AMC 8 · 2006 · #11
Grade 3 number-theoryProblem
How many two-digit numbers have digits whose sum is a perfect square?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Among all two-digit numbers (the integers from $10$ to $99$), count how many have a digit sum that is a perfect square.
Givens: Each number has a tens digit $t \in \{1, 2, \ldots, 9\}$ and a units digit $u \in \{0, 1, \ldots, 9\}$; The "digit sum" of the number is $t + u$; A perfect square is a number of the form $k^2$ for some non-negative integer $k$: $0, 1, 4, 9, 16, 25, \ldots$; Answer choices: (A) $13$, (B) $16$, (C) $17$, (D) $18$, (E) $19$
Unknowns: The total count of two-digit numbers whose digit sum $t + u$ is a perfect square
Understand
Restated: Among all two-digit numbers (the integers from $10$ to $99$), count how many have a digit sum that is a perfect square.
Givens: Each number has a tens digit $t \in \{1, 2, \ldots, 9\}$ and a units digit $u \in \{0, 1, \ldots, 9\}$; The "digit sum" of the number is $t + u$; A perfect square is a number of the form $k^2$ for some non-negative integer $k$: $0, 1, 4, 9, 16, 25, \ldots$; Answer choices: (A) $13$, (B) $16$, (C) $17$, (D) $18$, (E) $19$
Plan
Primary tool: #13 Count Systematically
Secondary: #7 Break Into Subproblems, #14 Try the Extreme Case
Tool #14 (Try the Extreme Case) shows that digit sums only run from $1$ up to $18$, so the perfect squares we need to hit are just $1, 4, 9, 16$ — that's the whole list. Tool #7 (Break Into Subproblems) splits the question into four small counting tasks, one per square. Tool #13 (Count Systematically) then handles each subproblem the same way: list every pair $(t, u)$ with $t + u$ equal to the target, where $t \in \{1, \ldots, 9\}$ and $u \in \{0, \ldots, 9\}$. Adding the four counts gives the answer.
Execute — Answer: C
3.OA.C.7 Step 1 - Bound the digit sum.
- The tens digit is between $1$ and $9$, the units digit is between $0$ and $9$, so the smallest digit sum is $1$ and the largest is $9 + 9 = 18$.
- The perfect squares inside the range $[1, 18]$ are $1, 4, 9, 16$.
💡 Listing the perfect squares $1, 4, 9, 16, 25, \ldots$ is a Grade 3 multiplication-facts move ($1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4$). Anything from $25$ on is too big to be a digit sum.
2.NBT.A.1 Step 2 - Count numbers with digit sum $1$.
- We need $t + u = 1$ with $t \ge 1$.
- The only choice is $t = 1, u = 0$, giving the number $10$.
💡 Grade 2 place value: a two-digit number is $10t + u$, so picking the digits picks the number. With $t \ge 1$ forced and $u = 1 - t$, only $t = 1$ works.
3.OA.D.9 Step 3 - Count numbers with digit sum $4$.
- List every pair $(t, u)$ with $t \in \{1, \ldots, 9\}$, $u \in \{0, \ldots, 9\}$, and $t + u = 4$: $(1,3), (2,2), (3,1), (4,0)$.
💡 Grade 3 patterns: as $t$ goes up by $1$, $u$ goes down by $1$. Walk $t$ from $1$ to $4$; once $t = 5$ we would need $u = -1$, which isn't a digit.
3.OA.D.9 Step 4 - Count numbers with digit sum $9$.
- The same walk works: $t$ runs from $1$ to $9$ and $u = 9 - t$ is always a valid digit between $0$ and $8$.
- Every value of $t$ gives a number.
💡 The pattern is the same — but this time none of the $9$ choices for $t$ get cut off, because $9 - t$ stays in $\{0, \ldots, 8\}$ for every $t$.
3.OA.D.9 Step 5 - Count numbers with digit sum $16$.
- Now $u = 16 - t$ must still be at most $9$, so $t \ge 7$.
- The valid pairs are $(7, 9), (8, 8), (9, 7)$.
💡 Same walk, but the units digit caps the count this time. For $t = 6$ we would need $u = 10$, which isn't a single digit, so $t$ starts at $7$.
2.OA.B.2 Step 6 - Add the four counts.
- There are $1 + 4 + 9 + 3 = 17$ two-digit numbers whose digit sum is a perfect square.
💡 Grade 2 addition fluency: the four subproblem counts add to $17$, which matches choice (C).
3.OA.C.7 Bound the digit sum. The tens digit is between $1$ and $9$, the units digit is b 2.NBT.A.1 Count numbers with digit sum $1$. We need $t + u = 1$ with $t \ge 1$. The only c 3.OA.D.9 Count numbers with digit sum $4$. List every pair $(t, u)$ with $t \in {1, \ldo 3.OA.D.9 Count numbers with digit sum $9$. The same walk works: $t$ runs from $1$ to $9$ 3.OA.D.9 Count numbers with digit sum $16$. Now $u = 16 - t$ must still be at most $9$, s 2.OA.B.2 Add the four counts. There are $1 + 4 + 9 + 3 = 17$ two-digit numbers whose digi Review
Reasonableness: Each case can be re-checked by reading the list out loud — $10$; $13, 22, 31, 40$; $18, 27, 36, 45, 54, 63, 72, 81, 90$; $79, 88, 97$ — and these are clearly different numbers, so no double-counting. The total $1 + 4 + 9 + 3 = 17$ matches answer (C). It also makes sense that the middle square $9$ produces the most numbers: $9$ is right in the middle of the digit-sum range $[1, 18]$, so neither the lower bound ($t \ge 1$) nor the upper bound ($u \le 9$) trims any cases away.
Alternative: Tool #2 (Make a List): write the two-digit numbers $10, 11, 12, \ldots, 99$ in a $9 \times 10$ grid, compute each digit sum, and circle the ones that equal $1, 4, 9,$ or $16$. Counting the circled cells gives $17$. The grid approach is slower but a good check, because it makes any miscount visible by inspection.
CCSS standards used (min grade 3)
2.NBT.A.1Understand place value: a two-digit number represents amounts of tens and ones (Treating each two-digit number as a tens digit $t$ paired with a units digit $u$, so counting numbers reduces to counting digit pairs.)2.OA.B.2Fluently add and subtract within $20$ (Adding the four case counts $1 + 4 + 9 + 3$ to get the final total $17$, and computing each digit sum $t + u$ along the way.)3.OA.C.7Fluently multiply and divide within $100$; know products from memory (Recognizing the small perfect squares $1, 4, 9, 16, 25$ as $1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4, 5 \times 5$, and seeing that $25$ is already too big to be a digit sum.)3.OA.D.9Identify arithmetic patterns and explain them using properties of operations (Sweeping the tens digit $t$ through its allowed range and reading off $u = (\text{target}) - t$ — the same pattern repeated for each target sum.)
⭐ Digit sums for two-digit numbers only run from $1$ to $18$, so the perfect squares to chase are $1, 4, 9, 16$. Count each case, add them up, and you get $17$.
⭐ Digit sums for two-digit numbers only run from $1$ to $18$, so the perfect squares to chase are $1, 4, 9, 16$. Count each case, add them up, and you get $17$.