Discussion

Explanation:

Let the integers be a, ar and ar2, then a + ar + ar2 = 111
∴ a(1 + r + r2) = 111
We note that (1 + r + r2) > 0 , hence a > 0
∴ a(1 + r + r2) = (1 × 111), or (3 × 37), or (37 × 3) or (111 × 1)

So, we have the following four cases:

Case 1: When a = 1
1 + r + r2 = 111
∴ r2 + r − 110 = 0
∴ (r + 11)(r − 10) = 0
∴ r = 10 or −11
So, we get two sets of solutions (1, 10, 100) and (1, −11, 121).

Case 2: When a = 3
1 + r + r2 = 37
∴ r2 + r − 36 = 0
This has no integral solution.

Case 3: When a = 37
1 + r + r2 = 3
∴ r2 + r − 2 = 0
∴ (r − 1)(r + 2) = 0
∴ r = 1 or −2
So, we get two sets of solutions (37, 37, 37) and (37, −74, 148). The first set is to be rejected as the integers are not distinct.

Case 4: When a = 111
1 + r + r2 = 1
∴ r2 + r  = 0
∴ r(r + 2) = 0
∴ r = 0 or −1
So, we get two set of solutions (111, 0, 0) and (111, −111, 111). Both are to be rejected as the integers are not distinct.

∴ From the above four cases, we have 3 distinct valid sets.

Hence, 3.

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