Question: In how many ways can 111 be written as the sum of three distinct integers in geometric progression?
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.