superprismatic

Nice job! I like your analysis. It's easy to follow. I wish I had thought of this way to do it.

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

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!

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.

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

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. 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.

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?

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 ....".

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

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.

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.

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".

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, 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.

No.