document.write( "Question 823707: Please help me fast for this problem!
\n" ); document.write( "how many couples of (p,q) If known LCM of p and q is 3300?
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Algebra.Com's Answer #495727 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "3300 = 2×2×3×5×5×11\r\n" );
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document.write( "3300 has 2 factors each of 2 and 5, and 1 factor each of 3 and 11\r\n" );
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document.write( "So for p and q to have LCM = 3300, the following must be true:\r\n" );
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document.write( "A. One or the other of p and q must have 2 factors of 2 and the\r\n" );
document.write( "other have 0,1, or 2 factors of 2.  The same must be true about\r\n" );
document.write( "factors of 5.  \r\n" );
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document.write( "Also,\r\n" );
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document.write( "B. One or the other of p and q must have 1 factor of 3 and the\r\n" );
document.write( "other have 0 or 1 factor of 3.  The same can be said about\r\n" );
document.write( "factors of 11.\r\n" );
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document.write( "So let's list the cases:\r\n" );
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document.write( "Factors of 2:\r\n" );
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document.write( "Case 1: both p and q have 2 factors of 2.\r\n" );
document.write( "Case 2: p has 2 factors of 2 and q has 0 factors of 2.\r\n" );
document.write( "Case 3: p has 2 factors of 2 and q has 1 factor  of 2.\r\n" );
document.write( "Case 4: p has 0 factors of 2 and q has 2 factors of 2.\r\n" );
document.write( "Case 5: p has 1 factor  of 2 and q has 2 factors of 2.\r\n" );
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document.write( "That's 5 ways the factors of 2 can be had by p and q.\r\n" );
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document.write( "Factors of 3:\r\n" );
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document.write( "Case 1: both p and q have 1 factor of 3.\r\n" );
document.write( "Case 2: p has 1 factor  of 3 and q has 0 factors of 3.\r\n" );
document.write( "Case 3: p has 0 factors of 3 and q has 1 factor  of 3.\r\n" );
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document.write( "That's 3 ways the factors of 3 can be had by p and q.\r\n" );
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document.write( "Factors of 5: Same as for factors of 2\r\n" );
document.write( "That's 5 ways the factors of 5 can be had by p and q.\r\n" );
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document.write( "Factors of 11: Same as for factors of 11\r\n" );
document.write( "That's 3 ways the factors of 5 can be had by p and q.\r\n" );
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document.write( "Answer: 5×3×5×3 = 225\r\n" );
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document.write( "Note: The 225 counts, for instance, (44,75) and (75,44) as two\r\n" );
document.write( "different couples.  Suppose you want to avoid such duplicates\r\n" );
document.write( "as that and you want to count only the cases where say p≦q,\r\n" );
document.write( "Then since the only case of p=q is (3300,3300), we can subtract\r\n" );
document.write( "that case and divide by 2.  That is 225-1=224, and then divide\r\n" );
document.write( "that by 2, getting 112.  Then we add the couple (3300,3300) that\r\n" );
document.write( "we subtracted, and that would give 113 couples where p≦q.\r\n" );
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document.write( "Edwin
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