AMC 8 · 2013 · #17
Grade 6 arithmeticalgebraPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Six consecutive integers have a clean pattern: their sum is always 6 times the average, and the average sits exactly in the middle of the list. Tool #5 (Look for a Pattern) lets us turn the sum directly into an average without algebra. Tool #7 (Identify Subproblems) splits the work into two easy pieces — first find the average, then count up to the largest. We avoid Tool #13 (Algebra) because the pattern lets a young solver finish using only division and a small addition.
For consecutive integers, sum = count × average, so 2013 ÷ 6 gives the average 335.5.
Treating the sum as 6 × average is the Grade 6 idea that the mean is the "balance point" of the data.
6.SP.B.5Look For A PatternWith an even count the average lands between the 3rd and 4th numbers, so the 3rd is 335 and the 4th is 336.
A decimal like .5 landing between two whole numbers is the Grade 5 "halfway point" reading of a decimal.
5.NBT.B.7Look For A PatternThe largest is the 6th number, three steps past the 3rd: 335 + 3 = 338.
Adding 3 to the 3rd term to reach the 6th term is a Grade 3 multi-step word-problem add.
3.OA.A.3Identify SubproblemsMatch 338 to the answer choices — it is option (B).
Closing the final subproblem by comparing your value to the listed options is a Grade 3 multi-step move.
3.OA.A.3Identify SubproblemsSix numbers in a row add up to 6 times their average — once you find the average (2013 ÷ 6 = 335.5), the largest is just three steps past the middle.