document.write( "Question 618067: A positive integer n has exactly 4 positive divisors that are perfect fifth powers, exactly 6 positive divisors that are perfect cubes, and exactly 12 positive divisors that are perfect squares. Find the least possible number of possible integers that are divisors of n. \n" ); document.write( "
Algebra.Com's Answer #388771 by Edwin McCravy(20054)\"\" \"About 
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document.write( "If n is of the form p^r*q^s where p and q are primes, then all its divisors\r\n" );
document.write( "will be of the form p^a*q^b where 0<=a<=r and 0<=b<=s\r\n" );
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document.write( "To get 4 perfect 5th power divisors, we choose a and b to be multiples of 5,\r\n" );
document.write( "including 0.  Then the smallest possible 4 perfect 5th power divisors would be: \r\n" );
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document.write( "p^0*q^0, p^0*q^5, p^5*q^0, and p^5*q^5\r\n" );
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document.write( "To get 6 perfect cube divisors, we choose a and b to be multiples of 3,\r\n" );
document.write( "including 0.  Then the smallest possible 6 perfect cube divisors would be \r\n" );
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document.write( "p^0*q^0, p^0*q^3, p^0*q^6, p^3*q^0, p^3*q^3, p^3*q^6\r\n" );
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document.write( "To get 12 perfect square divisors, we choose a and b to be multiples of 2,\r\n" );
document.write( "including 0.  Then the smallest possible 12 perfect square divisors would be \r\n" );
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document.write( "p^0*q^0, p^0*q^2, p^0*q^4, p^0*q^6, p^2*q^0, p^2*q^2, p^2*q^4, p^2*q^6,\r\n" );
document.write( "p^4*q^0, p^4*q^2, p^4*q^4, p^4*q^6\r\n" );
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document.write( "The smallest way to get all those divisors is by chosing r=5 and s=6, i.e.,\r\n" );
document.write( "n = p^5*q^6.\r\n" );
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document.write( "Since all divisors of p^5*q^6 are of the form p^a*q^b where \r\n" );
document.write( "0<=a<=5 and 0<=b<=6, there are 6 choices for a and 7 choices for b.\r\n" );
document.write( "That makes 6*7 or 42 divisors of n\r\n" );
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document.write( "Edwin
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