Question: Help I have this puzzle that I cant
figure out. "The Spider and the FLY"
A room is 30ft long, 12 feet wide, and 12 feet high. At
one end of the room, 1 foot from the floor, and midway
from the sides, is a fly. At the other end, 11 feet from
the floor, and midway from the sides, is a spider.
Determine the shortest by way of the floor, ends, sides,
and ceiling, the spider can take to capture the fly. How
long in feet, is this path?
CAN YOU HELP?
Thank You, Brandi
Answer: We've included such a puzzle with the complete solution to it
PuzzlePLAYGROUND sector. It can be found
We thank you, Brandi, for your message. It has inspired us
to publish this puzzle on our site.
Reply: Hi, I am glad you could use it. It is a fun puzzle.
Question: HI, we are student's and learning computer programmation. right now we are
learning the .net language. we have a project to programme, the rush hour
game. so we wanted to know if u know where we could get help. By the way,
in which language, is this game written???
thanx in advance for answering.
there are many versions of Rush Hour® programmed in different languages
and available on the Web at the moment. More information on this can be
found in our Rush Hour section at http://www.puzzles.com/products/rushhour.htm.
By the way, several more links to other Rush Hour® versions on the Web can
be found there as well.
Please please please help me this puzzle is driving me my family and a few
The pharaoh's daughter has 45 precious stones, one ruby, two emeralds,
three diamonds, four sapphires, five opals, six amethysts, seven pearls,
eight moonstones, nine garnets, she has a jewellery box with nine
compartments can you help her to arrange the jewels in the box so that
there is only one kind of jewel per compartment and so that each
horizontal, vertical and diagonal adds up to the same number of jewels????
Hope you can help many thanks J C.
puzzle can be solved quite easy if to solve it in accordance with a Magic
Square pattern. In this case the Magic Sum for the Square is 15. First it
is required to put the five opals into the central compartment and then
add to each pair of the adjacent compartments (horizontally, vertically
and diagonally) such a number of jewelers which could add up to 15 with
those opals in the center.
One of the solutions (and there is a lot of them) is as follows.
Top row: 2 emeralds, 7 pearls, and 6 amethysts.
Middle row: 9 garnets, 5 opals, and 1 ruby.
Bottom row: 4 sapphires, 3 diamonds, and 8 moonstones.
Question: hello, i am a student that is just starting to learn about the tools used to create interactive web pages. could you please tell me how you created the scrolling news link and what tools you used. thank you for your time.
Answer: Actually we at Puzzles.COM use the Macromedia Studio, specifically Macromedia Flash utility. This tool is employed to create the entire home page and also other interactive puzzles and features which can be found all over our site.
Regarding the scrolling news - it is not quite a standard element. Therefore, it can't be easily found directly in the Flash pack. In fact this scrolling news bar unites an animation, a dynamic text and a text file which holds the entire news within it.
Question: Our professor gave us this word puzzle for fun and I cannot figure it out. I would appreciate any help you can offer:
"One morning, exactly at sunrise, a Buddhist monk began to climb a tall mountain. A narrow path, no more than a foot or two wide, spiraled around the mountain to a glittering temple at the summit. The monk ascended at varying rates of speed, stopping many times along the way to rest and eat dried fruit he carried with him. He reached the temple shortly before sunset. After several days of fasting and meditation he began his journey back along the same path, starting at sunrise and again walking at variable speeds with many pauses along the way. His average speed descending was, of course, greater than his average climbing speed. Prove that there is a spot along the path that the monk will occupy on both trips at precisely the same time of day"
Any help would be greatly appreciate...there is no grade for this, it's just bothering me that I cannot figure it out
Answer: We've included this puzzle with the complete solution to it into our PuzzlePLAYGROUND sector. It can be found here.
We thank you, Julia, for the inspiration to publish this puzzle on our site.
You have a matchstick puzzle where the matches are arranged to show " VII=I ".
The Puzzle is to move 1 matchstick to make this equation correct. Your solution is to move the first 1 in 7 to make a square root sign.
