Brain Teaser

Author: ethang5 ,

Topic's posts

Posts in total: 89
  • ethang5
    ethang5 avatar
    Debates: 1
    Forum posts: 4,457
    3
    3
    6
    ethang5 avatar
    ethang5
    1. Can you write a 6 digit number where...

    2. Digits cannot all be in ascending order ex. (134579)

    3. Digits cannot all be in descending order ex. (964321)

    4. No digit can be zero.

    5. No digit can be a multiple of a digit beside it. ex. (374852) 4 and 8 are side by side, and 4 is a multiple of 8.

    6. No digit can be used more than once. ex. (375594) 5 is used twice.

    7. No digit can be 1 more or 1 less than the digit before or after it. ex. (253894) 9 is one more than 8.

    8. No digits beside each other can share a multiple. ex. (386947) 8 and 6 are beside each other, and 2 is a multiple of both. 6 and 9 are beside each other, and 3 is a multiple of both.

    9. The six digit number cannot have more than 3 consecutive odd numbers.
    ex. (495372) 953 and 7 are all consecutive and all odd numbers.

    10. Finally, neither the first or last 3 digits of your 6 digit number should have a sum of more than 15
    (352947) 9+4+7=20
    ------------------------------------------
    Can you do it?

  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    914725

    If you are a stickler for the rules this may technically break rule 5 because the number one is technically a multiple of all other digits. This would mean that the number one cannot be used at all however so I chose to interpret rule number 5 as not applying to the digit one.

    Methodology:

    Choose a completely random digit to type. List all digits that may follow without breaking any rules. Select one at random to use as next digit. If all possible next digits break some rule then start over.

    Time taken: About 15 minutes

    Notes:

    The first idea that came to mind was to select digits non-randomly from lowest to highest and upon entering a fail state simply going back one digit and choosing the next higher possible digit (rather than starting over completely). This idea was discarded after reading rule 10 because rule 10 would likely cause a lower-to-higher trend to result in a lot of fail states at the last one or two digits. Rule 10 encourages even distribution of high and low numbers thus making random selection probably better than non-random selection (I never did actually try the non-random selection method though so that might be incorrect).
  • ethang5
    ethang5 avatar
    Debates: 1
    Forum posts: 4,457
    3
    3
    6
    ethang5 avatar
    ethang5
    --> @Discipulus_Didicit
    Good work. Your method was impressive too. I'm not surprised it was you first.

    But 1 does break rule 5 as it is a multiple of every whole number.

    So kudos for an impressive try, and for being so quick, but 914725 violates rule #5. Try again?


  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    --> @ethang5
    But 1 does break rule 5 as it is a multiple of every whole number.

    Okay fine by me. I'll give it another shot tonight maybe.
  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    Actually this just popped into my head a minute or two after I said "I will look at it later" and I wish to post it now so I do not forget.

    So, firstly there is a sort of hidden rule that we can... interpret I guess, for lack of a better word... from rule 10. I will call this 'rule 11' and it says "All six digits together must add up to 30 or less"

    Because we cannot use 0 or 1 we can use rule 11 to vastly narrow the possible options because 2+3+4+5+6+7=27, very close to 30.

    I will use this as a first stepping stone when I come back to the problem later but now I really must be going to do other things.

    I am no longer certain that this problem is solveable but even if it is not solveable I should be able to use this as a starting point for constructing a proof of that fact.
  • ethang5
    ethang5 avatar
    Debates: 1
    Forum posts: 4,457
    3
    3
    6
    ethang5 avatar
    ethang5
    --> @Discipulus_Didicit
    My, you are bright!

    I am pretty certain you will solve it. I was taken aback by how many correct numbers there were. I'm still trying to figure out the relationship between them.
  • zedvictor4
    zedvictor4 avatar
    Debates: 12
    Forum posts: 2,368
    3
    2
    3
    zedvictor4 avatar
    zedvictor4
    --> @ethang5
    473852
  • ethang5
    ethang5 avatar
    Debates: 1
    Forum posts: 4,457
    3
    3
    6
    ethang5 avatar
    ethang5
    --> @zedvictor4
    Correct! How did you arrive at it? And how long did it take you?

    You get extra kudos because answers with three even numbers including 4 and 8 are the hardest to figure out. Good work!

