document.write( "Question 670019: Sum of digit of the smallest number by which 1440 be multiplied so that it becomes a perfect cube. \n" ); document.write( "
Algebra.Com's Answer #416741 by Edwin McCravy(20055)\"\" \"About 
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document.write( "We break 1440 down into its prime factorization:\r\n" );
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document.write( "1440 = 2×720 = 2×2×360 = 2×2×2×180 = 2×2×2×2×90 = 2×2×2×2×2×45 =\r\n" );
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document.write( "2×2×2×2×2×3×15 = 2×2×2×2×2×3×3×5 = 25×32×51\r\n" );
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document.write( "The smallest cube larger than that is the smallest one in which\r\n" );
document.write( "the exponents of the primes 2,3, and 5 are all multiples of 3.\r\n" );
document.write( "That would be 26×33×53.\r\n" );
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document.write( "So \r\n" );
document.write( "we need 21 to multiply by the 25 to give 26\r\n" );
document.write( "we need 31 to multiply by the 32 to give 33\r\n" );
document.write( "we need 52 to multiply by the 51 to give 53\r\n" );
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document.write( "So we need to multiply 1440 by\r\n" );
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document.write( "21×31×52 = 150 so that it will become\r\n" );
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document.write( "26×33×53, and have all the exponents of\r\n" );
document.write( "prime numbers being multiples of 3.\r\n" );
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document.write( "The sum of the digits of 150 is 1+5+0 = 6.\r\n" );
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document.write( "Edwin
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