Question 1150758
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Let  m+1, m+2, m+3, . . . . , n be a series of 567 consecutive integers with the sum of 567, with the first term (m+1),

where m is a positive integer, and the last term n,  n > m.


Then 

    {{{(n*(n+1))/2}}} - {{{(m*(m+1))/2}}} = {{{k^3}}}

using the formula of the sum of the first positive integer numbers, with some integer positive k.


The formula (1) implies

    {{{n^2+n}}} - {{{(m^2+m)}}} = {{{2k^3}}}

    {{{n^2-m^2}}} + {{{(n-m)}}} = {{{2k^3}}}

    (n-m)*(n+m) + (n-m)         = {{{2k^3}}}

    (n-m)*(n+m+1)               = {{{2k^3}}}.    (2)


Notice that  n-m = 567.  Thus the formula (2) becomes 

    567*(n+m+1)                 = {{{2k^3}}}.    (3)


The number 567 has the prime decomposition  567 = {{{3^4*7}}}.


Therefore, in order for equation (3) be true with lowest possible value of (n+m-1), it should be

    n + m + 1 = {{{2*3^2*7^2}}} = 882.


Thus we have two equations

    n - m     = 567,      (4)

    n + m + 1 = 882.      (5)


By adding equations, you get  

    2n + 1 = 567 + 882 = 1449,

    2n = 1449-1 = 1448,

     n = 1448/2 = 724.


Then from equation (4),  m = 724 - 567 = 157.


Thus the sequence is

    158, 159, 160, . . . , 724.


Its sum is {{{k^3}}} = {{{3^6*7^3}}} = {{{(3^2*7)^3}}} = {{{63^3}}} = 250047.


<U>ANSWER</U>.  The smallest possible positive sum for this series is {{{63^3}}} = 250047.
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