gift.

Tryagain, this budding gem is for you. I hope you like it, it was all I could afford.

TAX CODE
I've only been able to come up with instances where the four non-zero people are at the far right. I'm not yet sure if that is the only possibility.
Anyway, in that case, the answer is:
SUM(i=3 to s, C(i,3))
= (s^4 - 2s^3 - s^2 + 2s) / 24

Here are a couple of variations:

How many (and what are they) solutions are there to the following:

ABC = A^3 + B^3 + C^3 (A, B, and C are 1-digit numbers)

ABCDEF = AB^3 + CD^3 + EF^3 (AB, CD, and EF are 2-digit numbers)

Paula eloquently writes, "Are you saying,...Francis is like a baby in a topless bar, and I accidentally started dancing, with said baby?"


That may well be the case. However, I could not possibly comment. Where is the suspect? Sorry, I mean the great man. Or, how do you say in English, he should answer the charge, or is it the deposition of ?'a cradle snatcher'.

On a far happier note; I am not worthy of your most gracious gift, although it is warmly received and will look dazzling in my cave. I have to fess up to my knowledge being all smoke and mirrors.
(I still get to keep the ?'ice'… right?)




Mark:
8-DIGIT NUMBER
94122353



We have N=10000B+C and N=B^2+C^2.
So 10000B+C=B^2+C^2, 10000B-B^2 = C^2-C.
Also B must lie between 1000 and 9999, in order for N to be an eight-digit number. For each trial value of B in that range, compute f(B)=10000B-B^2.

Let c be the square root of f(B), rounded up to the nearest integer. If our equation is satisfied, we will have f(B)=c^2-c.
(The square root of f(B) will be about C-1/2, so c=C.) So we only have to do about 9000 trials to find the lone answer: B=9412, C=2353, N = 94122353 = 9412^2 + 2353^2. 2. A few readers reminded us how to solve this problem algebraically, without exhaustive search.

This write-up follows S V Nagaraj's solution.
Continue to let "^" denote exponentiation.
The solution is obtained by considering 10000B + C = B^2 +C^2,
Rewriting this as 10000^2 +1 = (2*B-10000)^2 +(2*C-1)^2,
we just look at the factorization of 10000^2 + 1 = 10^8 +1 = 17 *5882353.
Express each prime factor as the sum of squares:
17 = 4^2 + 1^2 = w^2 + x^2;
5882353 = 588^2 + 2353^2 = y^2 + z^2.
Then combine them according to the usual laws (related to the multiplication of complex numbers):
(w^2 + x^2)(y^2 + z^2) = (wy+xz)^2 + (wz-xy)^2
(w^2 + x^2)(y^2 + z^2) = (wy-xz)^2 + (wz+xy)^2
The second equation gives an uninteresting solution. The first gives us 4705^2 + 8824^2. Since 2C-1 is odd and 2B-10000 is even, we know which is which. 2C-1=4705 so that C=2353, and 2B-10000=8824 (or -8824) so that B is 9412 (or 588).
B=588 is ruled out by demanding a strict 8-digit number. So the answer is B=9412, C=2353, N=94122353.

Mark, your variations will require more than a bit of thought.


Mark:
TAX CODE
I've only been able to come up with instances where the four non-zero people are at the far right. I'm not yet sure if that is the only possibility.
Anyway, in that case, the answer is:
SUM(i=3 to t-1, C(i,3))

I did not actually expect anyone to attempt to answer such a question.


In the first year the government collects a net of qrst/24.
In subsequent years it collects a net of 0.

In general, if after n years the only nonzero amounts were at the n+1 positions 0, b1, b2, ..., bn, then the government's net take during the first year is b1*b2*...*bn/n!.

Consider the polynomial f(x) whose coefficient f of x^i is the original net worth of the resident at position i;
here and throughout, "x^i" denotes exponentiation.

