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General Puzzles, Problems & Riddles! (View 1st Post BEFORE You Answer or Ask!)

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  • Close answer, but not the one I expect...

    You now know the values of I and M. The final part comes combinatorics and some multiplication. I suggest you to list the possible values for each set.

    Is there a rule that ALL letters must represent only 1-digit integer ?

    P.S. I'm a bit offended that you included Raoul in the solvers list. A solution must be backed by a proof, that is the whole pupose. I'd be pleased and encouraged if you remove his name from the list if he fails to submit a proof within next 3 days.
  • _raoul_raoul Posts: 280 Pool Pro
    Close answer, but not the one I expect...

    You now know the values of I and M. The final part comes combinatorics and some multiplication. I suggest you to list the possible values for each set.

    Using a similar approach to Satya, I get 64 for problem 5.
  • _raoul_raoul Posts: 280 Pool Pro
    Close answer, but not the one I expect...

    You now know the values of I and M. The final part comes combinatorics and some multiplication. I suggest you to list the possible values for each set.

    Yes, after figuring out I=8 am M=4 always, I just listed out all the possible values for the rest. Each of these gives eight solutions:

    O=0/7 N=7/0 H=2 K=5 G=1/3/6/9
    O=0/7 N=7/0 H=3 K=6 G=1/2/5/9
    O=0/7 N=7/0 H=6 K=9 G=1/2/3/5
    O=1/6 N=6/1 H=0 K=3 G=2/5/7/9
    O=1/6 N=6/1 H=2 K=5 G=0/3/7/9
    O=2/5 N=5/2 H=0 K=3 G=1/6/7/9
    O=2/5 N=5/2 H=3 K=6 G=0/1/7/9
    O=2/5 N=5/2 H=6 K=9 G=0/1/3/7

  • *Aurora**Aurora* Sorry, I'm a lady...Posts: 6,864 Pool Forum VIP
    _raoul wrote: »
    Close answer, but not the one I expect...

    You now know the values of I and M. The final part comes combinatorics and some multiplication. I suggest you to list the possible values for each set.

    Yes, after figuring out I=8 am M=4 always, I just listed out all the possible values for the rest. Each of these gives eight solutions:

    O=0/7 N=7/0 H=2 K=5 G=1/3/6/9
    O=0/7 N=7/0 H=3 K=6 G=1/2/5/9
    O=0/7 N=7/0 H=6 K=9 G=1/2/3/5
    O=1/6 N=6/1 H=0 K=3 G=2/5/7/9
    O=1/6 N=6/1 H=2 K=5 G=0/3/7/9
    O=2/5 N=5/2 H=0 K=3 G=1/6/7/9
    O=2/5 N=5/2 H=3 K=6 G=0/1/7/9
    O=2/5 N=5/2 H=6 K=9 G=0/1/3/7

    You're doing it wrong. I'm working on a solution, I'll get back to you when I finish... earliest after Christmas...
  • Michael HuangMichael Huang Posts: 4,114 mod
    edited October 2016
    Two members obtained the incorrect answers.. If less than 40% of the members guessed the number of solutions correctly before I come to this thread, the solution will be posted. Otherwise, if @M V Satya Mohan finally answered the question correctly before any wrong answer appear, I won't post the solution.

    Besides logic and mathematics, Problem 5 does not necessarily involve extreme manipulation of multiple expressions (as in calculus and number theory). Solving cryptograms is simply about understanding the mechanism of the given mathematical puzzle.

    Some hints for Problem 5:
    1. Three digit-sums involve the total of most numbers, ending with the digit 0.
    2. Create the list of possible number candidates for the variables based on what you know so far after finding the values of I and M. This eliminates the same integers for other variables. For multiple variables that exist in the certain digit-sum, the candidates were chosen pair-wise.

      For instance, if I want to find the possible distinct candidates for the following...
      kfn5sfn553ft.gif

      Then, I can see that the possibilities are...
      hth3bfd2ettr.gif

      If you know that another variable is 7, the second set is eliminated as well as A = 7.
    3. When counting, keep in mind that (1) G does not depend on the rest of the variables, and that (2) keeping one candidate eliminates some sets of variables.

    If you have done some number and/or math puzzles, like Kakuro, you should know that there exists unidentified candidates for each box. Inputting one value in a box eliminates at least a pair of candidates, containing that number. If no candidate exists in any pair possibility, there must be the mistake in the puzzle. This is different from cryptogram, but that is how puzzles are.

    I have more problems to put up, but I am going to wait for the correct responses.

    Good luck to you all!
  • _raoul_raoul Posts: 280 Pool Pro
    *Aurora* wrote: »
    _raoul wrote: »

    You're doing it wrong. I'm working on a solution, I'll get back to you when I finish... earliest after Christmas...

    Oops, you are right... i had some cases with a leading 0. The answer should be 48:

    O=0/7 N=7/0 H=2 K=5 G=1/3/6/9
    O=0/7 N=7/0 H=3 K=6 G=1/2/5/9
    O=0/7 N=7/0 H=6 K=9 G=1/2/3/5
    O=1/6 N=6/1 H=2 K=5 G=0/3/7/9
    O=2/5 N=5/2 H=3 K=6 G=0/1/7/9
    O=2/5 N=5/2 H=6 K=9 G=0/1/3/7
  • omgsunomgsun Posts: 2,239 Pool Champion
    michael, u need to set a rule for not using computational devices
  • omgsunomgsun Posts: 2,239 Pool Champion
    Close answer, but not the one I expect...

    You now know the values of I and M. The final part comes combinatorics and some multiplication. I suggest you to list the possible values for each set.

    Is there a rule that ALL letters must represent only 1-digit integer ?

