Let X and Y be the sets of all positive divisions of400 and 1000 respectively (including 1 and the number). Then `n(X cap Y)` is T, then `(T)/(4)` is-

Answer

To solve the problem, we need to find the number of common positive divisors of the numbers 400 and 1000, and then divide that count by 4.

### Step 1: Find the divisors of 400
First, we need to find the prime factorization of 400.
- 400 can be expressed as:
\[ 400 = 4 \times 100 = 4 \times 10^2 = 2^2 \times (2 \times 5)^2 = 2^4 \times 5^2 \]

### Step 2: Find the divisors of 1000
Next, we find the prime factorization of 1000.
- 1000 can be expressed as:
\[ 1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3 \]

### Step 3: Find the common divisors (intersection of sets X and Y)
The common divisors of two numbers can be found using the greatest common divisor (GCD). We first find the GCD of 400 and 1000 using their prime factorizations:
- For 400: \(2^4 \times 5^2\)
- For 1000: \(2^3 \times 5^3\)

The GCD is obtained by taking the minimum power of each prime factor:
- For \(2\): minimum of \(4\) and \(3\) is \(3\)
- For \(5\): minimum of \(2\) and \(3\) is \(2\)

Thus, the GCD is:
\[ \text{GCD}(400, 1000) = 2^3 \times 5^2 = 8 \times 25 = 200 \]

### Step 4: Find the number of divisors of the GCD
Now we need to find the number of positive divisors of 200.
- The prime factorization of 200 is:
\[ 200 = 2^3 \times 5^2 \]
- The number of divisors can be calculated using the formula \((e_1 + 1)(e_2 + 1)\) where \(e_1, e_2, \ldots\) are the powers of the prime factors.
- Here, \(e_1 = 3\) (for \(2\)) and \(e_2 = 2\) (for \(5\)):
\[ \text{Number of divisors} = (3 + 1)(2 + 1) = 4 \times 3 = 12 \]

### Step 5: Calculate \(T/4\)
Let \(T\) be the number of common divisors, which we found to be 12.
Now we calculate:
\[ \frac{T}{4} = \frac{12}{4} = 3 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{3} \]