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 ?

Writing everything in prime powers, $$4^3 = 2^6$$ and $$6^7 = 2^7 \cdot 3^7$$, so the number is $$2^{5+6+7} \cdot 3^{6+7} \cdot 5^3 = 2^{18} \cdot 3^{13} \cdot 5^3$$. A perfect square needs every exponent even, and here the exponents of 3 and 5 are odd. Multiplying by $$3 \times 5 = 15$$ fixes both, so the least such integer is 15.