AMC 8 · 2013 · #9
Easy mode Grade 6Problem
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,…
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.
AMC 8 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: 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?
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
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
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.
💡 Spotting the rule "each term is double the last" is exactly the Grade 4 "generate a number pattern following a given rule" standard.
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 $n = 2^{n-1}$.
💡 Writing $1, 2, 4, 8$ as $2^0, 2^1, 2^2, 2^3$ is a Grade 6 "whole-number exponents" expression.
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.
💡 Doubling one entry at a time in order is the Tool #2 systematic-list move applied to powers.
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 $10$.
💡 Checking which value in the list first satisfies $2^{n-1} > 1000$ is the Grade 6 "find values that make an inequality true" idea.
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 $n = 11$.
💡 Mapping "position in the pattern" to "jump number" is just running the Grade 4 pattern rule backward.
4.OA.C.5 Write the first few jump distances and look at how they grow. Each one is double 6.EE.A.1 Rewrite each distance as a power of $2$ to get a formula for the $n$-th jump. Ju 6.EE.A.1 List the powers of $2$ in order until one passes $1{,}000$. Keep doubling and wa 6.EE.B.5 $2^9 = 512$ is still under $1{,}000$, but $2^{10} = 1024 > 1{,}000$. So the expo 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.5Generate 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.1Write 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.5Understand 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!
⭐ This AMC 8 problem only needs Grade 6 "powers of $2$" expressions you already know!
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