Question 1150758
A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series. 
<pre>Yet, ANOTHER method!!
Sum of an A.P.:	{{{matrix(1,3, S[n], "=", (n/2)(2a[1] + (n - 1)d))}}}
                {{{matrix(1,3, S[567], "=", (567/2)(2a[1] + (567 - 1)1))}}} ------ Substituting 567 for n, and 1 for d
                {{{matrix(1,3, S[567], "=", 567(2a[1] + 566)/2)}}} ======> {{{matrix(1,3, S[567], "=", 567(2)(a[1] + 283)/2)}}} ======> {{{matrix(1,3, S[567], "=", 567cross((2))(a[1] + 283)/cross(2))}}} =====> {{{matrix(1,3, S[567], "=", 567(a[1] + 283))}}}
		{{{matrix(1,3, S[567], "=", (3^4)(7)(a[1] + (3^4)(7) * 283))}}} ------- Substituting PRIME FACTORS
                {{{matrix(1,3, S[567], "=", (3^4 * 7)(a[1] + 283))}}} ------ Factoring out GCF, 3<sup>4</sup> * 7
From above, it can be seen that a PERFECT CUBE of base 3 would be 3<sup>6</sup>, so ANOTHER 3<sup>2</sup> is needed (to be MULTIPLIED), and a PERFECT CUBE of base 7 would be 7<sup>3</sup>,
and so, ANOTHER 7<sup>2</sup> is needed (to be MULTIPLIED) also.
Therefore, for the SMALLEST CUBE, we need to have: 3<sup>6</sup> * 7<sup>3</sup>, or {{{highlight_green(matrix(1,6, the, SMALLEST, POSSIBLE, SUM, of, "250,047"))}}}, which is actually {{{highlight_green(matrix(1,7, 3^6 * 7^3, "=", (3^2)^3 * 7^3, "=", (3^2 * 7)^3, "=", 63^3))}}}.