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Maths olympiad question · P5–P6★★★★ Expert

Tree planting: olympiad question 9

An expert question on tree planting for P5–P6, with a step-by-step solution, diagrams and the answer. Common in the Hua Xia Cup's P2 to P5 papers. Sawing logs, striking clocks and lining up lamp posts are all the same problem.

The question

Both sides of a straight road were planted with trees: each side from one end to the other, a tree at both ends, the trees evenly spaced a whole number of metres apart, from 3 to 30 m.

Both sides together have 36 trees.

Now the road is widened: one side's trees all go; the other side is respaced to one tree every 9 m, still with a tree at both ends.

Old trees already on a new spot stay; the rest go; every new spot without a tree gets a new one.

17 new trees are planted.

How long is the road, in metres?

Show the hints (3)
  1. Hint 1: 36 trees on both sides: how many on each side, and how many gaps? How is the road's length tied to the old spacing?
  2. Hint 2: A kept tree stands at a multiple of both the old spacing and 9 m. New trees = new spots − kept trees.
  3. Hint 3: 18 trees and 17 gaps a side. The road must be a multiple of 9, so the old spacing can only be 9, 18, 27 m. Work out the new trees for each.

Worked solution

Have a go first, then read the solution step by step.

  1. 1.Each side has 36 ÷ 2 = 18 trees, with trees at both ends, so 17 gaps: road = 17 × the old spacing.
  2. 2.The new side has trees at both ends, so the road is a multiple of 9; there are road ÷ 9 + 1 spots. Kept trees stand at common multiples of the old spacing and 9: road ÷ LCM + 1 of them. New = road ÷ 9 − road ÷ LCM.
  3. 3.17 × spacing must be a multiple of 9, so the spacing can only be 9, 18, 27 m. Each in turn: 9 m → road 153 m, 0 new; 18 m → road 306 m, 17 new ✓; 27 m → road 459 m, 34 new.
  4. 4.Only 18 m needs 17 new trees, so the road is 17 × 18 = 306 m.
  5. 5.Check: 306 ÷ 9 + 1 = 35 spots; LCM of 18 and 9 is 18, so 306 ÷ 18 + 1 = 18 kept; 35 − 18 = 17 new ✓
  6. ⚠️Common mistake: taking 36 as one side's trees, or counting kept trees in the middle only and missing the two at the ends (the ends are always multiples of both spacings).

Answer: 306 m

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