AMC 10 · 2015 · #7
Easy mode Grade 4Start at 13 and keep adding 3: you get 13, 16, 19, and so on, all the way up to 70 and then 73. How many numbers are in this list?
Pick an answer.
AMC 10 2015 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: Count how many terms are in the list $13, 16, 19, \dotsc, 70, 73$, where each term is $3$ more than the one before it.
Givens: The sequence starts at $13$ and ends at $73$; Each term is $3$ larger than the previous term; Answer choices: (A) $20$, (B) $21$, (C) $24$, (D) $60$, (E) $61$
Unknowns: How many terms the sequence has
Understand
Restated: Count how many terms are in the list $13, 16, 19, \dotsc, 70, 73$, where each term is $3$ more than the one before it.
Givens: The sequence starts at $13$ and ends at $73$; Each term is $3$ larger than the previous term; Answer choices: (A) $20$, (B) $21$, (C) $24$, (D) $60$, (E) $61$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern
Counting an unfamiliar list like $13, 16, 19, \dotsc$ is hard to eyeball, so Tool #9 (Solve an Easier Related Problem) reshapes it into a list anyone can count at a glance. First Tool #5 (Look for a Pattern) reveals the constant step of $3$; then shifting every term down and dividing by that step turns the sequence into $0, 1, 2, \dotsc, 20$, where the count is obvious. Reshaping never changes how many terms there are.
Execute — Answer: B
4.OA.C.5 Step 1 Find the constant step
- Look at how the terms grow: $16 - 13 = 3$ and $19 - 16 = 3$.
- Every term is exactly $3$ more than the one before, so this is an arithmetic sequence with a step of $3$.
💡 A sequence is easiest to handle once you know the single fixed amount it jumps by each time.
4.NBT.B.4 Step 2 Shift the list to start at 0
- Subtract $13$ from every term.
- The endpoints become $13 - 13 = 0$ and $73 - 13 = 60$, so the list turns into $0, 3, 6, \dotsc, 60$.
- Sliding every term down by the same amount does not add or remove any terms.
💡 Shifting every number by the same amount keeps the count the same, just like moving a row of chairs sideways.
4.NBT.B.6 Step 3 Divide every term by 3
- Now divide each term of $0, 3, 6, \dotsc, 60$ by the step $3$.
- The endpoints become $0 \div 3 = 0$ and $60 \div 3 = 20$, so the list becomes $0, 1, 2, \dotsc, 20$.
- Scaling every term by the same factor still keeps the same number of terms.
💡 Squeezing the gaps to size $1$ makes the terms line up with the counting numbers you already know.
4.NBT.B.4 Step 4 Count the whole numbers from 0 to 20
- The reshaped list is every whole number from $0$ up to $20$.
- Counting from $0$ to $20$ gives $20 - 0 = 20$ steps, but the starting number $0$ is also a term, so there is one more term than there are steps: $20 + 1 = 21$.
- The answer is (B).
💡 Counting fence posts, not gaps: a row with $20$ gaps between posts has $21$ posts because both ends count.
4.OA.C.5 Look at how the terms grow: $16 - 13 = 3$ and $19 - 16 = 3$. Every term is exact 4.NBT.B.4 Subtract $13$ from every term. The endpoints become $13 - 13 = 0$ and $73 - 13 = 4.NBT.B.6 Now divide each term of $0, 3, 6, \dotsc, 60$ by the step $3$. The endpoints bec 4.NBT.B.4 The reshaped list is every whole number from $0$ up to $20$. Counting from $0$ t Review
Reasonableness: Check the off-by-one is right by testing a tiny version: $13, 16, 19$ has step $3$, gap $19 - 13 = 6$, so $6 \div 3 = 2$ steps and $2 + 1 = 3$ terms — which matches the three terms we can see. The same rule on the full list gives $21$, so (D) $60$ and (E) $61$ (the raw gap and gap-plus-one) are traps, and (A) $20$ forgets to count the first term.
Alternative: Use the arithmetic-sequence term formula directly: the $n$-th term is $13 + (n-1)\times 3$. Set it equal to the last term $73$: $13 + (n-1)\times 3 = 73$ gives $(n-1)\times 3 = 60$, so $n - 1 = 20$ and $n = 21$ — the same answer (B).
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing the constant step of $3$ that defines the arithmetic sequence.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Shifting the endpoints with $73 - 13 = 60$ and finishing the count with $20 + 1 = 21$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing the shifted endpoint $60 \div 3 = 20$ to rescale the sequence to $0, 1, \dotsc, 20$.)
⭐ To count an evenly-spaced list, divide the gap from first to last by the step, then add one for the term you started on.
⭐ To count an evenly-spaced list, divide the gap from first to last by the step, then add one for the term you started on.
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