Susan (circle), Ahmad (triangle), and Sam (star) are each walking in an octagonal park on a Sunday morning. At 10:00, the following figure depicts their positions. Susan moves 3 vertices clockwise in each hour, Ahmad moves 1 vertex clockwise each hour, while Sam moves 2 vertices anticlockwise each hour. Will all three ever meet on a vertex at the same instant?

Explanation

We are given:An octagonal park (8 vertices)Initial positions at 10:00 AM:Susan (circle) at vertex ASam (star) at vertex BAhmad (triangle) at vertex C (Assuming A, B, C are adjacent clockwise vertices just to denote initial positions)Movements per hour:Susan: 3 vertices clockwiseAhmad: 1 vertex clockwiseSam: 2 vertices anticlockwiseLet's define the 8 vertices as positions numbered from 0 to 7.Assume:Susan starts at position 0Sam at position 1Ahmad at position 2All three positions repeat every 8 hours (cycle repeats).From the table:Ahmad and Sam meet at hour 5 (position 7)But Susan is at position 7 only at hour 5 + 3 = 8 hours later (hour 13)Let's now check up to hour 23 (full LCM cycle)We check for any hour h such that:(3h mod 8) = (1h + 2) mod 8 = (-2h + 1) mod 8This is a modular system:Let's solve:Let all be equal to xSo,Susan: x ≡ 3h mod 8Ahmad: x ≡ h + 2 mod 8Sam: x ≡ -2h + 1 mod 8Now solve:3h ≡ h + 2 (mod 8) ⇒ 2h ≡ 2 (mod 8) ⇒ h ≡ 1 (mod 4)3h ≡ -2h + 1 (mod 8) ⇒ 5h ≡ 1 (mod 8)Find h such that 5h ≡ 1 mod 8 → Try h = 5: 5×5 = 25 ≡ 1 mod 8 So h = 5 satisfies both conditions.Check if it also satisfies h ≡ 1 mod 4: 5 ≡ 1 mod 4 Yes, all three will meet at the same vertex after 5 hours, i.e., at 3:00 PM.