document.write( "Question 1207623: Let $m$ and $n$ be positive integers. If $m$ has exactly $10$ positive divisors, $n$ has exactly $16$ positive divisors, and $mn$ has exactly $21$ positive divisors. How many divisors does $m^2 n^2$ have?
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Algebra.Com's Answer #845595 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "Let p and q represent unknown prime numbers.

\n" ); document.write( "Using the method for finding the number of positive divisors of a number, if mn has 21 positive divisors, then it is either $\"p%5E20\"$ or $\"%28p%5E6%29%28q%5E2%29\"$ . So the product of m and n contains factors of at most two prime numbers.

\n" ); document.write( "If mn is $\"p%5E20\"$ , then m and n are both powers of p, but that would make m = $\"p%5E9\"$ and n = $\"p%5E15\"$ , in which case mn would be $\"p%5E24\"$ .

\n" ); document.write( "So mn is of the form $\"%28p%5E6%29%28q%5E2%29\"$ .

\n" ); document.write( "m has 10 positive divisors, and its prime factorization can have at most 2 different primes, and no prime factor can occur more than 6 times. The only possibility is then for m to be of the form $\"%28p%5E4%29%28q%5E1%29\"$ .

\n" ); document.write( "Similarly, n has 16 positive divisors, and its prime factorization can have at most 2 different primes, and no prime factor can occur more than 6 times. The only possibility is then for n to be of the form $\"%28p%5E3%29%28q%5E3%29\"$ .

\n" ); document.write( "But if mn is of the form $\"%28p%5E6%29%28q%5E2%29\"$ , n can't be of the form $\"%28p%5E3%29%28q%5E3%29\"$ .

\n" ); document.write( "So, while the problem could have been a good one for using math and logical reasoning to find the solution, the given information is inconsistent -- it is impossible for m to have 10 positive divisors, n to have 16, and mn to have 21.

\n" ); document.write( "ANSWER: NO SOLUTION

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