AMC 8 · 2013 · #9

Easy mode Grade 6

Problem

The Incredible Hulk doubles his jump distance every time he jumps. His first jump is $1$ meter, his second is $2$ meters, his third is $4$ meters, his fourth is $8$ meters, and so on.

So the distance keeps doubling: $1, 2, 4, 8, 16, 32, \ldots$

Picture the jumps getting longer and longer. We want to know when his jump first goes over $1$ kilometer, which is the same as $1{,}000$ meters.

On which jump does he first travel more than $1{,}000$ meters?

Pick an answer.

(A)

$9^ ext{th}$

(B)

$10^ ext{th}$

(C)

$11^ ext{th}$

(D)

$12^ ext{th}$

(E)

$13^ ext{th}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: The Hulk's jumps double each time: $1, 2, 4, 8, \dots$ meters. On which jump does the distance first exceed $1{,}000$ meters?

Givens: $1$st jump $= 1$ m; Each jump is twice the previous one (geometric, common ratio $2$); Threshold to beat: $1$ km $= 1{,}000$ m; Answer choices: (A) $9$th, (B) $10$th, (C) $11$th, (D) $12$th, (E) $13$th

Unknowns: The smallest jump number $n$ such that the $n$-th jump distance is greater than $1{,}000$ m

Understand

Restated: The Hulk's jumps double each time: $1, 2, 4, 8, \dots$ meters. On which jump does the distance first exceed $1{,}000$ meters?

Plan

Primary tool: #5 Look for a Pattern

Secondary: #2 Make a Systematic List

The jumps are a clean doubling sequence, so Tool #5 (Look for a Pattern) is the natural fit: spot that the $n$-th jump is $2^{n-1}$. Tool #2 (Systematic List) then turns the search into bookkeeping — list the powers of $2$ in order until one breaks past $1{,}000$. No algebra or logarithms are needed; just keep doubling and stop at the first value over $1{,}000$.

Execute — Answer: C

#5 Look for a Pattern 4.OA.C.5 Step 1

Write the first few jump distances and look at how they grow.
Each one is double the one before, so the sequence is $1, 2, 4, 8, 16, \dots$ — a doubling pattern.

$$\text{Jump } 1{:}\;1,\;\text{Jump } 2{:}\;2,\;\text{Jump } 3{:}\;4,\;\text{Jump } 4{:}\;8$$

💡 Spotting the rule "each term is double the last" is exactly the Grade 4 "generate a number pattern following a given rule" standard.

#5 Look for a Pattern 6.EE.A.1 Step 2

Rewrite each distance as a power of $2$ to get a formula for the $n$-th jump.
Jump $1 = 2^0$, jump $2 = 2^1$, jump $3 = 2^2$, so jump $n = 2^{n-1}$.

$$\text{Jump } n = 2^{n-1}$$

💡 Writing $1, 2, 4, 8$ as $2^0, 2^1, 2^2, 2^3$ is a Grade 6 "whole-number exponents" expression.

#2 Make a Systematic List 6.EE.A.1 Step 3

List the powers of $2$ in order until one passes $1{,}000$.
Keep doubling and watch when the value crosses the threshold.

$$2^7 = 128,\;2^8 = 256,\;2^9 = 512,\;2^{10} = 1024$$

💡 Doubling one entry at a time in order is the Tool #2 systematic-list move applied to powers.

#2 Make a Systematic List 6.EE.B.5 Step 4

$2^9 = 512$ is still under $1{,}000$, but $2^{10} = 1024 > 1{,}000$.
So the exponent that first wins is $10$.

$$512 < 1000 < 1024 \;\Rightarrow\; \text{exponent} = 10$$

💡 Checking which value in the list first satisfies $2^{n-1} > 1000$ is the Grade 6 "find values that make an inequality true" idea.

#5 Look for a Pattern 4.OA.C.5 Step 5

Convert the exponent back to a jump number.
Since jump $n$ has distance $2^{n-1}$, an exponent of $10$ means $n - 1 = 10$, so $n = 11$.

$$n - 1 = 10 \;\Rightarrow\; n = 11 \;\Rightarrow\; \textbf{(C)}$$

💡 Mapping "position in the pattern" to "jump number" is just running the Grade 4 pattern rule backward.

[1] #5 4.OA.C.5 Write the first few jump distances and look at how they grow. Each one is double

[2] #5 6.EE.A.1 Rewrite each distance as a power of $2$ to get a formula for the $n$-th jump. Ju

[3] #2 6.EE.A.1 List the powers of $2$ in order until one passes $1{,}000$. Keep doubling and wa

[4] #2 6.EE.B.5 $2^9 = 512$ is still under $1{,}000$, but $2^{10} = 1024 > 1{,}000$. So the expo

[5] #5 4.OA.C.5 Convert the exponent back to a jump number. Since jump $n$ has distance $2^{n-1}

Review

Reasonableness: The $11$th jump is $2^{10} = 1024$ m, which is just barely above $1{,}000$ m, and the previous jump $2^9 = 512$ m is just barely below. That "just barely" feel is exactly what a doubling sequence should produce near a threshold, since each step jumps over a $2\times$ gap. The answer (C) sits in the middle of the choices, also consistent with the way AMC distractors flank a correct answer.

Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: just compute the jump distance for each. (A) $9$th $= 2^8 = 256$, (B) $10$th $= 2^9 = 512$, (C) $11$th $= 2^{10} = 1024$, (D) $12$th $= 2048$, (E) $13$th $= 4096$. The first one over $1{,}000$ is (C). Choices (D) and (E) also exceed $1{,}000$ but they are not the *first* — eliminate them on the "smallest $n$" requirement.

CCSS standards used (min grade 6)

4.OA.C.5 Generate a number or shape pattern following a given rule (Spotting the doubling rule $1, 2, 4, 8, \dots$ and mapping a pattern position back to a jump number.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Rewriting each jump distance as a power of $2$ ($2^{n-1}$) and evaluating $2^7, 2^8, 2^9, 2^{10}$.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Identifying the first exponent for which $2^{n-1} > 1000$ holds, namely $n - 1 = 10$.)

⭐ This AMC 8 problem only needs Grade 6 "powers of $2$" expressions you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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