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 solve the problem, we need to find the least number \( A \) such that \( 10A \) is a perfect square and \( 35A \) is a perfect cube. We will break down the steps systematically.

### Step 1: Analyze the conditions for \( 10A \) to be a perfect square
The number \( 10 \) can be factored into its prime factors:
\[ 10 = 2^1 \times 5^1 \]
For \( 10A \) to be a perfect square, all the exponents in its prime factorization must be even. Let’s denote \( A \) in terms of its prime factorization:
\[ A = 2^x \times 5^y \times 7^z \times \ldots \]
Then,
\[ 10A = 2^{x+1} \times 5^{y+1} \times 7^z \times \ldots \]
To ensure \( 10A \) is a perfect square, we need:
- \( x + 1 \) is even
- \( y + 1 \) is even
- \( z \) is even (and so on for other primes)

This implies:
- \( x \) is odd
- \( y \) is odd
- \( z \) is even

### Step 2: Analyze the conditions for \( 35A \) to be a perfect cube
The number \( 35 \) can be factored as:
\[ 35 = 5^1 \times 7^1 \]
For \( 35A \) to be a perfect cube, all the exponents in its prime factorization must be multiples of 3. Thus,
\[ 35A = 5^{y+1} \times 7^{z+1} \times 2^x \times \ldots \]
We need:
- \( x \) is a multiple of 3
- \( y + 1 \) is a multiple of 3
- \( z + 1 \) is a multiple of 3

### Step 3: Determine the values of \( x, y, z \)
From the conditions derived:
1. \( x \) is odd and a multiple of 3. The smallest odd multiple of 3 is \( 3 \).
2. \( y \) is odd and \( y + 1 \) is a multiple of 3. The smallest odd number that satisfies this is \( 1 \) (since \( 1 + 1 = 2 \) which is not a multiple of 3, but \( 3 + 1 = 4 \) is not odd, so we try \( 1 \)).
3. \( z \) is even and \( z + 1 \) is a multiple of 3. The smallest even number that satisfies this is \( 2 \) (since \( 2 + 1 = 3 \) which is a multiple of 3).

Thus, we have:
\[ x = 3, \quad y = 1, \quad z = 2 \]

### Step 4: Calculate \( A \)
Now substituting back into the expression for \( A \):
\[ A = 2^3 \times 5^1 \times 7^2 \]

### Step 5: Calculate the number of positive divisors of \( A \)
The formula for the number of positive divisors \( d(n) \) of a number \( n = p_1^{e_1} \times p_2^{e_2} \times \ldots \) is given by:
\[ d(n) = (e_1 + 1)(e_2 + 1) \ldots \]
For \( A = 2^3 \times 5^1 \times 7^2 \):
- The exponent of \( 2 \) is \( 3 \) (so \( 3 + 1 = 4 \))
- The exponent of \( 5 \) is \( 1 \) (so \( 1 + 1 = 2 \))
- The exponent of \( 7 \) is \( 2 \) (so \( 2 + 1 = 3 \))

Thus, the number of positive divisors of \( A \) is:
\[ d(A) = (3 + 1)(1 + 1)(2 + 1) = 4 \times 2 \times 3 = 24 \]

### Final Answer
The number of positive divisors of \( A \) is \( 24 \).