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2 missing number riddles

  1. Feb 18, 2009 #1
    http://users.telenet.be/Edu/raadsel01.jpg


    http://users.telenet.be/Edu/raadsel02.jpg


    I have been looking for hours... pfff...
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 18, 2009 #2
    These are kind of silly, since they probably have dozens of possible solutions-- there aren't any stated rules, it's just whatever makes sense to you. Here's some quickie answers I got:

    [/URL]

    One answer could be 1725. If upper-left is A, lower-left is B, upper right is C, and lower-right is D, then you could have: |C-(|2A-2B|^D)|

    [/URL]

    I get 24-- if the labels are A, B, C (leftmost-to-rightmost), then one formula could be A+2B+2C+1. [edit]oops, should be 18, not 24[/edit]

    DaveE
     
    Last edited by a moderator: Apr 24, 2017
  4. Feb 18, 2009 #3
    Shouldn't that be
    18
    ?
     
  5. Feb 18, 2009 #4
    Oops, yep, apparently there was a bug in my math.

    DaveE
     
  6. Feb 19, 2009 #5
    Ok, some more info... You can only use the numbers in the riddle!
     
  7. Feb 19, 2009 #6
    Difference in right two - difference in left two? So -6?
     
    Last edited by a moderator: Apr 24, 2017
  8. Feb 19, 2009 #7
    If you use absolute differences you get 6.
     
  9. Feb 19, 2009 #8
    I found this too... But it's wrong... (a friend of mine gave me the riddles)

    For the second one 18 is wrong. He said you have to use squares...
     
  10. Feb 19, 2009 #9

    Gokul43201

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    Well, looks like you've found them both (the number riddles that were missing).
     
    Last edited by a moderator: Apr 24, 2017
  11. Feb 19, 2009 #10
    I can understand you saying that it is not the number that your friend was thinking of, but why do you say it is wrong? It meets all the conditions of the puzzle. For one thing, you should add all the conditions to the puzzle at the outset and not hand them out piecemeal. For another, you should add an extra condition to this puzzle: If the puzzle has more than one solution, then only the one my friend is thinking of counts.
     
  12. Feb 19, 2009 #11

    Gokul43201

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    I think the friend is handing out the conditions piecemeal.
     
  13. Feb 19, 2009 #12
    Yeah, the friend is inaccurate to declare the answer as wrong. Note that "He said you have to use squares" in the 2nd problem, and yet there isn't even a number 2 in the 2nd drawing. So raising something to the 2nd power is using a non-existent number already. You could admittedly GET a 2 if you do a 7-5, but then you've used up a 7 and 5, and can't get a 20 using only 2 and 6. And if you allow using numbers multiple times, then you could say that my earlier answer was correct by rewriting it:
    A+B+B+C+C+(A/A)
    .

    I wrote a program a while back to solve things like these that goes through all the permutations of multiplication, addition, subtraction, division, powers, and modulus-- maybe I can tweak it to show all the possible solutions... Hmmm...

    [edit]
    Ok, my original program worked for 4 input numbers, so I could verify that using addition, subtraction, multiplication, division, powers, logs, and modulus, and using each "outlying number" only once, that these are the available solutions for the 1st problem (assuming upper-left is A, lower-left is B, upper-right is C, lower-right is D):

    (((B-A)%C)+D)
    (D-((A-B)%C))
    (D+((B-A)%C))
    (((D%C)+B)-A)
    (((D%C)+B)%A)
    ((B+(D%C))-A)
    ((B+(D%C))%A)
    (((D%C)-A)+B)
    (B+((D%C)-A))
    (B%((D%C)-A))
    (B-(A-(D%C)))
    (B%(A-(D%C)))
    (((A+B)+D)%C)
    (((B+A)+D)%C)
    (((B-A)+D)%C)
    ((D+(A+B))%C)
    ((D-(A-B))%C)
    ((D+(B+A))%C)
    ((D+(B-A))%C)
    (((A+D)+B)%C)
    (((D+A)+B)%C)
    (((D-A)+B)%C)
    ((B+(A+D))%C)
    ((B-(A-D))%C)
    ((B%(A-D))%C)
    ((B+(D+A))%C)
    ((B+(D-A))%C)
    ((B%(D-A))%C)
    (((B+D)+A)%C)
    (((B+D)-A)%C)
    (((B+D)%A)%C)
    (((D+B)+A)%C)
    (((D+B)-A)%C)
    (((D+B)%A)%C)
    ((A+(B+D))%C)
    ((A+(D+B))%C)

    Obviously that doesn't include absolute value, trig functions, factorials, "square", "square root", "cube", "cube root", up-arrow notation, Choose, etc. So that's examining 32,928 possible formulas. If you included a "reasonable" amount of the other functions, it'd be on the order of 5 octillion possible formulas (really infinite), so I decided to skip that.
    [/edit]

    DaveE
     
    Last edited: Feb 19, 2009
  14. Feb 19, 2009 #13

    Gokul43201

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    Whe I first saw a bunch of Sengupta's cryptarithms, I was thinking about exactly this: that it wouldn't take too long to write a program (and much less time to run it) to wade through the 9! or so possibilities for all of those puzzles. Next, I wondered if that wasn't exactly what you had done. :biggrin:
     
  15. Feb 19, 2009 #14
    Yeah, depending on the question, sometimes I'll brute force my way through. But usually I'll try and see what mathemagical tricks I can use before I resort to that. The permutations listed above were actually for a much trickier problem:

    https://www.physicsforums.com/showthread.php?t=204239

    Basically, the inputs, outputs, and functions were all fair game, so brute force was the only way I could think of to hack it. And sometimes, you can sort of know in advance that it's the only way-- like when dealing with primes, which don't have a nice mathematical formula for solving them.

    DaveE
     
  16. Feb 20, 2009 #15
    Probably a stupid question, but what is % ? (i know it's the symbol for percentage, but what do you use it for here?)
     
  17. Feb 20, 2009 #16
    That's the modulus function-- it's basically the "remainder" function. So 12 % 5 = 2, 33 % 41 = 33, 11 % 2.5 = 1, etc.

    DaveE
     
  18. Feb 20, 2009 #17
    Thanks, I know it as 'rest' :)

    He said I was very close with "the difference between the sum of the left numbers and the sum of the right numbers" ... It has to be something this easy...

    (I hope you understand, I'm dutch-speaking ;) )
     
  19. Feb 20, 2009 #18
    I got the one with the triangles

    A=bottom left, B=top, C=bottom right

    (A^2+C^2) - B = middle
     
  20. Feb 26, 2009 #19
    The first one could be:

    Zero. You can make each of the five numbers by adding or subtracting two or more of the others
     
    Last edited by a moderator: Apr 24, 2017
  21. Feb 27, 2009 #20
    It's the sum of the even numbers - the sum of the odd numbers

    That stupid...
     
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