Let A be the least number such that 10A is a perfect square and 35 A is perfect cube. Then the number of positive divisors of A is :

Answer

To find the least number \( A \) such that \( 10A \) is a perfect square and \( 35A \) is a perfect cube, we can follow these steps:

### Step 1: Prime Factorization
First, we need to factor \( 10 \) and \( 35 \):
- \( 10 = 2^1 \times 5^1 \)
- \( 35 = 5^1 \times 7^1 \)

### Step 2: Conditions for Perfect Square and Perfect Cube
For \( 10A \) to be a perfect square, all the exponents in its prime factorization must be even. For \( 35A \) to be a perfect cube, all the exponents in its prime factorization must be multiples of 3.

Let’s denote the prime factorization of \( A \) as:
\[ A = 2^{x} \times 5^{y} \times 7^{z} \]

### Step 3: Analyze \( 10A \)
The prime factorization of \( 10A \) is:
\[ 10A = 2^{1+x} \times 5^{1+y} \times 7^{z} \]

For \( 10A \) to be a perfect square:
- \( 1 + x \) must be even
- \( 1 + y \) must be even
- \( z \) must be even

This gives us the following conditions:
1. \( x \) is odd (since \( 1 + x \) is even)
2. \( y \) is odd (since \( 1 + y \) is even)
3. \( z \) is even

### Step 4: Analyze \( 35A \)
The prime factorization of \( 35A \) is:
\[ 35A = 5^{1+y} \times 7^{1+z} \times 2^{x} \]

For \( 35A \) to be a perfect cube:
- \( 1 + y \) must be a multiple of 3
- \( 1 + z \) must be a multiple of 3
- \( x \) must be a multiple of 3

This gives us the following conditions:
1. \( y \equiv 2 \mod 3 \) (since \( 1 + y \equiv 0 \mod 3 \))
2. \( z \equiv 2 \mod 3 \) (since \( 1 + z \equiv 0 \mod 3 \))
3. \( x \equiv 0 \mod 3 \)

### Step 5: Solve the Conditions
Now we need to find the smallest non-negative integers \( x, y, z \) that satisfy all the above conditions.

1. **For \( x \)**:
- \( x \) is odd and \( x \equiv 0 \mod 3 \).
- The smallest odd number that is a multiple of 3 is \( 3 \).
- So, \( x = 3 \).

2. **For \( y \)**:
- \( y \) is odd and \( y \equiv 2 \mod 3 \).
- The smallest odd number that satisfies this is \( 5 \) (since \( 5 \equiv 2 \mod 3 \)).
- So, \( y = 5 \).

3. **For \( z \)**:
- \( z \) is even and \( z \equiv 2 \mod 3 \).
- The smallest even number that satisfies this is \( 2 \).
- So, \( z = 2 \).

### Step 6: Construct \( A \)
Now we can construct \( A \):
\[ A = 2^{3} \times 5^{5} \times 7^{2} \]

### Step 7: Calculate the Number of Positive Divisors
The number of positive divisors \( d(n) \) of a number \( n = p_1^{e_1} \times p_2^{e_2} \times p_3^{e_3} \) is given by:
\[ d(n) = (e_1 + 1)(e_2 + 1)(e_3 + 1) \]

For \( A = 2^{3} \times 5^{5} \times 7^{2} \):
- \( e_1 = 3 \) (for \( 2 \))
- \( e_2 = 5 \) (for \( 5 \))
- \( e_3 = 2 \) (for \( 7 \))

Thus, the number of positive divisors of \( A \) is:
\[ d(A) = (3 + 1)(5 + 1)(2 + 1) = 4 \times 6 \times 3 = 72 \]

### Final Answer
The number of positive divisors of \( A \) is \( 72 \).