document.write( "Question 978290: I am a proper fraction.The sum of my numerator and my denominator is a one-digit number.Their product is a cube.What am I?\r
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Algebra.Com's Answer #599749 by Edwin McCravy(20060) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "If the numerator is 1, then the product of it and the denominator will be that\r\n" );
document.write( "denominator, so choose it as the only digit greater than 1 which is a cube,\r\n" );
document.write( "which is 8. Then the sum of numerator and denominator will still be a one-digit\r\n" );
document.write( "number, 9. \r\n" );
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document.write( "So one solution is .\r\n" );
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document.write( "But there another solution because 8 is also the product of 2 and 4, and 2+4=6\r\n" );
document.write( "is a one digit number \r\n" );
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document.write( "The only other solution is .  I agree that's not reduced to lowest terms, but \r\n" );
document.write( "the instructions don't state anything about it having to be reduced.\r\n" );
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document.write( "There are no more solutions because the next cube after 8 is 27, but it has\r\n" );
document.write( "factors 3 and 9, whose sum is 12, which is not a 1-digit number.\r\n" );
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document.write( "Answers:  and \r\n" );
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document.write( "Edwin

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