AMC 10 · 2012 · #7
Easy mode Grade 5Small lights hang along a string, each one 6 inches from the next. The same block of five lights repeats over and over: red, red, green, green, green, then red, red, green, green, green, and so on. Count the red lights from the start of the string. How many feet apart are the 3rd red light and the 21st red light?
Note: 1 foot is equal to 12 inches.
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Lights hang in a row, each one $6$ inches from the next, repeating the block red, red, green, green, green forever. Find the distance in feet from the 3rd red light to the 21st red light.
Givens: Neighbouring lights are $6$ inches apart; The colour pattern repeats the block $RRGGG$: $2$ red then $3$ green; $1$ foot $= 12$ inches; Answer choices: (A) $18$, (B) $18.5$, (C) $20$, (D) $20.5$, (E) $22.5$
Unknowns: The distance, in feet, between the 3rd red light and the 21st red light
Understand
Restated: Lights hang in a row, each one $6$ inches from the next, repeating the block red, red, green, green, green forever. Find the distance in feet from the 3rd red light to the 21st red light.
Givens: Neighbouring lights are $6$ inches apart; The colour pattern repeats the block $RRGGG$: $2$ red then $3$ green; $1$ foot $= 12$ inches; Answer choices: (A) $18$, (B) $18.5$, (C) $20$, (D) $20.5$, (E) $22.5$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem, #8 Analyze the Units
The string repeats with period $5$, so Tool #5 (Look for a Pattern) turns "which light is the 21st red?" into a division question instead of a listing question. Tool #1 (Draw a Diagram) fixes the one thing that sinks most attempts: distance is counted in gaps between lights, not in lights. Tool #9 (Solve an Easier Related Problem) supplies the cross-check — work out the tiny case of moving forward by just two reds, then repeat it. Tool #8 (Analyze the Units) keeps the inches-to-feet conversion honest at the end.
Execute — Answer: E
4.MD.A.2 Step 1 Distance counts gaps, not lights
- Number the lights $1, 2, 3, \dots$ along the string.
- Light $1$ sits at $0$ inches, light $2$ at $6$ inches, light $3$ at $12$ inches, and so on, so light $n$ sits at $6(n-1)$ inches.
- Subtracting, the distance between the light in position $i$ and the light in position $j$ is $6|i-j|$ inches.
- The key point is that this counts the gaps between the two lights, and there is always one fewer gap than there are lights in the span.
- So the whole problem reduces to one question: what positions do the 3rd and 21st red lights occupy?
💡 Lights are fenceposts and the $6$ inches are the fence panels between them, so it is the panels you count.
4.OA.C.5 Step 2 Locate the two red lights
- The pattern repeats every $5$ lights, so cut the string into blocks of $5$.
- Block $m$ covers positions $5m-4$ through $5m$, and its two reds are the first two slots, positions $5m-4$ and $5m-3$.
- Every block gives exactly $2$ reds, so block $m$ holds red number $2m-1$ and red number $2m$.
- For the 3rd red, $3 = 2m-1$ gives $m = 2$, so it sits at position $5(2)-4 = 6$.
- For the 21st red, $21 = 2m-1$ gives $m = 11$, so it sits at position $5(11)-4 = 51$.
- Both are odd-numbered reds, so both are the *first* red of their block — the same slot in the pattern.
💡 Two reds per block of five means the red count and the block count run in lockstep, so you can jump straight to block $11$ instead of listing lights.
4.OA.C.5 Step 3 Cross-check the span of 45
- Positions $51$ and $6$ are $51 - 6 = 45$ gaps apart.
- Check that independently with the smallest version of the same move: going from any red to the red two later means moving to the same slot one block on, which is exactly $+5$ positions.
- Going from red $\#3$ to red $\#21$ is a jump of $21 - 3 = 18$ reds, and since each step covers $2$ reds that is $\frac{18}{2} = 9$ repetitions, giving $9 \times 5 = 45$ positions.
- This is why the span comes out clean: $21$ and $3$ differ by an *even* number, so the two reds land in the same slot of their blocks and the distance is a whole number of blocks.
- Had the two indices differed by an odd number, the span would have been $45 \pm 1$ instead.
