AMC 10 · 2021 · #11

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

modular-arithmeticplace-valuedivisibility-rulespolynomial-factoring identify-subproblemssystematic-enumeration ↑ Prerequisites: modular-arithmetic

📏 Medium solution 💡 2 insights

Problem

For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2021 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.

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Toolkit + CCSS Solution

Understand

Restated: Treat $2021_b$ and $221_b$ as base-$b$ numerals. For which choice of $b$ from $\{3, 4, 6, 7, 8\}$ is the subtraction $2021_b - 221_b$ NOT a multiple of $3$?

Givens: $2021_b$ is a 4-digit base-$b$ numeral (digits $2, 0, 2, 1$); $221_b$ is a 3-digit base-$b$ numeral (digits $2, 2, 1$); Candidate bases: $b \in \{3, 4, 6, 7, 8\}$; Question asks which $b$ makes the difference NOT divisible by $3$

Unknowns: The single base $b$ from the choices for which $3 \nmid (2021_b - 221_b)$

Understand

Restated: Treat $2021_b$ and $221_b$ as base-$b$ numerals. For which choice of $b$ from $\{3, 4, 6, 7, 8\}$ is the subtraction $2021_b - 221_b$ NOT a multiple of $3$?

Plan

Primary tool: #7 Identify Subproblems

Secondary: #9 Easier Related Problem, #3 Eliminate Possibilities, #6 Guess and Check

Tool #7 (Subproblems) breaks the question into two clean pieces: (a) write each base-$b$ numeral as a polynomial in $b$, (b) subtract and factor. Tool #9 (Easier Related Problem) lets us check one base by hand first ($b=3$) to confirm the formula. Then Tool #3 (Eliminate) plugs every choice into the simplified expression and Tool #6 (Guess and Check) just reads off divisibility by $3$. No algebra heavier than expanding place values is needed.

Execute — Answer: E

#7 Identify Subproblems 5.NBT.A.1 Step 1

Expand each base-$b$ numeral using place values.
The digits of $2021_b$ from right to left are $1, 2, 0, 2$, so $2021_b = 2 \cdot b^3 + 0 \cdot b^2 + 2 \cdot b + 1$.
The digits of $221_b$ from right to left are $1, 2, 2$, so $221_b = 2 \cdot b^2 + 2 \cdot b + 1$.

$$2021_b = 2b^3 + 2b + 1,\quad 221_b = 2b^2 + 2b + 1$$

💡 Grade 5 place value: each digit's value is digit times power of the base.

#7 Identify Subproblems 6.EE.A.3 Step 2

Subtract.
The $+2b$ and $+1$ terms cancel cleanly, leaving only the cubic and quadratic terms.

$$2021_b - 221_b = (2b^3 + 2b + 1) - (2b^2 + 2b + 1) = 2b^3 - 2b^2$$

💡 Grade 6: combine like terms when subtracting polynomial-shaped expressions.

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

Factor the difference to see its divisibility structure.
Pull out the common factor $2b^2$.

$$2b^3 - 2b^2 = 2b^2(b - 1)$$

💡 Grade 6 equivalent expressions: factoring reveals what divides the value.

#7 Identify Subproblems 6.NS.B.4 Step 4

Ask: when is $2b^2(b-1)$ divisible by $3$?
The factor $2$ is not.
So we need $3$ to divide $b^2(b-1)$, which happens exactly when $3 \mid b$ (then $3 \mid b^2$) or $3 \mid (b-1)$ (i.e., $b \equiv 1 \pmod 3$).
In short: divisibility by $3$ holds unless $b \equiv 2 \pmod 3$.

$$3 \mid 2b^2(b-1) \iff b \equiv 0 \text{ or } 1 \pmod{3}$$

💡 Grade 6 GCF/divisibility: a product is a multiple of $3$ iff at least one factor is.

