How is this one different from the earlier question? The only difference I noticed was Y being addressed as "a real number" rather than "a positive integer" in the previous case. If the answer to the previous case was 25, why would it be any different now?

I'm not a programmer so may be I am missing something in the question!!

There is only one answer in the positive integers. But there may be more in the reals.

In fact, there is at least one other.

The numbers 1, Y, and Y^2, Y a real

number, are used as the basis for a

sequence in which every value after the

first three is the average of the

previous three numbers. This is

continued indefinitely and the sequence

is found to converge to the value 321.

Find all possible values for Y.

SUPERPRISMATIC NOTE: This is much more

of a challenge for the programming

solvers out there than Penney Puzzle

#7 from yesterday was!

The numbers 1, Y, and Y^2, Y a positive

integer, are used as the basis for a

sequence in which every value after the

first three is the average of the

previous three numbers. This is

continued indefinitely and the sequence

is found to converge to the value 321.

Determine Y.

SUPERPRISMATIC NOTE: Not all numbers

in this sequence are integers! So,

programmers beware!

My pleasure. Again, that is a nice puzzle that you posted.

Here's a detailed approach

Let c = [ A, B, C, D, E, F]', and let C = [ 36.90, 20.05, 24.70, 37.75]'

Let A =

[ 2 5 10 7 8 1
1 9 3 1 0 4
6 1 3 2 4 5
3 2 5 12 8 1]

So far the information we have can be summed in the matrix equation A c = C. if we need to solve for c, we can't do it because we have 4 equations for 6 variables. Fortunately, we only need to find a sum of the variable, which requires less information.

Assume that there exist 4 row multipliers (a,b,c,d), such that

a* (2 5 10 7 8 1) + b*(1 9 3 1 0 4 ) + c*(6 1 3 2 4 5) + d*(3 2 5 12 8 1) = (1 1 1 1 1 1)

We can rewrite this as a matrix equation

(a b c d ) * A = (1 1 1 1 1 1)

This time, we have enough information for solve for (a, b, c, d), since we have 6 equations with 4 variables. The quick way is to do the QR decomposition. The slow way is to do the Gauss-seidel business. The solution for (a,b,c,d) is ( 0.015625, 0.078125, 0.125000, 0.046875 ). Multiply this with the prices and sum, and there you have it.

Yeah, that's the straightforward linear algebra approach I meant. I don't know why it didn't hit me when I made up the problem.

It took Kriil to jog the old cobwebs from the old thinkin' machine. Thanks to both of you.

Nice puzzle. We don't have enough information to determine the price of each type of candy, but we do have enough information for a sum.

7 dollars

Do you mind telling us how you approached the problem? Is it the straightforward linear algebra approach?

:Throws his linear algebra book at superprismatic:

Is there a way to solve this without using determinants? I almost gave myself an embolism when I wrote down the system of equations and started to solve using Kramer's rule. Nice problem.

Well, I know the answer because I made up the problem. To be honest, I don't know a good way to go about solving it. Part of the reason I put the problem out was to find out if anyone comes up with a nice approach to it.

Basically, the problem asks if (1,1,1,1,1,1) is in the vector space spanned by the 4 basis vectors and, if so, what linear combination of those basis vectors gives you (1,1,1,1,1,1).

Ain't Linear Algebra grand?

Sorry, I know an easy way to do it. It just didn't occur to me until now.

A store is selling 6 new kinds of candy

bars, call them A,B,C,D,E,and F. My

4 friends, bought some of the bars.

Each friend remembered how many of each

kind of bar he bought and the total

price he paid (there's no sales tax

in this ficticious state). Of each of

A,B,C,D,E,and F, my first friend bought

2,5,10,7,8,and 1 at a cost of $36.90;

my second friend got 1,9,3,1,0,and 4 at

a cost of $20.05; my third friend got

6,1,3,2,4,and 5 for $24.70; my fourth

friend got 3,2,5,12,8,and 1 for $37.75.

I'm more frugal than any of my friends,

so I would like to buy 1 of each of the

six bars to see which I like best.

How much would that cost me?

Here's a cute little word problem which, although

not difficult, is very hard for most people to parse:

"The ship is twice as old as the boiler was when the

ship was as old as the boiler is now. What are the

relative ages of the ship and the boiler?"

It can melt the neurons on some teenagers!

On a TV stunt show, a couple is given the

sequence of numbers 8,7,2,1,4,10,6,11,12,9,3,5

and required to straighten them out in the

following way. The husband is to interchange

pairs of numbers, never touching the same

number twice. His wife is then to take the

sequence her husband leaves and also make

interchanges of numbers, again without touching

the same number twice. When she finishes, the

numbers are to be in order from 1 to 12. The

husband begins by switching 1 and 8; how

should he continue?

