AMC 10 · 2021 · #13

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

base-conversionsystems-of-equationsplace-valuepolynomial-roots convert-to-algebraidentify-subproblems ↑ Prerequisites: place-valuesystems-of-equations

📏 Medium solution 💡 3 insights

Problem

Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$

Pick an answer.

(A)

~10

(B)

~11

(C)

~13

(D)

~15

(E)

~16

View mode:

Toolkit + CCSS Solution

Understand

Restated: Find a positive integer base $n$ and a digit $d$ so that (i) the three-digit numeral $\overline{32d}$ read in base $n$ equals $263$ in base $10$, and (ii) the three-digit numeral $\overline{324}$ read in base $n$ equals the four-digit numeral $\overline{11d1}$ read in base $6$. Then report $n + d$.

Givens: $\overline{32d}_n = 3 \cdot n^2 + 2 \cdot n + d = 263$; $\overline{324}_n = 3 \cdot n^2 + 2 \cdot n + 4$; $\overline{11d1}_6 = 1 \cdot 216 + 1 \cdot 36 + d \cdot 6 + 1 = 253 + 6d$; Answer choices for $n + d$: $10, 11, 13, 15, 16$

Unknowns: Base $n$ (positive integer); Digit $d$ (single digit, $0 \le d \le \min(n, 6) - 1$)

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems, #6 Guess and Check, #3 Eliminate Possibilities

Tool #13 (Algebra): translate each numeral by its place-value formula. Tool #7 (Subproblems) sets two equations $3n^2 + 2n + d = 263$ and $3n^2 + 2n + 4 = 253 + 6d$. Tool #7 again: subtract one from the other to kill $3n^2 + 2n$, leaving a linear equation in $d$ that gives $d = 2$. Tool #6 (Guess and Check): plug small bases into $3n^2 + 2n = 261$; $n = 9$ works on the first try. Tool #3 (Eliminate) matches $n + d = 11$ to the choices.

Execute — Answer: B

#13 Convert to Algebra 5.NBT.A.1 Step 1

Write each numeral in expanded base form.
$\overline{32d}$ in base $n$ has place values $n^2, n^1, n^0$, so it equals $3n^2 + 2n + d$.
Similarly $\overline{324}_n = 3n^2 + 2n + 4$.
And $\overline{11d1}$ in base $6$ has place values $6^3, 6^2, 6^1, 6^0$, so it equals $216 + 36 + 6d + 1 = 253 + 6d$.

$$\overline{32d}_n = 3n^2 + 2n + d, \quad \overline{324}_n = 3n^2 + 2n + 4, \quad \overline{11d1}_6 = 253 + 6d$$

💡 Grade 5 place value: each digit's value is digit $\times$ (base)$^{(\text{position})}$.

#7 Identify Subproblems 6.EE.B.6 Step 2

Form two equations from the two conditions.
Condition (i): $3n^2 + 2n + d = 263$ — call this $(E_1)$.
Condition (ii): $3n^2 + 2n + 4 = 253 + 6d$ — call this $(E_2)$.

$$(E_1):\;\; 3n^2 + 2n + d = 263 \qquad (E_2):\;\; 3n^2 + 2n + 4 = 253 + 6d$$

💡 Grade 6 expressions with variables: turn each numeral condition into an equation.

#7 Identify Subproblems 6.EE.B.7 Step 3

Subtract $(E_2)$ from $(E_1)$ to eliminate the $3n^2 + 2n$ term.
$(E_1) - (E_2)$: $(d) - (4) = 263 - (253 + 6d) = 10 - 6d$.
So $d - 4 = 10 - 6d$, giving $7d = 14$ and $d = 2$.

$$d - 4 = 10 - 6d \;\Rightarrow\; 7d = 14 \;\Rightarrow\; d = 2$$

💡 Grade 6 one-variable equation: subtraction cancels the $n$ terms, leaving $d$ alone.

#6 Guess and Check 6.EE.B.5 Step 4

Substitute $d = 2$ into $(E_1)$: $3n^2 + 2n + 2 = 263$, so $3n^2 + 2n = 261$.
Guess-and-check with small bases.
$n = 8$: $3(64) + 16 = 208$, too small.
$n = 9$: $3(81) + 18 = 243 + 18 = 261$.
✓.
$n = 10$ would give $320$, too big.
So $n = 9$.

$$n = 9: \; 3 \cdot 9^2 + 2 \cdot 9 = 243 + 18 = 261 \;\checkmark$$

💡 Grade 6 testing values: small base candidates check the equation quickly.

