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Thread: Obedient figures/numbers

  1. #1

    Obedient figures/numbers

    Lisa calls a natural number obedient if she can be written as a product of the two figures following on each other. Example: The number 20 is obedient, because 20 = 4∙5.
    Points: To every obedient number there is at least one second obedient number which proves an obedient number with the first one multiplied again.
    Example: 2∙3=6 and 3∙4=12 and 6∙12=72 which is: 8∙9
    Is the sentence true??Why?? Is there a proof?
    Last edited by mmm4444bot; 11-14-2017 at 10:49 PM. Reason: Removed 52 links to external dictionary

  2. #2

    Obedient numbers

    Lara calls a natural number obedient if she can be written as a product of the two figures following on each other. Example: The number 20 is obedient, because 20 = 4∙5. Points: To every obedient number there is at least one second obedient number which proves an obedient number with the first ones multiplied again. 2 times 3 is 6 // 3 times 4 is twelve // 6 times twelve is 72 and 72 is 8 times 9 // Is there an Abstract proof for this sentence?
    Last edited by mmm4444bot; 11-14-2017 at 10:50 PM. Reason: Removed another 53 links to external dictionary

  3. #3
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    I'm not at all sure I understand the question.

    As I understand it, an "obedient number" is one that is the product of two consecutive integers, n(n+1) for some integer n.

    But how is the second number supposed to be related to the first? What does "with the first one multiplied again" mean?

    Possibly you mean that if m = n(n+1) is an obedient number, then m times something else will be an obedient number as well; but that is obviously true, since m(m+1) will be obedient. Your example looks like you must mean something more than that.

  4. #4
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    "two figures following each other"

    First, this is impossible without a closed path (maybe a circle) -- Better definition, please.

    Second, (Dog Stick Figure) => (Cat Stick Figure) -- (running in a circle)

    Demonstrate this multiplication.

    Seriously improved definition and problem statement, please.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5

    Obedient numbers

    Quote Originally Posted by Dr.Peterson View Post
    I'm not at all sure I understand the question.

    As I understand it, an "obedient number" is one that is the product of two consecutive integers, n(n+1) for some integer n.

    But how is the second number supposed to be related to the first? What does "with the first one multiplied again" mean?

    Possibly you mean that if m = n(n+1) is an obedient number, then m times something else will be an obedient number as well; but that is obviously true, since m(m+1) will be obedient. Your example looks like you must mean something more than that.
    Hello Dr. Peterson!
    Thank you for your answer! The exercise wants me to prove the following:
    Each obedient number m=n(n+1) can be multiplied with at least one other obedient number like o=p(p+1) or in different words for each m exists at least one other obedient number like for example o which than, multiplied with the first obedient number, will produce a third obedient number. So m times o is y and y=x(x+1)
    How the second obedient number is related to the first one is not really defined, only with: there is at least one second obedient number which will - multiplied with the first obedient number - produce a third obedient number.
    I just found an example that matches with 2 times 3 is 6 and 3 times 4 is 12 Those two obedient numbers: 6 times 12 produce 72 which is a third obedient number whereas 8 times 9 is 72.
    I hope this shows the Problem in a better way!?
    Sincerly, Yours Enoimreh
    Last edited by enoimreh7; 11-14-2017 at 02:02 PM.

  6. #6

    dog and cat

    Quote Originally Posted by tkhunny View Post
    "two figures following each other"

    First, this is impossible without a closed path (maybe a circle) -- Better definition, please.

    Second, (Dog Stick Figure) => (Cat Stick Figure) -- (running in a circle)

    Demonstrate this multiplication.

    Seriously improved definition and problem statement, please.
    They both know karate, will stop to run and decide that they do not need a circle, because of their respect for each other.

  7. #7
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    Nice try, but they are still different species and multiplication is unlikely.

    So, m(m+1) * n(n+1) = q(q+1)

    Well, there are only 13 choices for q. Check them all out. Can the LHS be so factored?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  8. #8
    Quote Originally Posted by tkhunny View Post
    Nice try, but they are still different species and multiplication is unlikely.

    So, m(m+1) * n(n+1) = q(q+1)

    Well, there are only 13 choices for q. Check them all out. Can the LHS be so factored?
    Thank you for trying. I followed your idea till I found the twenteeth possibility.

    m(m+1) * n(n+1) = x = q(q+1)

    01*02=02 02*03=06 12=03*04
    02*03=06 03*04=12 72=08*09
    03*04=12 04*05=20 240=15*16
    20xxxxxxx 05*06=30 600=24*25
    30xxxxxxx 06*07=42 1260=35*36
    42xxxxxxx 07*08=56 2352=48*49
    56xxxxxxx 08*09=72 4032=63*64
    72xxxxxxx 09*10=90 6480=80*81
    90xxxxxxx 10*11=110 9900=99*100
    110 * 132= 14520=120*121
    132 * 156= 20592=143*144
    156 * 182= 28392=168*169
    182 * 210= 38220=195*196
    210 * 240= 50400=224*225
    240 * 272= 65280=255*256
    272 * 306= 83232=288*289
    306 * 342=104652=323*324
    342 * 380=129960=360*361
    380 * 420=159600=399*400
    420 * 462=194040=440*441

    What I can see is, that if
    m(m+1) * n(n+1) = x = q(q+1)
    than
    m(m+1) is related to q because q is:
    first m(m+1) plus 1
    second m(m+1) plus 2
    third m(m+1) plus 3
    and so forth.
    twenteeth possibility I showed q is m(m+1) plus 20

    I think that continues ..... and is infinite
    but how do I put this in a proof????????

    And also this is only one way to find the second obedient number, it could also be found in a different way like I don't know how, but I have to exclude it in a proof.

    Could you please help me to find the equotation with one or even two unknown ????
    Thank you very much, Sincerly, E.

    And what does LHS mean?
    Last edited by enoimreh7; 11-14-2017 at 08:53 PM.

  9. #9
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    Quote Originally Posted by enoimreh7 View Post
    And what does LHS mean?
    LHS = Left Hand Side

    Not a proof. One must show them ALL.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  10. #10
    Quote Originally Posted by tkhunny View Post
    LHS = Left Hand Side

    Not a proof. One must show them ALL.
    First an equation, than use the equation. Do you really mean there is no equation? Sincerly, E.

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