💡 Every two reds you pass costs exactly one full block of five lights, so nine repeats of that small move covers the whole trip.
4.NBT.B.5 Step 4 Turn 45 gaps into inches
- Each of the $45$ gaps is $6$ inches wide, so the total distance is $45 \times 6 = 270$ inches.
- Note that this multiplies the gap count, not the light count: the span from position $6$ to position $51$ contains $46$ lights but only $45$ gaps, and using $46$ here would overshoot by half a foot.
💡 Equal gaps mean one multiplication replaces adding $6$ inches forty-five times.
5.MD.A.1 Step 5 Convert to feet and answer
- The question asks for feet, and $1$ foot is $12$ inches, so divide: $270 \div 12 = 22.5$ feet.
- Watch the block index here — if you place the 21st red in block $10$ instead of block $11$, you get position $46$, a span of $40$ gaps, and $240$ inches $= 20$ feet, which is exactly the trap choice (C).
- The correct answer is $22.5$ feet, choice (E).
💡 Feet are bigger units than inches, so the number has to shrink — dividing by $12$ is the only move that makes sense.
4.MD.A.2 Number the lights $1, 2, 3, \dots$ along the string. Light $1$ sits at $0$ inche 4.OA.C.5 The pattern repeats every $5$ lights, so cut the string into blocks of $5$. Bloc 4.OA.C.5 Positions $51$ and $6$ are $51 - 6 = 45$ gaps apart. Check that independently wi 4.NBT.B.5 Each of the $45$ gaps is $6$ inches wide, so the total distance is $45 \times 6 5.MD.A.1 The question asks for feet, and $1$ foot is $12$ inches, so divide: $270 \div 12 Review
Reasonableness: Spot-check the position rule by listing the start of the string: positions $1,2$ are red, $3,4,5$ green, $6,7$ red, $8,9,10$ green, $11,12$ red. That makes the reds fall at positions $1, 2, 6, 7, 11, 12, \dots$, so red $\#3$ really is at position $6$, matching the block formula. The final size is also sensible: $45$ gaps at half a foot each is $22.5$ feet, and since the two reds are $45$ lights apart on a string where lights sit $\frac{1}{2}$ foot apart, anything near $22$ or $23$ feet is the right ballpark. Every wrong choice is smaller than $22.5$, which is the signature of undercounting — either dropping a block (choice (C), $20$) or counting the $6$ inches as red-to-red spacing rather than light-to-light spacing.
Alternative: Use density instead of block indices. Red lights occur at a fixed rate of $2$ per $5$ consecutive positions, so advancing by $18$ reds advances by $18 \times \frac{5}{2} = 45$ positions. That single line replaces all the block bookkeeping, and it is exact precisely because $18$ is even: an even number of reds is a whole number of blocks, so no leftover partial block can shift the count. As a formula, an odd-numbered red $\#k$ sits at position $\frac{5k-3}{2}$, giving $\frac{5(21)-3}{2} - \frac{5(3)-3}{2} = 51 - 6 = 45$ gaps, hence $\frac{45 \times 6}{12} = 22.5$ feet again.
CCSS standards used (min grade 5)
4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Modelling the string as evenly spaced points so that distance is $6$ inches times the number of gaps, not the number of lights.)4.OA.C.5Generate a number or shape pattern following a given rule (Using the period-$5$ repeat to place red $\#3$ at position $6$ and red $\#21$ at position $51$, and to justify the $+5$ positions per $+2$ reds step.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Converting the $45$ gaps into a length: $45 \times 6 = 270$ inches.)5.MD.A.1Convert among different-sized standard measurement units within a given system (Changing $270$ inches into feet by dividing by $12$, as the question demands feet.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Handling the non-whole quotient $270 \div 12 = 22.5$ and comparing it against the decimal answer choices.)
⭐ Number the lights first, then count the gaps between the two you care about — with a block that repeats every $5$ lights and $2$ reds inside it, jumping ahead $18$ reds always means jumping ahead exactly $45$ lights.
⭐ Number the lights first, then count the gaps between the two you care about — with a block that repeats every $5$ lights and $2$ reds inside it, jumping ahead $18$ reds always means jumping ahead exactly $45$ lights.
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