Another solution is:
Move the second 1 in 7 and place it at a diagonal over the equals sign to make an equation " VI doesn't equal I ". This makes the equation True.
Answer: Actually the tricks of crossing the equality signs with a matchstick are quite common and good ones in this type of puzzles... but they make the equality disappear. The expression becomes the inequality instead. In such a way the trick violates the very condition of the puzzle where the goal is to "make the equation correct". In the case of the crossing of the equality sign we cannot say "the equation correct" - there is no equation anymore and what we can really say is the "true inequality" instead. And this doesn't satisfy the rules of the puzzle.
Do you have a solution as to how to work this? This is my first experience with cryptarithm.
Answer: Actually the expression itself can be considered as the true one. According to the Roman numerals the expression consists of:
XLIV or 44, because X (10) to the left of L (50) gives 40 and the IV gives 4 or 44 in total;
X or 10;
CDXL or 440 because C (100) to the left of D (500) gives 400, XL gives 40 (see above) and in total it makes 440.
As the result we have:
44 x 10 = 440, which is true.
If to convert the expression to the Arabic numerals then the solution can be obtained in the following logical way.
Known that X is 2, we have:
2LIV x 2 = CD2L
Let's start with the V.
1 can not replace V because multiplying by 2 it gives 2 behind L in the CDXL number and that is impossible - 2 is hidden behind X.
2 can not replace V as well. The same reason - 2 is behind X already.
3 can not replace V. If it is there then L in the CDXL number is 6. And the I in the XLIV number has to be 1 in order that XL in the CDXL is 26 (X is 2 and L is 6). At the same time L in the XLIV is 6 too. Now we have all the digits in XLIV (2, 6, 1 and 3), or it is 2613. Multiplying 2613 by 2 we get 5226 or 2 behind D - and that is impossible, because 2 is behind X.
Now let's see at digits 5, 6, 7, 8 and 9. If any of them is behind V in the CDXL number than multiplied by 2 they give us the two digit number each time: 5 gives 10, 6 - 12, 7 - 14, 8 - 16 and 9 gives 18. That means we must add the 1 (the first digit of the 10, 12, 14, 16 or 18 numbers) to the second digit (from the right) of the CDXL. In other words to get 2 in place of X in CDXL we will need to have behind I in XLIV a digit which would give us 1 when multiplied by 2. But none of the digits can give an odd number when multiplied by 2. It means that none of the 5 - 9 digits can go behind V in XLIV.
Thus the only number left to be placed instead of V in XLIV is 4. When V is 4 then multiplied by 2 it gives 8 in place of L in CDXL. Automatically 8 is in place of L in XLIV as well. To have 2 instead of X in CDXL it has to be 1 instead of I in XLIV. Now as we have all the digits for XLIV (X - 2, L - 8, I - 1 and V - 4) we know the entire XLIV number. It is 2814. Multiplied by 2 it gives 5628.
Thus the solution is:
X - 2,
L - 8,
I - 1,
V - 4,
C - 5,
D - 6.
Question: I’m not sure exactly how it’s supposed
to work, but I’ve tried numerous times to do the “Three
Dice Trick”, but it doesn’t seem to be working right.
That or the explanation on the trick’s page was lacking
in detail. Here is how I read it to work:
1. I rolled 3 dice and got; 2, 4, 5.
2. I added all three numbers; 11.
3. I took a die, 2, and added it to what I already
4. I rolled the die I took, 2, and added it to the
previous sum; 3,4,5 – 12
5. I added 7 to the sum;19.
3 six-sided dice only go up to 18.
I think that more clarification is needed...
Answer: The first two
steps in this interpretation are correct. Unfortunately
the 3rd and the 4th steps contain mistakes in them. This
is where the problem with the trick comes from.
If we understand it right, the die 2 is in fact the die
which showed 2 with the initial roll. In this case it is
exactly the bottom (!) number of that die which has to
be added to the previous sum, not the top number! Since
the top number was 2, the bottom (opposite) number was 5
(7 - 2). In other words step 3 should provide the result
of 16 (11 + 5), not 13.
