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Guest Message by DevFuse

Perhaps check it again

Member Since 02 Dec 2013
Offline Last Active Mar 18 2015 04:35 PM

#341899 Orange square

Posted by Perhaps check it again on 12 March 2015 - 11:47 PM

My sincerest apologies...for offending his fragile sensibilities.


-1 for being an actual non-apology


-2 for deflecting onto me





As a token of apology, ...


-3 for continuing to be an actual non-apology


-4 for continuing to deflect onto me



 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -




What I, gavinksong, should have stated is, if I had the ability to,

I would have deleted my bragging post which prompted

Perhaps check it again to reply to in the first place.

  • -1

#341881 Orange square

Posted by Perhaps check it again on 11 March 2015 - 04:49 PM

"I'm getting quite a collection of bonanova stars. (emoticon with big grin)



Well, if you don't toot your own horn, others probably won't come along and do it for you.


Someone should send you some ointment to put on your back where you have been frequently patting it.

  • -3

#341478 Polygons make a line

Posted by Perhaps check it again on 30 January 2015 - 10:58 PM

I stated "No line can pass through all the interiors of the the three triangles."


BMD, you stated "Why cant [sic] a line pass through [triangle] ABE and the common point of

[triangles] ACD and EDF?"


In my attempt to work on the problem, I misunderstood "crossing a polygon" from the

original post and changed the subject.


But, with you asking me that question (in the quotes just above), you changed the subject. 

I never addressed/denied that situation.  It's not a line passing through the interiors

of three triangles anyway.

  • -2

#341466 Polygons make a line

Posted by Perhaps check it again on 28 January 2015 - 11:30 PM

I claim it's not true, and I show it with a counterexample.


In the xy-plane, let the following points be labelled as such:


A = (0, 1)


B = (1, 1)


C = (-1, 0)


D = (0, 0)


E = (1, 0)


F = (0, -1)



Triangle ABE shares point A with triangle ACD.


Triangle ACD shares point D with triangle EDF.


Triangle ABE shares point E with triangle EDF.



No line can pass through all the interiors of the three triangles.

  • 0

#341378 Maybe factoring a polynomial completely

Posted by Perhaps check it again on 12 January 2015 - 08:16 PM

Regarding post #2, you have a certain amount of doubt regarding correctness and/or

justification of certain steps, as evidenced by question marks (as a for instance).



I am looking for a solution by anyone made with complete confidence, and where

the steps are justified from the prior ones.  I'm not wanting any steps that come about

through luck/inspection/trial and error.

  • -1

#341376 Maybe factoring a polynomial completely

Posted by Perhaps check it again on 11 January 2015 - 09:37 PM

Can the following be completely factored over integer coefficients, where the highest degrees

on x and y are both one?



2x + 3xy - 2y2 - x + 3y - 1



If not, state that.


If so, then show the steps leading up to and the final factored form.

  • 1

#340804 Six sixes to make 1,000 (modified version)

Posted by Perhaps check it again on 27 October 2014 - 02:06 AM

Purposely repetitively being argumentative and failing to follow instructions repeatedly, gets your past

and any possible future posts/contributions in this thread ignored by me, DejMar.


The puzzle is still open.

  • -1

#340787 Six sixes to make 1,000 (modified version)

Posted by Perhaps check it again on 26 October 2014 - 05:30 PM

No, the standard for these types of puzzles is the solver is permitted to only concatenate the original

digits given in the puzzle at the outset, giving numbers to work with.  The resulting numbers are then 

operated on.  Any resulting numbers in the expression are not allowed to be concatenated with each



Post # 4 was a clarification and emphasis.  Please do not make a post like # 5 again, which goes

against post #4, or the rest of your posts/solutions on here will be ignored, and credit will be

given to other users who post any correct solutions that match the ones that you have already

posted in this thread.


Neither solutions in post #5 are correct due to their form.


The solution given in post # 6 is one of the correct ones.

  • -1

#339530 Total number of equilateral triangles

Posted by Perhaps check it again on 15 August 2014 - 06:08 PM

*             *              *             *    


       *              *             *             *


               *             *             *            *


                      *              *


              *              *




The 16 points above lie in a plane on an equilateral triangular lattice.


Certain sets of three points of the figure correspond to the vertices of equilateral triangles.


Suppose you were to form all of the equilateral triangles possible, such that, for any given


equilateral triangle, it must have its three vertices coincide with three of the 16 points.



How many total equilateral triangles can be formed this way in the figure?



  • 1

#338639 Four numbers equals two dozen

Posted by Perhaps check it again on 16 June 2014 - 04:48 PM

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.


Spoiler for - - - Here is a common form of the solution:



Unfortunately/fortunately, it is relatively easy to google this problem and find

multiple sites with the solution.

  • -4

#338560 License Plate Number Problem

Posted by Perhaps check it again on 04 June 2014 - 07:47 AM

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

  • -2

#338263 The King takes a long walk

Posted by Perhaps check it again on 05 May 2014 - 02:53 AM

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.

  • -2