  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    --> @ethang5
    My, you are bright!

    Thanks.
  • SupaDudz
    SupaDudz avatar
    Debates: 29
    Forum posts: 10,248
    5
    8
    11
    SupaDudz avatar
    SupaDudz
    352749
  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    You get extra kudos because answers with three even numbers including 4 and 8 are the hardest to figure out. Good work!

     While working on this I thought I had a proof worked out that no solution could contain a 9 but this statement is making me doubt this (after eliminating all possible sets containing a 9 I was left with only sets that had three even numbers including a 4 and 8). I am having trouble finding any flaw in my proof though. Can you confirm whether any solutions contain a 9 and if so provide an example? This would help me identify the flaw in my reasoning if there is any.
  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    --> @SupaDudz
    Breaks rule 10
  • SupaDudz
    SupaDudz avatar
    Debates: 29
    Forum posts: 10,248
    5
    8
    11
    SupaDudz avatar
    SupaDudz
    --> @Discipulus_Didicit
    947253
  • SupaDudz
    SupaDudz avatar
    Debates: 29
    Forum posts: 10,248
    5
    8
    11
    SupaDudz avatar
    SupaDudz
    385927

  • SupaDudz
    SupaDudz avatar
    Debates: 29
    Forum posts: 10,248
    5
    8
    11
    SupaDudz avatar
    SupaDudz
    295834
  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    Using just rules 6 and 11 we see all solutions are some combination of the following sets:

    234567
    234568
    234569
    234578
    234579
    234589
    234678

    However a little basic arithmetic reveals that it is impossible to have any number higher than 5 on the same 'side' of the solution*

    That leaves us with:

    234567
    234568
    234569
    234578
    234579

    We can further rule out any solution containing a 9*

    Thus our possibilities are narrowed to:

    234567
    234568
    234578

    My hope was that by sufficiently limiting the sets I would be able to churn out solutions easily by simply rearranging digits from the limited sets at random until a solution was found. After doing this for about five minutes however I was unable to come up with any solutions. Back to the drawing board, let's see if we can eliminate any more sets.

    And it turns out that we can actually eliminate set 234567*

    This may in part explain my failure from a moment ago since that was naturally the set I worked with most during those five minutes since it was at the top of the list.

    This means that unless I screwed something up (not at all unlikely, see post 11) the only possible sets for solutions are:

    234568
    234578

    The only confirmed solution in this thread is a combination of one of these sets so that is promising. I am now ready to try brute force on one of these sets.

    Annnd presto, after about two minutes of executing the non-random method described in my first post using only numbers of this limited sequence I am finally able to find a solution (though it took well over two minutes to narrow the possibilities down to these two sets... Nearly an hour and a half working on and off. Call it about an hour).

    257438

    *Explanation for any of these claims available upon request.
  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    --> @SupaDudz
    All three of those break rule 10 lol
  • SupaDudz
    SupaDudz avatar
    Debates: 29
    Forum posts: 10,248
    5
    8
    11
    SupaDudz avatar
    SupaDudz
    --> @Discipulus_Didicit
    Oh shoot I thought it was 20
  • SupaDudz
    SupaDudz avatar
    Debates: 29
    Forum posts: 10,248
    5
    8
    11
    SupaDudz avatar
    SupaDudz
    947253
  • SupaDudz
    SupaDudz avatar
    Debates: 29
    Forum posts: 10,248
    5
    8
    11
    SupaDudz avatar
    SupaDudz
    OH WAIT FIRST 3 TOO
  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    --> @SupaDudz
    That also breaks rule 10.

    9 plus 7 alone equals more than 15

    You are just typing random answers at this point lol.
  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    Supa supa supa... Lmao
  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    --> @SupaDudz
    Read my long ass post for a few tips and if you have any questions ask. I have a lot of weaknesses but math and logic is not one of them.
  • SupaDudz
    SupaDudz avatar
    Debates: 29
    Forum posts: 10,248
    5
    8
    11
    SupaDudz avatar
    SupaDudz
    492735

  • Discipulus_Didicit
    Discipulus_Didicit avatar
    Debates: 9
    Forum posts: 3,943
    3
    4
    10
    Discipulus_Didicit avatar
    Discipulus_Didicit
    --> @SupaDudz
    LMAO

    Rule 9.