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Math Help - help plz...

  1. #11
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    Wink Thank you

    Quote Originally Posted by CaptainBlack
    Right, I can solve this, but the solution is does not qualify as high school
    maths. Are you sure that you don't already have a short cut for this problem
    in your notes or textbook?

    Also its a lot of typing, I will start typing the solution in about 3 hours.

    RonL
    I saw the first problem in a magazine not in my textbook,so you can work it out in any ways.And thank you for your later typing.

    The second one is from Cambridge College Math Institute and the answer to this problem will not be given untill several months later. Anyone who works it out can send email to puzzlemaster@cambridgecollege.edu with your solution.

    Thank you , thank you for your thinking and typing,
    and words fail me to say anything others.
    Last edited by yiyayiyayo; February 4th 2006 at 07:25 PM.
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  2. #12
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    The problems were very nice problems.
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  3. #13
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    I was further thinking about the second problem you gave. I believe I have a simplified version other than I posted previously. I think (not completely proven but close to it) that all number which CANNOT be expressed as the sum of consecutives is of the form 2^n.

    I will not post my solution now, but later on in a few hours, I do not have enough time now.
    Last edited by ThePerfectHacker; February 5th 2006 at 01:51 PM.
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  4. #14
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    Let me show my proof to problem 2.

    Notice that a number expressed, as m consecutives (odd number), then we have,
    n+(n+1)+(n+2)+...+(n+m-1),
    Thus,
    mn+\frac{m(m-1)}{2},
    Now m-1 is even, thus,
    mn+mj
    where j is an integer thus,
    m(n+j).
    Thus, if a number can be expressed as a sum of m (odd) consecutives, then it is divisible by m.

    Now we prove, the converse (not exactly a converse you will see). That if a number is divisible by m then, we can express it as a sum of consecutives (I did not say m consecutives rather I said it can be expressed as consecutives!). If m|n then there exists k such as mk=n. Thus, instead of k write m(j+s)=n where j=(m-1)/2 (as before), thus we can find such an s. The problem is that s can be a negative number. Thus, the number can be expressed as:
    n=s+(s+1)+(s+2)+...+(s+m-1).
    The following is the foundation of the proof; observe that if a number is expressed as a sum of consecutives starting from a negative number then it still can be expressed as a sum of consecutives! Because they cancel each other out. Observe,
    (-4)+(-3)+(-2)+(-1)+0+1+2+3+4+5+6+7
    Now follow with me. A number is expressed as a sum of consecutives from a negative number. Then they cancel each other out, and you are left with,
    5+6+7,
    still a sum of consecutives.
    The problem is of course if they cancel each other out in such a way then you a left with a single number. For example,
    (-1)+0+1+2, becomes,
    2.
    But 2 is not expressing 2 as a sum of consecutives by just 2 (that is trivial).
    But to show this does not happen over here is because since m is an odd number, and zero is part of this expansion, thus there are a total of even numbers. Thus, either the number of negatives is more than positives (which is not possible). They are equal (which is not possible). Thus, the number of positives must overtake the number of negatives by and even amount! Thus at least two which is considered a sum of consecutives.

    Now we have that ANY number divisible by any odd number CAN be expressed as a sum of consecutives. Thus, if a number IS NOT expressable as a sum of consecutives it must have the form 2^n.

    Finally, we prove the converse, that if a number has the form 2^n then it cannot be expressed as a sum of consecutives, then we have that,
    2^n=a+(a+1)+...+(a+b-1)
    Thus,
    2^n=ab+\frac{b(b-1)}{2}
    Thus,
    2^{n+1}=2ab+b(b-1)
    Thus,
    2^{n+1}=b(2a+b-1).
    Since of equality and the fact we are using integers we have that b cannot have odd factors thus,
    b=2^{m},
    Thus,
    2^{n+1}=2^m(2a+2^{m+1}-1)
    But the right factor of the LHS is odd,
    Thus an impossibility.
    Thus, all and only thus of the form 2^n. Cannot be expressed as a sum of consecutives.
    Q.E.D.

    I hope I did not make a mistake in proof. I think this is a nice problem and I had fun solving it.
    Last edited by ThePerfectHacker; December 18th 2007 at 02:13 PM.
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  5. #15
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    You're very kind,and I really hope too much typing didn't trouble you. Though I want to say something,my poor English frustrated me, sorry.
    In one word, your proof is wonderfull.Thank you.
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  6. #16
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    Quote Originally Posted by yiyayiyayo
    You're very kind,and I really hope too much typing didn't trouble you. Though I want to say something,my poor English frustrated me, sorry.
    In one word, your proof is wonderfull.Thank you.
    It took me a quick time to type that, besides I had fun solving this problem. Where are you from?
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  7. #17
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    I am from China and I'm a student studying in Senior Middle School.
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