A bit of geometry is what’s challenging the team this week.
Back from a foreign trip, the coach reports an interesting pricing policy at a certain country’s post office. A package is charged according to the total of length + width + height of the package, where the three dimensions are whole numbers. The coach had considered a box with integers for each of the dimensions but found by changing each dimension by no more than one unit, the sum could be reduced while retaining the same volume.
If the volume of the coach’s package was between 300 and 400, what were the dimensions of the package?
The coach has invited the team to a pool party at his house. The pool is circular and the largest possible for the coach’s yard. It just touches the corner of a 2 foot by 9 foot walkway nestled in the lower left corner of the yard as shown.
How big is the coach’s yard?
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Solutions to week 145
Find m and n answer explained:
Rewrite the equation as 615 = 4n – m2. Then 615 = 3 × 5 × 41 = (2n – m) × (2n + m). Note that the sum of the two factors of 615 must be 2n+1 and the only factors giving this are 5 and 123. This produces m = ±59 and n = 6 as the only answers.
Escape Room answer explained:
Let G = the number of people in the larger group, P = the initial price per minute, and T = the time limit. If the larger group spent a time, t1, between T/2 and T, then their total charge was PGt1 and the total charge for the group of three was 3P[T + (2t1 – T) × 3/2]. If these are equal, then t1 = 1.5T/(9 – G) and the total charge for the two groups is C1 = 3GPT/(9 – G).
Links to all of the puzzles and solutions are on the Complete Varsity Math page.
Come back next week for answers and more puzzles.