Perhaps check it again

Pickett, can you come up with a different set with smaller magnitudes (as in absolute values) for a, b, c, and d?

Suppose a, b, c, and d belong to the set of nonzero integers. Let P(x) = (x4 + ax3 + bx2 + cx + d)2. Determine one of the sets of values of a, b, c, and d, such that when P(x) is expanded into individual terms of an 8th degree polynomial, that polynomial will have the fewest number of nonzero terms possible. Bonus: Write P(x) in its expanded form.

nana77, I don't see that number of yours coming up. Maybe if you shared even some nonspecific thoughts as to why you think it is the number you stated, it would lead me to give (possibly needed) clarifications. I am open to some other possible answers other than the one I have in mind, but I want to know if someone is making some more/different assumptions about the problem than the ones I have stated in the original post. The problem is still open.

Consider concave octagons in the plane that are non-self-intersecting. What is the minimum number of diagonals possible for one of these octagons if it is required that each of the diagonals lie entirely within the octagon?

bonanova stated: "plasmid should get the solve. I only arranged his examples in symmetric pairs." ----------------------------------------------------------------------- No, plasmid's posts are not recognized on here by me for any credit. plasmid already knows he/she has to state to me a retraction of a post that plasmid made to me elsewhere that was/is out of line. And plasmid already knows that there are no conditions on the retraction, regardless that he/she is trying to place any on the retraction. And none of plasmid's other posts will be recognized by me for any credit on any other puzzle challenge threads that I start unless a retraction is made by plasmid to me in the forum.

bonanova, for your #6 category that was up for debate for three more possible hexagons in your last spoiler, they won't fall under the convex/concave types. That node of degree four that you pointed out will have to disqualify them. Every convex/concave polygon that is not self-intersecting must have a degree of two for all of its nodes/vertices.

In the plane you get to choose six points. By choosing these six points in one of an infinite number of appropriate configurations, what is the maximum number of combined convex and concave (but not self-intersecting) hexagons that can be formed, if each of the six points is to be one of the vertices for every hexagon so formed? For reference sake, the points could be labeled A, B, C, D, E, and, F. I am not asking for any coordinates. However, if you want to volunteer a set of them to illustrate your work/solution, then that would be fine. ---------------------------------------------------------------------------------------------------------------------- Here is an example with four points: A, B, C, and, D. Place them in the plane. You choose where. What is the maximum number of combined convex and concave (but not self-intersecting) quadrilaterals that can be formed, if each of the four points is to be one of the vertices for every quadrilateral so formed? If you do not choose all of the points to be distinct and/or you place at least three of them on the same lime, you won't get any quadrilaterals formed. If you place the four points so that they are the vertices of a convex quadrilateral, then you will get one total (convex) quadrilateral. If you place three of the points as vertices of a triangle and the fourth point as an interior point of that triangle, then three total (concave) quadrilaterals can be formed. "Three" is the answer, unless there is some other general configuration for four points that has been overlooked with a higher total number of quadrilaterals. . . .

gmgm, m00li's approach is a *possible* one, but there are an infinite number of possibilities for the twelfth term. m00li assumed a polynomial formula.

bonanova, please spell out my username when addressing me. For several seconds at first, I didn't know what "Pcia" was. Yes, so Board Guideline 6 suggests being positive, and I was. I was positive, thorough, accurate, honest, assertive, and to the point. Don't mistake my style for not being positive. I just happened to express a concern about someone using Google. Now let's look at your comment about buying a puzzle book that happens to have an answer section. No, I don't agree with you. Don't speak for me. One of the main points for me *is* the answer section of the puzzle book. I don't trust the author to know what he or she to know what the answer is without the alleged solution given in the back. I'm looking for errors in the answers. I'm looking for missing additional solutions (and I don't mean typos). I'm looking for errors such as faulty wording in the statements of the original problems. I don't have to consider at all the reasons (See? They are plural?) why I joined this site. I have total confidence in my background of decades of exposure to various puzzles and certain branches of mathematics and the challenge of new puzzles that I encounter here, just as a for instance. I see you repeated to a user from my own post what he/she couldn't use for his/her solution. And "Thanks" back at you.

A little more care has to be taken when presenting puzzles like this. Instructions need to made as explicit as possible. In this (original) puzzle, you are supposed to be told that you must use both of the threes and both of the eights, but no other digits. And you must use them in their original orientation. For example, you cannot flip a "3" around and meld it to a second "3" to form an "8." Concatenation, for example forming "33," should not be used unless stated. You can choose only from some combination of addition, subtraction, division, and multiplication. (An aside: Is the "-" permissible as a negative sign?) And unlimited use of parentheses is allowed. Notice how parentheses may be used for grouping and/or multiplication, depending on the circumstances of their placement in a given expression. Any of those alleged solutions in this thread that made use of the square root don't count, because that is not one of the four basic operations listed above. Unfortunately/fortunately, it is relatively easy to google this problem and find multiple sites with the solution.