SUPERPRISMATIC NOTE: No information is either

provided or implied about the number of swaps

each makes. Also, "Touching" a number means

using it in a swap. When the wife's turn begins,

she is allowed to touch any or all of the numbers

of the sequence she inherits from her husband's

turn, but no more than once each.

Nope.

Consider a unit cube with its vertices at (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), and (1,1,1).

Two of the diagonals in question are between the vertices of (0,0,0) and (1,1,1) and (0,0,1) and (1,1,0). Those four vertices define a rectangle with a height of 1, and a width of sqrt(2), so those diagonals are the diagonals of this rectangle as well. Since this rectangle is not a square, its diagonals are not perpendicular.

You are correct, but there's a much, much simpler explanation!

The diagonals of a square are mutually perpendicular. Is the same true for the spacial diagonals of a cube?

And it is now possible to determine that P can only be 5, which solves Sengupta's original puzzle.

We have already found from previous posts that, if

2^M - 3P = N^2, then M must be even so that, putting M = 2m

(2^m - N)(2^m + N) = 3P, an equation which can only be solved if

(a) 2^j - N = 3 and 2^j + N = P so that 2^(j+1) = P + 3

and/or

(b) 2^k - N = 1 and 2^k + N = 3P so that 2^(k+1) = 3P + 1.

If the problem as stated possesses two solutions for the same prime number P, then

3 * 2^(j+1) - 2^(k+1) = 8

i.e.

3 * 2^(j-2) - 2^(k-2) = 1.

Clearly the only solution to this Diophantine equation is j = 2 and k = 3.

This immediately yields P = 5.

Nice finish, Jerbil! Your completion also means (see my previous observations) that the only prime

of the form 1+sum{i=1 to n}(4^i) is 5. Now, we can have a real number theory type proof of this

fact starting with "For a given prime P, consider multiple solutions of 2^M - 3P = N^2 ....".

I cannot refute the methodology corresponding to the foregoing proof. However, P=5 seems to satisfy all the conditions of the given problem due to the following reasons:

The equation becomes 2^M – 15 = N^2 at P = 5. Reducing this to mod 3, it can easily be shown that: M must be even, so that: (2^M/2 + N) (2^M/2 - N) = 15.

Since, M and N are positive integers it follows that: 2^M/2 + N > 2^M/2 – N, and accordingly:

(2^M/2 + N, 2^M/2 - N) = (15, 1), (5, 3), giving: (2^M/2, N) = (8, 7), (4, 1), so that:

(M, N) = (6, 7), (4, 1).

Consequently, it seems that there exists at least one value of P that satisfies all the given conditions.

Although I came across this problem in an old mathematics periodical, the said magazine does not have the answer to this problem.

Till date, I have not yet been able to come up with a proof giving all the value(s) of P and I'm still looking for the same.

Thanks, you helped me find the kink in my reasoning!

Thanks, you have found the flaw in my argument! I didn't consider one other case, namely, that (2^(M/2)-N)=1 and (2^(M/2)+N)=3*P. So, when P=5, you get the (8,7) answer and from the (2^(M/2)-N)=3 and (2^(M/2)+N)=P possibility, you get the (4,1) answer. So, that means that this pair of possibilities are the only way you can get 2 answers (the other case, when [(2^(M/2)-N)=P and (2^(M/2)+N)=3], is actually inconsistent because P must be 2 -- the only prime less than 3). Anyway, in my overlooked case, (2^(M/2)-N)=1 and (2^(M/2)+N)=3*P, P must be of the form 1+2^2+2^4+2^6+...+2^(2n) for some n. 5 is the case when n=1. If there are any other primes of this form, then there are other answers with 2 distinct solutions in (M,N). I checked up to n=20 with no luck. If I find another, I'll be sure to let you know. Thanks for the wonderful problem and for giving me the hint to the flaw in my "proof"!

Each of M and N is a positive integer. Determine all possible value(s) of a positive prime constant P such that the equation: 2^M – 3*P = N² has precisely two distinct solutions in (M, N).

There are none.

N can't be 0 mod 3 as that would imply that 2^M is divisible by 3. If N is 1 mod 3, then 3 must divide 2^M-1 and so M must be even (it is easy to show that 3 divides 2^X-1 iff X is even). If N is 2 mod 3, then 3 must divide 2^(M-2)-1 and again M must be even. So, M is even. Then, we can rearrange and factor to get: (2^(M/2)-N)*(2^(M/2)+N)=3*P So, either [(2^(M/2)-N)=P and (2^(M/2)+N)=3] or [(2^(M/2)-N)=3 and (2^(M/2)+N)=P]. In either case, setting P determines both N and M uniquely. Therefore, there is no P which will give two distinct solutions in (N,M).