Each year the effect of the taxation is to multiply f(x) by (1-x), since the new value of the coefficient f is the old value of f - f[i-1]. After four years, the new polynomial is g(x)=f(x)*(1-x)^4. We are told that g(x) has only five nonzero coefficients, including g[0]=1.

Let the nonzero coefficients be g[q]=b, g[r]=c, g[s]=d, g[t]=e. Because g(x) is divisible by (1-x)^4, we know that g and its first three derivatives all vanish at x=1: g(1)=g'(1)=g''(1)=g'''(1)=0.
The fourth derivative of g at 1 is equal to 24 times f(1), and in turn f(1) is the sum of the coefficients of f,
that is, the total initial wealth of the residents.
After one year, the total wealth of the residents is:
(f(x)*(1-x)) evaluated at x=1, that is, 0. The government essentially collects all the wealth in the first year.

So we have the linear equations (in b,c,d,e):
1 + b + c + d + e = 0
qb + rc + sd + te = 0
q(q-1)b + r(r-1)c + s(s-1)d + t(t-1)e = 0
q(q-1)(q-2)b + r(r-1)(r-2)c + s(s-1)(s-2)d + t(t-1)(t-2)e = 0
q(q-1)(q-2)(q-3)b + ... + t(t-1)(t-2)(t-3)e = 24*f(1).

Or, after subtracting constant multiples of some equations from
others, we get the simpler set of equations:

1 + 1*b + 1*c + 1*d + 1*e = 0
0 + q*b + r*c + s*d + t*e = 0
0 + q^2*b + r^2*c + s^2*d + t^2*e = 0
0 + q^3*b + r^3*c + s^3*d + t^3*e = 0
0 + q^4*b + r^4*c + s^4*d + t^4*e = 24*f(1).

The coefficients on the left-hand form a matrix M.
Applying M^(-1) to the vector (0,0,0,0,24*f(1))
we will recover 1 as well as the unknowns b,c,d,e.
We are interested in the upper right-hand entry of M^(-1)
(that is, its (1,5) entry).

By Cramer's rule, this is given by the ratio of two determinants:
in the numerator, the determinant of the upper right-hand 4x4
submatrix of M (up to sign),
1............ 1.............. 1............. 1
q............ r.............. s............. t
q^2........ r^2.......... s^2......... t^2
q^3........ r^3.......... s^3......... t^3

and in the denominator, the determinant of M itself, which is the same as the determinant of its lower right-hand 4x4
submatrix:
q............. r.............. s............. t
q^2......... r^2.......... s^2......... t^2
q^3......... r^3.......... s^3......... t^3
q^4......... r^4.......... s^4......... t^4


The latter differs from the former in that the first column has been multiplied by q, the second by r, and so on.
Putting it all together, we find
1 = (1/(q*r*s*t))(24*f(1)), or
f(1) = q*r*s*t/24 = the government's first-year profit.

Adapted from the 1986 Putnam Examination, problem A-6.



Anyone got a map?

Throughout, we will assume the earth is a sphere with a circumference of 25,000 miles.

Part 1:

Start at the equator where it crosses the prime meridian.
Head northwest, and continue to travel northwestuntil you reach the north pole.

How far will you travel


Part 2:

Where will you be when you first cross the prime meridian
(How far from the north pole?)


Part 3:

Where would you first cross the prime meridian if you always headed north-northwest (NNW, i.e. 22.5 degrees away from north) instead of northwest

Tryagain wrote:

"(I still get to keep the ?'ice'… right?)"

Of course you can keep it. Wear it well.

MAP
1) The duration of the trip is infinite. Unless I screwed something up, the distance is also. Here's what I got for the distance:

D = (12500/pi) * Integral(from 0 to infinity, sqrt(x^2 + 2)/(x^2 + 1)dx)

2) 627.990 miles from the pole (latitude = 80.957)
3) 261.925 miles from the pole (latitude = 86.228)

<edit: I'm pretty sure this is all wrong>