#13 Convert to Algebra 5.NBT.A.1 Step 5

Verify both original conditions.
(i) $\overline{32d}_9 = \overline{322}_9 = 3 \cdot 81 + 2 \cdot 9 + 2 = 243 + 18 + 2 = 263 \checkmark$.
(ii) $\overline{324}_9 = 243 + 18 + 4 = 265$, and $\overline{1121}_6 = 216 + 36 + 12 + 1 = 265 \checkmark$.
Both hold.
Also check digit constraints: $d = 2 \le 5$ (valid base-$6$ digit), and $n = 9 > 4$ (digits $0$–$8$ valid).

$$\overline{322}_9 = 263,\;\; \overline{324}_9 = 265 = \overline{1121}_6$$

💡 Grade 5 place value: convert each numeral back to base $10$ and confirm equality.

#3 Eliminate Possibilities 4.NBT.B.4 Step 6

Compute $n + d = 9 + 2 = 11$.
Match to choices: (B).

$$n + d = 9 + 2 = 11 \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 4 add multi-digit numbers: $9 + 2 = 11$ matches choice (B).

[1] #13 5.NBT.A.1 Write each numeral in expanded base form. $\overline{32d}$ in base $n$ has place

[2] #7 6.EE.B.6 Form two equations from the two conditions. Condition (i): $3n^2 + 2n + d = 263$

[3] #7 6.EE.B.7 Subtract $(E_2)$ from $(E_1)$ to eliminate the $3n^2 + 2n$ term. $(E_1) - (E_2)$

[4] #6 6.EE.B.5 Substitute $d = 2$ into $(E_1)$: $3n^2 + 2n + 2 = 263$, so $3n^2 + 2n = 261$. Gu

[5] #13 5.NBT.A.1 Verify both original conditions. (i) $\overline{32d}_9 = \overline{322}_9 = 3 \c

[6] #3 4.NBT.B.4 Compute $n + d = 9 + 2 = 11$. Match to choices: (B).

Review

Reasonableness: All three numerals come out integer and consistent: $\overline{322}_9 = 263$ matches the given $263$. $\overline{324}_9 = 265 = \overline{1121}_6$ — both four- and three-digit numerals agree. Digit constraints hold: $d = 2$ is a legal digit in base $6$ and base $9$; the digits $3, 2, 4$ in $\overline{324}_9$ are all less than $9$. The base $n = 9$ is reasonable: it sits comfortably between $5$ (the smallest base where $\overline{32d}$ makes sense) and $10$ (where $\overline{32d}_n = 263$ would force $d = 263 - 320 < 0$).

Alternative: Tool #6 (Guess and Check) directly on small bases. Try $n = 5, 6, 7, 8, 9, 10$ in $(E_1)$, computing $263 - 3n^2 - 2n$ and checking whether it gives a valid digit $d$ between $0$ and $\min(n-1, 5)$. Only $n = 9$ yields a usable $d = 2$. Then verify $(E_2)$ holds: $265 = 265$. Same answer $n + d = 11$, (B). Slightly faster but less rigorous than the algebra path.

CCSS standards used (min grade 6)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Final $9 + 2 = 11$ to match the answer choices.)
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right (Expanding each numeral by place value into base-$10$ form.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Testing $n = 8, 9, 10$ in $3n^2 + 2n = 261$ to pin down $n$.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $n, d$ be unknowns and writing the two place-value equations.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving $7d = 14$ from the subtracted equation to get $d = 2$.)

⭐ This AMC 10 problem only needs Grade 6 one-variable equations you already know! Write each numeral by place value: $\overline{32d}_n = 3n^2 + 2n + d = 263$ and $\overline{324}_n = 253 + 6d$. Subtract to kill the $n$ terms — you get $7d = 14$, so $d = 2$. Then $3n^2 + 2n = 261$ checks at $n = 9$. So $n + d = 11$, answer (B).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

More like this