A patio was to be laid in a design with one tile in the

first row, two tiles in the second row, three

tiles in the third row, and so on. Mr. Tong had 60

tiles to use. How many tiles should be placed

in the bottom row to use the most tiles?

10

Step-by-step explanation:

The number of tiles in the design is 1 + 2 + 3 + ...

We can model this as an arithmetic series, where the first term is 1 and the common difference is 1. The sum of the first n terms of an arithmetic series is:

S = n/2 (2a₁ + d (n - 1))

Given that a₁ = 1 and d = 1:

S = n/2 (2 (1) + n - 1)

S = n/2 (n + 1)

Since S ≤ 60:

n/2 (n + 1) ≤ 60

n (n + 1) ≤ 120

n must be an integer, so from trial and error:

n ≤ 10

Mr. Tong should use 10 tiles in the final row to use the most tiles possible.