The total number of positive integral solutions (x, y, z) such that xyz = 24 is :

Solution

Maths sir · 4 steps

Break 24 into primes, then share each prime's power among x, y and z separately.

1. Prime factorise. 24 = 2³ × 3¹. Every positive divisor of 24 is some power of 2 times some power of 3, so each of x, y, z looks like 2ᵃ3ᵇ.
2. Share the 2s. The powers of 2 must satisfy a₁ + a₂ + a₃ = 3 with each aᵢ ≥ 0. By stars and bars this has C(3+2, 2) = C(5, 2) = 10 solutions.
3. Share the 3. The powers of 3 must satisfy b₁ + b₂ + b₃ = 1. The single 3 goes to exactly one of x, y, z, so there are C(3, 2) = 3 ways.

   C(5, 2) × C(3, 2) = 10 × 3 = 30

4. So. The two primes are shared independently, so multiply. There are 30 ordered triples, which is option (d).

The idea to keep. For xyz = N, write N = pᵃ qᵇ … and share each exponent among the variables by stars and bars, C(a+r-1, r-1). Then multiply over the primes.

Answer from the NTA final answer key. Solution by PrepWiser.