#3 Eliminate Possibilities 4.OA.B.4 Step 5

Sort each answer choice into its residue class mod $3$.
$3 \equiv 0$, $4 \equiv 1$, $6 \equiv 0$, $7 \equiv 1$, $8 \equiv 2$.
Only $b = 8$ lands in the bad class ($\equiv 2$), so only $b = 8$ makes the difference NOT divisible by $3$.

$$3, 6 \equiv 0;\;\; 4, 7 \equiv 1;\;\; 8 \equiv 2 \pmod 3$$

💡 Grade 4 multiples and divisibility: sorting integers by remainder mod $3$.

#6 Guess and Check 4.OA.B.4 Step 6

Verify directly for $b = 8$ (Tool #6, Guess and Check).
$2 \cdot 8^2 \cdot (8-1) = 2 \cdot 64 \cdot 7 = 896$.
And $896 / 3 = 298.67\ldots$ — not an integer.
Confirmed: choice (E) is the answer.

$$b = 8:\; 2 \cdot 8^2 \cdot 7 = 896,\; 896 \bmod 3 = 2 \Rightarrow \textbf{(E)}$$

💡 Grade 4: one quick numerical check confirms the abstract result.

[1] #7 5.NBT.A.1 Expand each base-$b$ numeral using place values. The digits of $2021_b$ from rig

[2] #7 6.EE.A.3 Subtract. The $+2b$ and $+1$ terms cancel cleanly, leaving only the cubic and qu

[3] #7 6.EE.A.3 Factor the difference to see its divisibility structure. Pull out the common fac

[4] #7 6.NS.B.4 Ask: when is $2b^2(b-1)$ divisible by $3$? The factor $2$ is not. So we need $3$

[5] #3 4.OA.B.4 Sort each answer choice into its residue class mod $3$. $3 \equiv 0$, $4 \equiv

[6] #6 4.OA.B.4 Verify directly for $b = 8$ (Tool #6, Guess and Check). $2 \cdot 8^2 \cdot (8-1)

Review

Reasonableness: Sanity-check a 'good' base too. Try $b = 4$: $2 \cdot 16 \cdot 3 = 96 = 3 \cdot 32$, divisible by $3$. Try $b = 6$: $2 \cdot 36 \cdot 5 = 360 = 3 \cdot 120$, divisible by $3$. The only failure among the five choices really is $b = 8$, so (E) holds up. The structure also matches intuition: a factor of $(b-1)$ means whenever $b$ leaves remainder $1$ when divided by $3$, the difference is automatically a multiple of $3$; and a factor of $b^2$ catches the $b \equiv 0$ case.

Alternative: Tool #6 (Guess and Check) brute force: compute the actual difference for each base and divide by $3$. For $b = 3$: $2021_3 = 54+0+6+1 = 61$, $221_3 = 18+6+1 = 25$, difference $36 = 3 \cdot 12$. For $b = 8$: $2021_8 = 1024+0+16+1 = 1041$, $221_8 = 128+16+1 = 145$, difference $896$, and $896 = 3 \cdot 298 + 2$, not divisible. Same answer (E).

CCSS standards used (min grade 6)

4.OA.B.4 Find factor pairs; determine whether a whole number is a multiple of a given one-digit number (Sorting each candidate base $b$ by its remainder mod $3$ to decide divisibility.)
5.NBT.A.1 Recognize that in a multi-digit number, a digit in one place represents 10 times what it represents in the place to its right (Generalizing place-value to any base $b$ to write $2021_b$ and $221_b$ as polynomials in $b$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Subtracting the two polynomial expressions and factoring $2b^3 - 2b^2 = 2b^2(b-1)$.)
6.NS.B.4 Find the greatest common factor of two whole numbers; use the distributive property (Reading divisibility of $2b^2(b-1)$ by $3$ from its factored form.)

⭐ This AMC 10 problem only needs Grade 6 place value and factoring you already know! Once you write $2021_b - 221_b = 2b^2(b-1)$, the question becomes 'which $b$ gives a non-multiple of $3$?' — that's just sorting $3, 4, 6, 7, 8$ by remainder mod $3$. Only $8$ leaves remainder $2$, so the answer is (E).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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