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Guest Message by DevFuse
 

Perhaps check it again

Member Since 02 Dec 2013
Offline Last Active Yesterday, 05:37 AM
-----

#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

other.

 

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