AMC 10 · 2013 · #4
Grade 4 countingPick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question literally asks for a position in a list read from the end, so Tool #11 (Work Backwards) fits directly: start at 201 and count down until 53 is reached. Tool #7 (Identify Subproblems) supplies the one reusable fact this rests on — how many whole numbers lie in a stretch from one value to another, inclusive. Tool #5 (Look for a Pattern) gives a fast cross-check, since a forward position and a backward position in the same list always add to a fixed total.
Count-inclusive rule
From a up to b, both ends counted, there are b - a + 1 numbers. Test it: forward, 53 - 3 + 1 = 51, the stated 51st.
Subtracting the endpoints counts the gaps between numbers; the +1 adds back the starting number itself so both ends are included.
Subtracting the endpoints counts the gaps, so one is added back to include the starting number.
▸ Why?
The numbers march with a single fixed step, so gaps and steps line up exactly.
▸ Why?
Each number after the first pairs with one gap, leaving the first number needing its own count.
Reframe the backward count
Backward the count starts at 201 and steps down to 53, so n is just how many numbers run from 201 down to 53, both ends in.
Reading the list from the top, 53's rank is just the size of the block of numbers from the top end 201 down to 53.
2.OA.A.1Work BackwardsCompute n
201 - 53 = 148, then +1 for the endpoint 53 itself gives 149. Counting backward, 53 is the 149th number.
The single multi-digit subtraction 201 - 53 does all the work; the +1 is the same include-both-ends correction as before.
4.NBT.B.4Work BackwardsTo find a spot in a backward list, count how many numbers sit from the top end down to yours: 201 - 53 + 1 = 149, using include-both-ends counting you already know.
- Count-inclusive rule
- Reframe the backward count
- Compute n
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