AMC 10 · 2011 · #4
Grade 4 countingPick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Adding up 46 numbers twice would be slow and error-prone. Tool #5 (Look for a Pattern) spots the real structure instead: each number in Y is exactly 2 more than the matching number in X, so the whole difference is just (number of terms) times 2. Tool #7 (Identify Subproblems) handles the one supporting job — counting how many terms there are. Tool #16 (Change Focus) gives a fast cross-check by lining the two sums up and cancelling the block they share.
Match the lists term by term
Line the two sums up and pair them term by term: every term of Y is exactly 2 more than its partner in X.
Both lists climb by the same step, so shifting the start by one step makes every paired number differ by exactly that step.
Both lists climb by the same step, so shifting the start makes every paired number differ by exactly that shift.
▸ Why?
Each list adds the same fixed amount each time, so the two run in lockstep.
▸ Why?
Shifting both terms of a difference by the same amount leaves that difference unchanged.
Count how many terms there are
Count X's terms: (100-10) ÷ 2 = 45, plus 1 for the starting number, so 46 terms — and Y has just as many.
To count evenly-spaced numbers, see how many steps fit between the ends, then add one for the number you started on.
4.OA.A.3Identify SubproblemsAdd up the differences
All 46 pairs are ahead by 2, so Y-X is just the total of those gaps: 46 × 2 = 92, choice (A).
If every pair is ahead by the same amount, the whole list is ahead by that amount times the number of pairs.
4.NBT.B.5Look For A PatternWhen two lists are the same except shifted over, don't add them up — just multiply how far each number moved by how many numbers there are.
- Match the lists term by term
- Count how many terms there are
- Add up the differences