AMC 8 · 2006 · #11

Easy mode Grade 3

Problem

Think about all the two-digit numbers, from $10$ up to $99$ .

For each number, add its two digits. For example, the digits of $23$ add up to $2 + 3 = 5$ .

We want only the numbers whose digit sum is a perfect square. (A perfect square is a number like $1, 4, 9, 16, 25, \dots$ )

How many two-digit numbers fit this rule?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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.

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

#14 Try the Extreme Case 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 $1, 4, 9, 16$.

$$1 \le t + u \le 18 \;\Rightarrow\; t + u \in \{1, 4, 9, 16\}$$

💡 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.

#13 Count Systematically 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 $10$.

$$t + u = 1: \;\; (1, 0) \;\Rightarrow\; \{10\} \;\Rightarrow\; 1 \text{ number}$$

💡 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.

#13 Count Systematically 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 $t + u = 4$: $(1,3), (2,2), (3,1), (4,0)$.

$$t + u = 4: \;\; \{13, 22, 31, 40\} \;\Rightarrow\; 4 \text{ numbers}$$

💡 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.

#13 Count Systematically 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.

$$t + u = 9: \;\; \{18, 27, 36, 45, 54, 63, 72, 81, 90\} \;\Rightarrow\; 9 \text{ numbers}$$

💡 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 $t$.

#13 Count Systematically 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 $(7, 9), (8, 8), (9, 7)$.

$$t + u = 16: \;\; \{79, 88, 97\} \;\Rightarrow\; 3 \text{ numbers}$$

💡 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 $7$.

#7 Break Into Subproblems 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.

$$1 + 4 + 9 + 3 = 17 \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 2 addition fluency: the four subproblem counts add to $17$, which matches choice (C).

[1] #14 3.OA.C.7 Bound the digit sum. The tens digit is between $1$ and $9$, the units digit is b

[2] #13 2.NBT.A.1 Count numbers with digit sum $1$. We need $t + u = 1$ with $t \ge 1$. The only c

[3] #13 3.OA.D.9 Count numbers with digit sum $4$. List every pair $(t, u)$ with $t \in {1, \ldo

[4] #13 3.OA.D.9 Count numbers with digit sum $9$. The same walk works: $t$ runs from $1$ to $9$

[5] #13 3.OA.D.9 Count numbers with digit sum $16$. Now $u = 16 - t$ must still be at most $9$, s

[6] #7 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.1 Understand 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.2 Fluently 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.7 Fluently 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.9 Identify 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 $17$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 3)

More like this