AMC 8 · 2004 · #8
Grade 6 countingProblem
Find the number of two-digit positive integers whose digits total 7.
Pick an answer.
AMC 8 2004 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: How many two-digit positive integers have digits that add up to $7$?
Givens: The integer is two digits, so it lies from $10$ to $99$; The tens digit plus the units digit must equal $7$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Unknowns: The count of two-digit integers whose digit sum is $7$
Understand
Restated: How many two-digit positive integers have digits that add up to $7$?
Givens: The integer is two digits, so it lies from $10$ to $99$; The tens digit plus the units digit must equal $7$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Plan
Primary tool: #2 Make a Systematic List
The unknown is a count, the search space is small, and the digits obey a simple rule ($t + u = 7$). Tool #2 (Make a Systematic List) lets us walk the tens digit $t$ from $1$ up to $7$, set $u = 7 - t$ each time, and read off every valid number. Walking $t$ in order guarantees we miss none and repeat none.
Execute — Answer: B
5.OA.B.3 Step 1 - Set up the digit equation.
- Let $t$ be the tens digit and $u$ be the units digit.
- A two-digit integer needs $t \geq 1$, and the digit-sum rule is $t + u = 7$.
💡 Writing the rule once at the top makes the list mechanical: pick $t$, and $u$ is forced.
6.EE.B.5 Step 2 - Find the range of $t$ that actually works.
- Since $u = 7 - t$ and we need $u \geq 0$, we get $t \leq 7$.
- Combined with $t \geq 1$, the tens digit runs from $1$ to $7$.
💡 Bounding $t$ first turns an open-ended hunt into a fixed checklist of seven cases.
5.OA.B.3 Step 3 - Walk $t$ in order and record each number.
- For each tens digit, the units digit is $7 - t$, which gives one valid two-digit integer.
💡 Going in order means no duplicates and no gaps — every digit sum is exactly $7$.
5.OA.B.3 Step 4 - Count the entries in the list.
- There are seven numbers, one per value of $t$, so the answer is $7$.
💡 Each tens digit produced exactly one number, and we had $7$ tens digits, so the count is $7$.
5.OA.B.3 Set up the digit equation. Let $t$ be the tens digit and $u$ be the units digit. 6.EE.B.5 Find the range of $t$ that actually works. Since $u = 7 - t$ and we need $u \geq 5.OA.B.3 Walk $t$ in order and record each number. For each tens digit, the units digit i 5.OA.B.3 Count the entries in the list. There are seven numbers, one per value of $t$, so Review
Reasonableness: Cross-check by units digit instead. If $u = 0$, then $t = 7$ (giving $70$); $u = 1 \to t = 6$ ($61$); ... up to $u = 7 \to t = 0$, but $t = 0$ is not allowed, so that case drops out. The valid units digits are $0$ through $6$, which is $7$ values. Same answer, reached from the other side — the count is solid.
Alternative: Tool #5 (Look for a Pattern): for a digit sum $s$ from $1$ to $9$, the count of two-digit integers is always $s$ itself (tens digit runs $1$ to $s$). For $s = 7$ that pattern gives $7$ directly, confirming (B).
CCSS standards used (min grade 6)
5.OA.B.3Generate two numerical patterns using two given rules (Walking the tens digit $t$ from $1$ to $7$ and listing the matching units digit $u = 7 - t$ to enumerate every valid two-digit number.)6.EE.B.5Understand solving an equation or inequality as a process of answering which values make it true (Using the constraints $t \geq 1$ and $u \geq 0$ to bound the tens digit to $1 \leq t \leq 7$, turning an open search into a finite checklist.)
⭐ A clean digit-sum rule plus a systematic list catches every two-digit number once — no Algebra needed, just careful counting from Grade 5 patterns and a Grade 6 bound.
⭐ A clean digit-sum rule plus a systematic list catches every two-digit number once — no Algebra needed, just careful counting from Grade 5 patterns and a Grade 6 bound.
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