Dim5678's Weekly Math Competition (WMC)

Discussion in 'Miscellaneous' started by Dim5678, Aug 26, 2015.

  1. What this is:
    A weekly competition containing three problems, with problem #1 being the easiest and problem #3 being the hardest. Only one problem can be answered per user. Only one answer is allowed per user. There is no exact answer! If I think your answer makes sense, you answer will be correct, no matter how crazy it is!
    How to enter:
    Copy and Paste this into the reply box:

    What will I win?
    Problem#3 correct: 700 rupees
    Problem#2 correct: 200 rupees
    Problem#1 correct: 100 rupees
    1.The winner is the first member that answers the corrosponding problem correct.
    2.Only one problem can be answered per member.
    3.Only one answer is allowed per member.
    4.There is no exact answer! If I think your answer makes sense, you answer will be correct, no matter how crazy it is!
    5.The winners for the contest, my answers to the contest, and all three problems for the next contest will be released on Sunday, and the contest will end at 23:59 on Saturday.
    For every 11th week, there will be a final, in which all and only the previous winners participate in one contest, in which the members must answer ALL THREE PROBLEMS correct. The winner of a final recives the following:
    10k Rupees
    10 iron blocks
    10 diamonds
    10 emeralds
    10 lapiz lazuli blocks
    1 "Dim5678 WMC Final #X" written book, signed by me.
    Thats it, Thank You.
    HelloKittyRo, 607 and khixan like this.
  2. First round starts on Aug. 29, 2015
  3. K, welcome to Dim5678's Weekly Math Competition (WMC) Round #1, rules listed above.
    Problem #1: Using the pattern from the first two groups of numbers, what number does X represent?
    5/12/6, 4/20/8, 8/30/X
    Problem #2: Separate this shape with one (1) straight line into two (2) triangles.
    Problem #3: There are 9 cushioned boxes that looks and weighs exactly the same. Jack puts one slip of paper in one of the boxes, then mixes the boxes up again. Using only a scale with no weighs or anything else, how can you find the box with the slip of paper in it while using the scale the least amount of times?
    Round one ends on Sept. 9 at 23:59 P.M. EMC time, so good luck and have fun!
  4. Problem#: 3
    Answer: Threaten to hit Jack with the scale repeatedly unless he tells you which box the slip of paper is in.
    IGN: Defne_b_ded
    princebee likes this.
  5. Bravo, genius way to do it
  6. 5/12/6
    first number: starts with +-1 and then it is [number +5] so the next will be 8+9=17
    second number: starts with +8, then [number +2] so the next will be 30+12=42
    third number: it starts with +0 and then it is [number +2] so X will be 8+4 = 12 and the next will be 12+6=18

    so the full thing is:
    607 likes this.
  7. Nope, Jack mixed the boxes up, he dosnt know which box it is himself.
  8. Problem#:2
    Answer:When I first saw it, I tried to visualize it and couldn't thing of a single way to do it. Then I realized there were too many vertices and that things wouldn't really work making it impossible which led me to google which led me to a clever answer. Basically, you take the shape and put it on a spherical plane. When put on the sphere, the two points made by the cut corner will overlap. From there you can draw a line down the middle to get two triangles.
    Source: https://www.quora.com/Can-you-draw-...lowing-figure-to-divide-it-into-two-triangles
    IGN: jkrmnj

    The answer technically works but I am wondering if that was the answer you had come up with when you made the problem.

    A quick question for the third one: is the slip of paper heavy enough to make the box with it weigh any different than the others?
  9. Do I get to guess again? :3
  10. Problem#:3
    Answer: Split the boxes randomly into 2 groups of 3 and 1 group of two. weigh the two groups of 3 against each other. If one of the groups of three then throw out the lighter group of three and the group of two you set aside. With three boxes left you can weigh two of the groups against each other, if one is heavier you have found your box if they weigh the same the third one you set aside is the box with the paper in it. If the two groups of three weigh the same then take the other two you didn't initially weigh and weigh them against each other and the heavier one has the paper in it. Solved it in 2 moves . If all the boxes weigh the same Jack is a dirty liar and you make him swallow all the boxes as punishment.
    607 likes this.
  11. Yes
  12. New rule, if it wasn't obvious enough the first time:
    Keep answering, the people who answered before you might have the wrong answer, so you still have a chance to be the winner!
  13. Yes, you can measure it on a scale.
  14. So was I right orrrrr?
  15. #3
    You shake each box to hear a rattling sound. The one that has the rattling sound has the paper in it. You need not even use the scale.
  16. Thats why the boxes are cushioned XD
    607 likes this.
  17. Not telling whos right until next sunday:)
  18. Another rule: if I replied to your thread telling you that your answer is wrong, you can submit another answer