i have a whole in my logic somewhere and help would be appreciated by whomever

so take this from the perspective of the largest number taken A

there are summation (0 to A-1)=A(A-1)/2 ways that this number can be the biggest number and no triangle be made

for example for 5

4 with 1

3 with (1,2)

2 with (1,2,3)

1 with (1,2,3,4)

this counts the points being chosen each way second and last

notice if there are two of the biggest number there are no bad triangles so this is all the bad triangles

now you multiply by three because the largest number can be chosen in any position (but you have already accounted for the other two being chosen at different times)

now integrate for all A

N^3/2-3N^2/4

so by my logic it should be all - bad

N^3-N^3/2-3N^2/4

divide by n^3 and this should be the probability but it doesnt work

anything?

Although I can't follow your argument completely, it is clear that you allow 2 of the three numbers to be the same. The problem says "Three different numbers...." Perhaps that's the flaw in your logic.

Three different numbers are chosen

at random from the integers 1 to N,

inclusive. The probability that

these could be the lengths of the

sides of a (nondegenerate) triangle

is 75/161. Find N.

What is the probability that the numbers one, three, four, six, eight are written down in random order and the resulting five digit number is a square

This can't be the problem because the statement in the problem would have ended "what the" instead of "the the" which it is.

The words of a problem are numbered in lexicographical

order. Then the first word of the problem is written

in the position denoted by 1, the second word in the

position denoted by 2, etc. The result is: "five

random order is eight that numbers six one square four

are the what a written digit is resulting number

probability and three in down the the." Solve the

(mathematical) problem.

SUPERPRISMATIC'S ATTEMPT AT CLARIFICATION:

Suppose the original (mathematical) problem

were "two plus three add to what number?"

first we label each of the 7 word positions

1,2,3, etc. in alphabetical order. Since

"add" is first alphabetically, we label it 1,

since "number" is second alphabetically, we

label it 2, etc. Writing the sentence above

the labels, we get

"two plus three add to what number"

6 3 4 1 5 7 2

So, we place the first word ("two") into

the position labelled 1, the second word

("plus") into the position labelled 2, etc.

Thus, our result is

6 3 4 1 5 7 2

"what three add two to number plus".

So, had the puzzle had this result instead

of "five random order is eight....", the

answer to the (mathematical) problem would

have been 5.

A quick calculation gives the radius of the innermost circle as Product{n=3 to infinity}cos(pi/n) which is about .115

one question before i really look at this. are the final prices exact or rounded. For example could the mark up actually lead to the price 19.714634555?

Rounded.

For haggling purposes, a curio dealer tagged

his wares with the cost price, enciphered by

a simple substitution of letters for numbers.

In order to improve his haggling position,

a collector attempted to break the system by

noting the letters on the tags of certain

items in the window and then asking the

prices therefor. Three items bore tags

CJ.GA, IB.DA, and FH.AE, and the prices asked

were, respectively, $19.71, $39.69, and

$53.46. He assumed (correctly) that these

prices represented the same percentage

mark-up. What were the cost prices and what

was the mark-up?

Superprismatic's remark: Note the use of

the nearly archaic term "therefor".

Here is the one for a 16 by 16 grid.

Anyone notice a pattern????

sixteen.bmp

Your post made me think. Forgetting about the 23 business, what is the number of ways to make coordinates which give you all n^2 values in a n by n grid? One set of coordinates should be increasing starting at 0 and the other set increasing starting at 1. Do you have any idea? For a 2 by 2 there are 2 ways: (0,1)(1,3) and (0,2)(1,2). For a 3 by 3 there are also 2 ways: (0,1,2)(1,4,7) and (0,3,6)(1,2,3). For 4 by 4 there are 6 ways......

Yeah that's exactly what I got.

Why do I waste my time doing these when someone already comes up with the answer just before I post.

Yeah, nuclearlemons got it right. But, It's such a well designed puzzle that it's a pleasure to work it out!

I'd still like to see a really elegant way to solve it. My method was very brute force. In particular,

I'd be interested in a simple algorithm which would be easy to program. My method is very messy. Why don't

you try the analogous puzzle for a 10x10 having entries from 1 to 100 also with a coordinate of 23? Give me

some direction on how you solved it.

do the coordinates all have to be different?

No.

One set of eight positive integers is written down the

side of an 8x8 square grid, another set of eight across

the top. The entry in any cell of the grid is the sum

of the corresponding side and top coordinates. These

entries are the numbers from 1 to 64 inclusive. If

one of the coordinates is 23, what are the others?

CORRECTION ON THE PROBLEM STATEMENT:

"positive integers" in the problem should be "non-negative integers".

Sorry, I copied this verbatim from my source (privately published

material) and I didn't notice that the problem can't be solved

without 0 as one of the coordinates.