Step 4 from this interpretation should be related to the
third step from the rules on the
site. But what step 3 from the rules states
is that when the selected die is rolled again the number
it shows on top (!) has to be added to the PREVIOUS sum.
In our case the previous sum, i.e. the sum from the
above mentioned steps 1-3 is 16.
Step 5 is correct as the approach.
We hope with these additional explanations the trick can
be performed quite easily.
We are sorry about the confusions the rules to our
puzzles can cause to some of our visitors from time to
time. Naturally, we are glad to explain the rules in
more detail in such cases - in order to clarify their
Question: i might have caught a mistake, if im wrong i apologize and maybe you can clarify something, but, for the puzzle named Three Dice you advise to complete 3 steps. first throw 3 dice count up the sum, then take any 1 dice and add the number on the bottom face, and then throw that dice and add the number to the previous sum. if you do that then shouldn't the third step be switched with the second step, otherwise its still a game of probability without a trick and it wouldn't work. try the trick with completing steps 1,2,3 and it won't work everytime, if you do steps 1,3 then 2, throw dice, take one dice throw it again and the sum, then take the number on the bottom face that should work, i apologize if im wrong, but this seems to be the case.
Answer: Actually the explanation to the Three Dice trick is correct. As you know the main thing the performer (say, it is you) has to do is to say the correct sum that the one who had thrown the dice counted in total. What this sum consists of? During the first step of the trick (throwing of the three dice) the one who throws dice gets three numbers, say x, y and z. Then (the second step) he/she takes one die and adds up the number on the bottom face of it to the number he/she has already got i. e. to the x+y+z. Let it be that he/she takes the z die, that means he/she adds to the existing sum its bottom number. As far as we know the bottom number in a die and the face
number add up to 7, then this number is 7-z. After that (the third step) he/she just throws the same z die and leaves it as it is with the rest of the dice, and this time the die he/she just thrown shows up another number, say w. When you, as the performer, turn around you see the following numbers: x, y and w. Adding them up and then adding 7 you get the required sum. Why? Because what is actually the sum that was counted? It is: x+y+z (the first step) plus 7-z (the second step) plus w (the third step), or it is x+y+z+7-z+w. Two z's are mutually eliminated and the sum is x+y+w+7, or the same what you've counted.
Can you please give me the link or direct me to a web
site where the puzzle of seven or more nails supported
by one nail in a wooden block is described.
Question: During our holidays at a campsite last month a neighbour camper showed me a puzzle involving 7 five inch spikes; the challenge was to balance 6 of these spikes on one. I solved the puzzle, with some help from him, by arranging 6 spikes in such a manner that they were intertwined and balanced. The resulting arrangement could be rested on the head of the 7th spike that was vertically driven into a piece of wood. My problem is that I can't do it again and it's driving me nuts. Can you help me solve this?
Here is my Puzzle. I couldn't find the solution for it. I kindly need your help to solve this. Pls let me know the solution for the below puzzle
"Can You Place TEN dots using just FIVE lines of FOUR dots."
Awaiting for reply ASAP
Thanks in Advance
Question: Question for you:
You have ten trees to plant. The orchard must consist of five rows of trees, and each row must contain four trees and be straight. One straight line of ten trees cannot be used. What template do you use for the planting?
Answer: We've included this puzzle with the complete solution to it into our PuzzlePLAYGROUND sector. It can be found here.
We thank you both, Sundaresan.K.R and Dan, for the inspiration to publish this puzzle on our site.
Question: Can you please help solve a puzzle which is driving me mad?
Using the four numbers 3,3,8, and 8 and the usual arithmetic operations plus, minus, multiply and divide, you have to make exactly 24. No tricks are allowed like powers, cube roots, or putting 8 and 3 to make 83. Just pure maths.
Any help to put me out of my misery would be appreciated!!
Answer: The author of this wonderful puzzle is Wei-Hwa Huang. And we've included this puzzle with the complete solution to it into our PuzzlePLAYGROUND sector. It can be found here.
We thank you, Kim, for the inspiration to publish this puzzle on our site.