Consider a rational number `a/b` in its lowesty form `a,b` are integers, with `0 lt (a)/(b) lt 1, b gt 1`. How many of these have `ab=15!`?

Answer

To solve the problem of finding how many rational numbers \( \frac{a}{b} \) in lowest terms exist such that \( ab = 15! \) and \( 0 < \frac{a}{b} < 1 \) with \( b > 1 \), we can follow these steps:

### Step 1: Factorial Representation
We start by noting that \( ab = 15! \). The prime factorization of \( 15! \) is essential for our calculations.

\[ 15! = 1 \times 2 \times 3 \times \ldots \times 15 \]

### Step 2: Prime Factorization of \( 15! \)
The prime factorization of \( 15! \) can be computed as follows:

- Count the powers of each prime number up to 15:
- \( 2^{11} \) (from multiples of 2)
- \( 3^{6} \) (from multiples of 3)
- \( 5^{3} \) (from multiples of 5)
- \( 7^{2} \) (from multiples of 7)
- \( 11^{1} \) (from multiples of 11)
- \( 13^{1} \) (from multiples of 13)

Thus, we can express \( 15! \) as:

\[ 15! = 2^{11} \times 3^{6} \times 5^{3} \times 7^{2} \times 11^{1} \times 13^{1} \]

### Step 3: Finding Pairs \( (a, b) \)
Since \( ab = 15! \), we can express \( a \) and \( b \) in terms of their prime factors:

Let \( a = 2^{x_1} \times 3^{x_2} \times 5^{x_3} \times 7^{x_4} \times 11^{x_5} \times 13^{x_6} \) and \( b = 2^{y_1} \times 3^{y_2} \times 5^{y_3} \times 7^{y_4} \times 11^{y_5} \times 13^{y_6} \).

From the equation \( ab = 15! \), we have:

\[ x_i + y_i = \text{power of prime } i \text{ in } 15! \]

for \( i = 1, 2, 3, 4, 5, 6 \).

### Step 4: Conditions for Lowest Terms
For \( \frac{a}{b} \) to be in lowest terms, \( a \) and \( b \) must not share any prime factors. This means that for each prime factor, either all of its power goes to \( a \) or all goes to \( b \).

### Step 5: Counting Combinations
Since there are 6 distinct prime factors, and for each prime factor, we have 2 choices (either it goes to \( a \) or \( b \)), the total number of combinations is:

\[ 2^6 = 64 \]

### Step 6: Adjusting for the Condition \( 0 < \frac{a}{b} < 1 \)
Since \( \frac{a}{b} < 1 \), we must have \( a < b \). Out of the 64 combinations, half will satisfy \( a < b \) and half will satisfy \( a > b \). Therefore, we have:

\[ \frac{64}{2} = 32 \]

### Step 7: Excluding Invalid Cases
However, we must exclude the cases where either \( a = 0 \) or \( b = 0 \) (which is not possible since \( b > 1 \)). Thus, we do not need to subtract any cases here.

### Final Answer
Thus, the total number of rational numbers \( \frac{a}{b} \) in lowest terms such that \( ab = 15! \) and \( 0 < \frac{a}{b} < 1 \) is:

\[ \boxed{32} \]