AMC 10 · 2013 · #4

Grade 4 counting

Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery

consecutive-integer-countsymmetry-argument work-backwardspattern-recognition ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

When counting from $3$ to $201$ , $53$ is the $51^{\text{st}}$ number counted. When counting backwards from $201$ to $3$ , $53$ is the $n^{\text{th}}$ number counted. What is $n$ ?

Pick an answer.

(A)

146

(B)

147

(C)

148

(D)

149

(E)

150

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.

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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

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

#7 Identify Subproblems 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.

$$\text{position (forward)} = 53 - 3 + 1 = 51 \;\checkmark$$

💡 Subtracting the endpoints counts the gaps between numbers; the $+1$ adds back the starting number itself so both ends are included.

#11 Work Backwards 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 $201 - 53 + 1$.

$$n = 201 - 53 + 1$$

💡 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 $53$.

#11 Work Backwards 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.

$$n = 201 - 53 + 1 = 148 + 1 = 149 \;\Rightarrow\; \textbf{(D)}$$

💡 The single multi-digit subtraction $201 - 53$ does all the work; the $+1$ is the same include-both-ends correction as before.

[1] #7 2.OA.A.1 First pin down the counting rule. The amount of whole numbers from a smaller val

[2] #11 2.OA.A.1 Counting backward starts at $201$ and steps down $201, 200, 199, \ldots$ until i

[3] #11 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.1 Solve 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.4 Fluently 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Count-inclusive rule

Reframe the backward count

Compute n

Review

CCSS standards used (min grade 4)

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