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    Stuck on this question, prove or disprove sigma(n) is less than or equal to n^2 for n>0 where sigma(n) is the sum of the divisiors of n, so sigma(12)=1+2+3+4+6+12

    I can tell by intuition this is true, but don't know how to prove it
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    (Original post by Jimmath)
    Stuck on this question, prove or disprove sigma(n) is less than or equal to n^2 for n>0 where sigma(n) is the sum of the divisiors of n, so sigma(12)=1+2+3+4+6+12

    I can tell by intuition this is true, but don't know how to prove it
    It might be helpful to consider the sum of the reciprocals of the divisors.

    Edit: Or better, consider what is the maximum number of divisors and what is the biggest they can be!
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    (Original post by Jimmath)
    Stuck on this question, prove or disprove sigma(n) is less than or equal to n^2 for n>0 where sigma(n) is the sum of the divisiors of n, so sigma(12)=1+2+3+4+6+12

    I can tell by intuition this is true, but don't know how to prove it
    As an extension, have this problem:

    Let  n\geq 2 be a positive integer with divisors  1=d_1<d_2<...<d_k=n.
    Prove that  d_1d_2+d_2d_3+d_3d_4...+d_{k-1}d_k<n^2

    As a further extension, determine when  d_1d_2+d_2d_3+d_3d_4...+d_{k-1}d_k is a divisor of  n^2
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    (Original post by Renzhi10122)
    As an extension, have this problem:

    Let  n\geq 2 be a positive integer with divisors  1=d_1<d_2<...<d_k=n.
    Prove that  d_1d_2+d_2d_3+d_3d_4...+d_{k-1}d_k<n^2
    Consider in relation to n,

    What is d_k ?
    What is the largest value d_{k-1} can be?

    Continue.

    edit: I assume you are saying you have that problem, rather than proposing it for the OP. If not, ignore this.
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    (Original post by ghostwalker)
    Consider in relation to n,

    What is d_k ?
    What is the largest value d_{k-1} can be?

    Continue.

    edit: I assume you are saying you have that problem, rather than proposing it for the OP. If not, ignore this.
    Nah, ive solved this one, OP's problem simply reminded me of this.
 
 
 
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