If x^3 is a positive factor of 10!*11!, how many possible integer values of x are there?
First, let's find the prime factorization of \(10! \cdot 11!\):
\(10! \cdot 11! = 11 \cdot 10 \cdot10 \cdot9 \cdot9 \cdot8 \cdot8 \cdot7 \cdot7 \cdot6 \cdot6 \cdot5 \cdot5 \cdot4 \cdot4 \cdot3 \cdot3 \cdot2 \cdot2 \cdot1 \cdot1 = 2^{16} \cdot3^{8} \cdot5^{4} \cdot7^{2} \cdot11^{1}\)
Next, let's test values for x (by using the prime factorizion, of course!):
x = 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, ...
I will let you finish the problem. Keep in mind that the highest x can be is \(2^{5}\cdot3^{2}\cdot5=1440\).
Caution: It is important to note that the question says "integers" instead of "positive integers." The cube of a negative number is still negative, so it will not change the answer to this particular problem. However, if x was squared or raised to any even power, we would have to include the negatives.
