“…Then, x 3 = (2/3)f 0 , x i = 2f 0 /i, hence, 8 = (i − 4) × 2f 0 /i − (2/3) × f 0 , i.e., f 0 = 6i/(i − 6). Therefore, (i, f 0 , x 3 , x i ) = (7,42,28,12), (8,24,16,6), (9,18,12,4), (10,15,10,13) as x i ≥ 3 for all i, and f 0 ≥ 12. So, [p n 1 1 , .…”