Your rules of what to do are vague. And I don't think you mean "...and subtract FROM the remaining 4th number." I think you don't mean for the word "from" to be in there. You're talking about "digits," not "numbers." The following sentence doesn't make sense and/or is redundant: "All four numbers [read: digits] cannot be [the] same[.] and not all can be zero[.]" A zero *is* already included in the first part of the sentence. Supposedly these would be examples of yours: 4321 ----> 4 + 3 + 2 + 1 = 10 0505 ----> 0 + 5 + 0 + 5 = 10 8642 ----> (8 X 6)/4 - 2 = 48/4 - 2 = 12 - 2 = 10

The statement of your problem would have been much clearer (and leaving no ambiguity that you wanted only one question answered) if you had phrased part of it along these lines: [insert the first original two sentences as given.] We'll assume all squares are one unit on a side, and we might ask, "What is the length of such a trip?" Wait. This is Brainden. You are all geniuses. Let's alter the problem.** [insert the remaining original text.] ** You're not just "adding a wrinkle." You're substantially changing the problem, mainly going from shortest path to maximal length path.

It should be: "..., but in another paragraph, you are asking for..."

In one paragraph you are asking for the shortest path for the complete tour, but in another tour, you are asking for the maximal length of a complete tour? How are these two plans not contradicting each other?

There is an example with a set of seven coordinates on the plane such that no three points are collinear and every pairwise distance is an integer. So, if you want a set of six coordinates, just omit one of the seven. The example is at the top of the second page:

You will use exactly twelve of the numbers 0.7 and exactly four of the numbers 0.6. Or, you can just type ".7" and ".6," respectively, if you wish. Concatenation of the sevens and sixes (mixed or matched) is not allowed). No other numbers are to be used. You can use unlimited plus signs. An unlimited number of parentheses can be used for both grouping and for multiplication. No other symbols or operations may be used. Put these together to form an expression that gives the maximum value. Here are two examples of the forms: Example 1) (0.7 + 0.7 + 0.7 + 0.7 + 0.7 + 0.7)(0.7 + 0.7 + 0.7 + 0.7 + 0.7 + 0.7)(0.6 + 0.6 + 0.6 + 0.6) = (4.2)(4.2)(2.4) = 42.336 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Example 2) (.7 + .7)(.7 + .7)(.7 + .7)(.7 + .7)(.7 + .7)(.7 + .7)(.6 + .6)(.6 + .6) = [(1.4)^6](1.2)(1.2) = 10.843 (approximately) (Let me know if I made any errors, please.)

phil1882, 1) The sequence starts off with "2," not "1," because 1 is not a prime number. 2) There are commas between the numbers in the sequence and a comma before the ellipsis. 3) You are missing the number 18. 18 = 2*3*3 (The number of prime factors is odd.) Therefore, the sequence is: 2, 3, 5, 7, 8, 11, 12, 13, 17, 18, 19, 20, 23, 27, 28, 29, 30, 31, ...

"you start off with a circle, and you select 3 points on the circle..." This means you are not choosing any points of the *interior* of the circle. The likelihood has a probability of zero. The largest triangle area-wise that can be formed with three points chosen *on* the circle is an equilateral one. And it's area is less than half of the area of the circle. If you want to use the word "on" an object for meaning picking random interior points, you better change the object to a *disc* instead. A circle doesn't have any interior points to choose from, but a disc does. That is, if you are told you are selecting points on a circle, then you are not selecting points in the interior of the circle.

bonanova posted: ------------------------------------------------------------------------------- I would guess "ways" means combinations. The permutations of 271 1's are not that interesting. ------------------------------------------------------------------------------- I must be a genius when it comes to making arguments, because I would never have typed that second sentence highlighted above. It is irrelevant. When I asked my question, I had already thought through about repetitions of certain numbers. But your second sentence, bonanova, is irrelevant. It's on purpose that I try really hard not to resort to sarcasm with another math forum user when I don't understand why they stated something that I otherwise would not have stated myself. And now it should be an insult for me to point out that the permutations of 271 1's doesn't exclude the possibility of interesting permutations of other than the sole use of 1's to sum to 271.

On one level, that number can be shown to be between one of the numbers 40! and 44!, inclusive. The number in the puzzle written out ends in nine zeroes, so it has exactly none factors of 10. Then it has exactly nine factors of 5. Start listing consecutive multiples of 5, beginning with 5, and end until you get to a multiple that contains a cumulative ninth factor of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ... 5, 2*5, 3*5, 4*5, 5*5, 6*5, 7*5, 8*5 ====> Stop. If you have gone to 45 inclusive, you have gone too far, because that will provide a tenth factor of 5.

Is this with regards or without regards to orderings of the summands?

Original poster, are the three white pieces to necessarily be moved before the queen even begins to make the first capture?

Options 1) You can't have ambiguity. You can't have "2x squared" meaning (2x) squared = 4x2 and have "4x squared" meaning 4x2 also. 2) "x" and "X" are different cases of the same letter, and therefore they represent different variables. So, it would be consistent for those two equations to exist, because you have two different variables in them.

The solution given by user skca was already given as its equivalent in post #2 by a different user.