document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #772201 by MathTherapy(10557)\"\" \"About 
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A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series.
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Yet, ANOTHER method!!
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document.write( "Sum of an A.P.:	\"matrix%281%2C3%2C+S%5Bn%5D%2C+%22=%22%2C+%28n%2F2%29%282a%5B1%5D+%2B+%28n+-+1%29d%29%29\"
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document.write( "                 ------ Substituting 567 for n, and 1 for d
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document.write( "                \"matrix%281%2C3%2C+S%5B567%5D%2C+%22=%22%2C+567%282a%5B1%5D+%2B+566%29%2F2%29\" ======> \"matrix%281%2C3%2C+S%5B567%5D%2C+%22=%22%2C+567%282%29%28a%5B1%5D+%2B+283%29%2F2%29\" ======>  =====> \"matrix%281%2C3%2C+S%5B567%5D%2C+%22=%22%2C+567%28a%5B1%5D+%2B+283%29%29\"
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document.write( "		 ------- Substituting PRIME FACTORS
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document.write( "                \"matrix%281%2C3%2C+S%5B567%5D%2C+%22=%22%2C+%283%5E4+%2A+7%29%28a%5B1%5D+%2B+283%29%29\" ------ Factoring out GCF, 34 * 7
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document.write( "From above, it can be seen that a PERFECT CUBE of base 3 would be 36, so ANOTHER 32 is needed (to be MULTIPLIED), and a PERFECT CUBE of base 7 would be 73,
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document.write( "and so, ANOTHER 72 is needed (to be MULTIPLIED) also.
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document.write( "Therefore, for the SMALLEST CUBE, we need to have: 36 * 73, or , which is actually . \n" );
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