AMC 10 · 2008 · #13
Grade 6 algebraPick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The average of a group equals its total divided by how many things are in it. So I give the running total a name, S_n, the sum of the first n terms. The average rule turns into a clean formula for S_n. Then a single term is just the jump from one running total to the next, which splits the task into a small subproblem I can compute.
Turn the average into a sum
Total = average × count, so the sum S_n of the first n terms is n × n = n².
If a pile of n numbers averages to n, the whole pile must add up to n copies of n.
6.SP.B.5Introduce A VariableOne term is the jump between totals
The 2008th term is what the total gains from S₂₀₀₇ to S₂₀₀₈, so it equals 2008² - 2007².
Take away the total of everything before it, and only that one new term is left.
6.EE.B.6Identify SubproblemsFactor the difference of squares
Use a² - b² = (a-b)(a+b): 2008² - 2007² = 1 × 4015 = 4015, which is choice (B).
The gap between two neighbouring squares is just the sum of the two numbers being squared.
The gap between two neighbouring squares is just the sum of the two numbers being squared.
▸ Why?
A difference of two squares is the two numbers added multiplied by the two subtracted.
▸ Why?
Neighbours differ by one, so that second factor is one and only the sum is left.
If the average of the first n terms is always n, the running total is n², so each term is the gap between two neighbouring squares.
- Turn the average into a sum
- One term is the jump between totals
- Factor the difference of squares