We have 300 = 2² × 3 × 5²

Let xyz = 2² × 3 × 5² ……..(i)

Now x will be of the form 2^a1 × 3^b1 × 5^c1, y will be of the form 2^a2 × 3^b2 × 5^c2 and z will be of the form

2^a3 × 3^b3 × 5^c3.

Hence (i) becomes

2^a1+a2+a3 × 3^b1+b2+b3 × 5^c1+c2+c3 = 2² × 3 × 5²

Hence,

a₁ + a₂ + a₃ = 2 which gives us ⁴C₂ = 6 solutions

b₁ + b₂ + b₃ = 1 which gives ³C₂ = 3 solutions

c₁ + c₂ + c₃ = 2 which gives ⁴C₂ = 6 solutions

Hence, total solutions are 6 × 3 × 6 = 108.

Now out of these 108 solutions, there are some solutions in which two integers are equal. These are

1 ×1 × 300, 2 × 2 × 75, 5 ×5 × 12, 10 × 10 × 3. Now all these solutions must have occurred 3 times in the above 108 solutions. Hence, there are 108 – 12 = 96 solutions in which x, y and z are distinct.

Now x, y and z can be arranged in 3! = 6 ways. Hence the number of distinct solutions are 96/6 = 16. Hence there are 4 solutions in which two integers are equal and 16 solutions in which all the integers are unequal.

Now in case of 4 solutions, where 2 integers are equal, we have the cases where two integers are negative. For example, corresponding to the solution, 1 ×1 × 300, there are two other cases like

– 1 ×- 1 × 300 and 1 × – 1 × – 300. Hence, there are 4 × 3 = 12 solutions where two of the integers are equal.

Again for every solution where all the three integers are unequal, we can have 2 negative integers. This can be done in ³C₁ or 3 ways. For example, corresponding to the solution say 4 × 3 × 25 we have – 4 × – 3 × 25, – 4 × 3 × – 25, 4 × – 3 × – 25.

Hence, corresponding to each solution where x, y and z are distinct, we have 3 more solutions.

Hence, in total we have 16 × 4 = 64 solutions where all the three integers are distinct.

Hence, the total number of solutions is 64 + 12 = 76.