What is the least positive integer by which $2^5 \cdot  3^6 \cdot  4^3 \cdot  5^3 \cdot 6^7$ should be multiplied so that, the product is a perfect square?
1. First, we need to express each factor in the given product in terms of its prime factors:
   \[
   2^5 \cdot 3^6 \cdot 4^3 \cdot 5^3 \cdot 6^7
   \]
   - \(4 = 2^2\), so \(4^3 = (2^2)^3 = 2^6\)
   - \(6 = 2 \cdot 3\), so \(6^7 = (2 \cdot 3)^7 = 2^7 \cdot 3^7\)

2. Substitute these into the original expression:
   \[
   2^5 \cdot 3^6 \cdot 4^3 \cdot 5^3 \cdot 6^7 = 2^5 \cdot 3^6 \cdot 2^6 \cdot 5^3 \cdot 2^7 \cdot 3^7
   \]

3. Combine the powers of the same base:
   \[
   2^{5+6+7} \cdot 3^{6+7} \cdot 5^3 = 2^{18} \cdot 3^{13} \cdot 5^3
   \]

4. For the product to be a perfect square, all the exponents in its prime factorization must be even. Currently, the exponents are:
   - \(2^{18}\) (even)
   - \(3^{13}\) (odd)
   - \(5^3\) (odd)

5. To make the exponents of \(3\) and \(5\) even, we need to multiply by \(3\) and \(5\) respectively:
   - Multiplying by \(3\) will change \(3^{13}\) to \(3^{14}\)
   - Multiplying by \(5\) will change \(5^3\) to \(5^4\)

6. Therefore, the least positive integer we need to multiply by is:
   \[
   3 \cdot 5 = 15
   \]

The final answer is \(\boxed{15}\)