AMC 10 · 2013 · #4
Easy mode Grade 4Write the whole numbers in order from 3 up to 201. Counting from the start, 53 is the 51st number. Now write the same numbers in the opposite order, from 201 down to 3. In this backward list, what position does 53 hold? Call that position n.
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: List the whole numbers from $3$ up to $201$. Counting forward, $53$ lands in the $51$st spot. Now list the same numbers in the opposite order, from $201$ back down to $3$. In this backward list, $53$ sits in the $n$th spot. Find $n$.
Givens: The numbers run consecutively from $3$ to $201$; Counting forward (from $3$), $53$ is the $51$st number; The backward list runs from $201$ down to $3$; Answer choices: (A) $146$, (B) $147$, (C) $148$, (D) $149$, (E) $150$
Unknowns: $n$, the position of $53$ when the list is counted backward from $201$
Understand
Restated: List the whole numbers from $3$ up to $201$. Counting forward, $53$ lands in the $51$st spot. Now list the same numbers in the opposite order, from $201$ back down to $3$. In this backward list, $53$ sits in the $n$th spot. Find $n$.
Givens: The numbers run consecutively from $3$ to $201$; Counting forward (from $3$), $53$ is the $51$st number; The backward list runs from $201$ down to $3$; Answer choices: (A) $146$, (B) $147$, (C) $148$, (D) $149$, (E) $150$
Plan
Primary tool: #11 Work Backwards
Secondary: #7 Identify Subproblems, #5 Look for a Pattern
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.
Execute — Answer: D
2.OA.A.1 Step 1 Count-inclusive rule
- First pin down the counting rule.
- The amount of whole numbers from a smaller value $a$ up to a larger value $b$, counting both ends, is $b - a + 1$.
- Check it against the given fact: forward from $3$, the position of $53$ should be $53 - 3 + 1 = 51$, which matches the stated $51$st.
- The rule is trustworthy.
💡 Subtracting the endpoints counts the gaps between numbers; the $+1$ adds back the starting number itself so both ends are included.
2.OA.A.1 Step 2 Reframe the backward count
- Counting backward starts at $201$ and steps down $201, 200, 199, \ldots$ until it reaches $53$.
- So $53$'s backward position is simply how many numbers get listed from $201$ down to $53$, counting both ends.
- By the same inclusive rule, that count is $201 - 53 + 1$.
💡 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 $53$.
4.NBT.B.4 Step 3 Compute n
- Do the arithmetic.
- $201 - 53 = 148$, then add $1$ to include the endpoint $53$ itself, giving $149$.
- So counting backward, $53$ is the $149$th number.
💡 The single multi-digit subtraction $201 - 53$ does all the work; the $+1$ is the same include-both-ends correction as before.
2.OA.A.1 First pin down the counting rule. The amount of whole numbers from a smaller val 2.OA.A.1 Counting backward starts at $201$ and steps down $201, 200, 199, \ldots$ until i 4.NBT.B.4 Do the arithmetic. $201 - 53 = 148$, then add $1$ to include the endpoint $53$ i Review
Reasonableness: The whole list holds $201 - 3 + 1 = 199$ numbers. A number's forward position and backward position in the same list must add up to $199 + 1 = 200$ (the $+1$ because the number itself is counted from both directions). Forward, $53$ was the $51$st, so backward it should be $200 - 51 = 149$. This agrees with (D). It is also sensible that a number near the small end sits near the far end of the backward list, i.e. a large position like $149$.
Alternative: Tool #5 (Look for a Pattern) on a tiny list: numbers $1,2,3,4,5$ counted backward are $5,4,3,2,1$. The number $2$ is $2$nd forward and $4$th backward, and $2 + 4 = 6 = 5 + 1 = \text{total} + 1$. The same 'forward $+$ backward $=$ total $+1$' pattern gives $n = 200 - 51 = 149$ instantly.
CCSS standards used (min grade 4)
2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Setting up the inclusive-count rule ($b - a + 1$) and reframing 'position when counting backward' as a count from $201$ down to $53$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Carrying out the multi-digit subtraction $201 - 53 + 1 = 149$ to get the backward position.)
⭐ To 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.
⭐ To 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.
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