    P.S. I'm a bit offended that you included Raoul in the solvers list. A solution must be backed by a proof, that is the whole pupose. I'd be pleased and encouraged if you remove his name from the list if he fails to submit a proof within next 3 days.

    i also agree
  • *Aurora**Aurora* Sorry, I'm a lady...Posts: 6,864 Pool Forum VIP
    edited October 2016
    c9j3ho59xe5b.jpg

    I would start from the fact that since O, N and G don't change from HONG to KONG, then these additions must end in zero: 7+O+6+N, 5+M+1+I, I+0. The two latter ones added with the digits you carry from the column right to them. Am I correct? H can not be bigger than 7.

    Leading digits can't be zero – does that mean the letters I, H and K?
  • Michael HuangMichael Huang Posts: 4,114 mod
    edited October 2016
    _raoul wrote: »

    The answer should be 48

    The answer is correct. I guess that I have to adjust the rules a bit since there aren't many members here. Anyway, I am going to post the solution for difficult problems that can't be easily solved by many members.
    *Aurora* wrote: »
    Leading digits can't be zero – does that mean the letters I, H and K?

    That is correct!
    omgsun wrote: »
    michael, u need to set a rule for not using computational devices

    I did that already.
    Post edited by Michael Huang on
  • Close answer, but not the one I expect...

    You now know the values of I and M. The final part comes combinatorics and some multiplication. I suggest you to list the possible values for each set.

    Oh, now I know where I made a mistake... I listed all the possible solutions for O=0/1/2 and N=7/6/5, but I miscounted them as 25 instead of 24... So when I doubled, the answer was more by 2. Anyway, I'm content that atleast I got the logic right.
  • Michael HuangMichael Huang Posts: 4,114 mod
    Close answer, but not the one I expect...

    You now know the values of I and M. The final part comes combinatorics and some multiplication. I suggest you to list the possible values for each set.

    Oh, now I know where I made a mistake... I listed all the possible solutions for O=0/1/2 and N=7/6/5, but I miscounted them as 25 instead of 24... So when I doubled, the answer was more by 2. Anyway, I'm content that atleast I got the logic right.

    Great to hear! Guess you earn a point for almost arriving to the correct answer, even though @_raoul found the correct one first!
  • Michael HuangMichael Huang Posts: 4,114 mod
    edited October 2016
    ** New Problems Updated **

    Let's start with non-mathematical puzzle....

    Problem 7 - Sudoku Time! (Hard)
    Time Limit: 1 week
    jrv9x16l4jgy.png
    Other than Forced Moves and Intersections, list the minimum number of strategies used to solve the sudoku above.

    Example: If I used X-Wing and XY-Wing for the sudoku, I list those as...

    X-Wing
    XY-Wing

    Important: Any method that involves trial-and-error and programming is not allowed. Using the solver that gives the exact strategies is also not allowed, even though it may not be accurate and efficient.

    Note: One of the lists to go for is BrainBashers.

    Note: The convention depends on how solvers look at strategies. One strategy may be in the different category from another, say Hidden Subsets and Naked Subsets. I will think about how the solutions will be graded.

    Problem 8 - Math in 8 Ball Pool: So Much Winning! (Easy)
    Time Limit: 1 week
    8nnbtznqbf6i.gif
    Each letter must be distinct positive integer. Solve the above cryptogram. Then, determine the sum of all permuted strings of B, E, R, L, I, N, P, Z (The number must be very high that no player has reached!).
    Post edited by Michael Huang on
  • *Aurora**Aurora* Sorry, I'm a lady...Posts: 6,864 Pool Forum VIP
    I have solved the sudoku puzzle, but I don't know what you require as an answer. I had a method of ruling out impossible numbers as I went along, but I have written over the numbers... There is only one solution? I started from top left middle square and figured it can only be 7.

  • omgsunomgsun Posts: 2,239 Pool Champion
    edited October 2016
    *Aurora* wrote: »
    I have solved the sudoku puzzle, but I don't know what you require as an answer. I had a method of ruling out impossible numbers as I went along, but I have written over the numbers... There is only one solution? I started from top left middle square and figured it can only be 7.

    u have to find how many ways u can reach that solution
  • Michael HuangMichael Huang Posts: 4,114 mod
    For Problem 7, list the methods you have taken to determine the solution. There is in fact the shortest number of methods involved. Can you figure what they are?

    For Problem 8, each letter must be distinct. I revised the problem. In case you are not familiar with the term "permuted string", 12 and 21 are permuted strings. Solve the puzzle and figure out the sum.
  • *Aurora**Aurora* Sorry, I'm a lady...Posts: 6,864 Pool Forum VIP
    Heh, I'm not going to go through it again to count. Someone else can have the pleasure. I'm good with having worked it out by myself. :)
  • Michael HuangMichael Huang Posts: 4,114 mod
    That is alright. But do you remember how you solved the sudoku? Pictures and past come to mind!
  • _raoul_raoul Posts: 280 Pool Pro
    Here's 6B.

    Since 2016=32*9*7, you can solve the problem mod 32, mod 9, and mod 7, then use the chinese remainder theorem to find mod 2016.

    For mod 32, even values of i are 0, odd ones are 1. Then 2016/2 mod 32 = 16.

    For mod 9, values of i divisible by 3 are 0, the rest are 1. Then 2016*2/3 mod 9 = 3.

    For mod 7, values of i divisible by 7 are 0, the rest are 1. Then 2016*6/7 mod 7 = 6.

    Then combine with CRT to get 48 as the solution for mod 2016.
  • Michael HuangMichael Huang Posts: 4,114 mod
    Yes! That is it! Chinese Remainder Theorem! You have finished both parts of the problem! XD
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