AMC 10 · 2022 · #6

Easy mode Grade 5

Problem

Look at this list of numbers: $121, 11211, 1112111, \ldots$

Each number in the list has a single $2$ in the middle, with the same number of $1$ s on the left and on the right. The first one has one $1$ on each side. The next has two $1$ s on each side. The next has three. And so on.

Take the first ten numbers in this list. How many of them are prime numbers?

$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Look at the sequence $121, 11211, 1112111, \ldots$, where the $n$-th term is a string of $n$ ones, then a $2$, then another $n$ ones. Among the first ten of these numbers, how many are prime?

Givens: First three terms: $a_1 = 121$, $a_2 = 11211$, $a_3 = 1112111$; $a_n$ has $n$ ones, then a $2$, then $n$ more ones; Range to check: $n = 1$ through $n = 10$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$

Unknowns: How many of $a_1, a_2, \ldots, a_{10}$ are prime

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems

Computing $a_{10}$ digit-by-digit and primality-testing a 21-digit number is hopeless. Tool #9 (Easier Problem) says: try the small cases $n=1, 2, 3$ first and look for shared structure. Tool #5 (Pattern) then notices that each small case splits the same way — a string of ones, shifted, plus a string of ones — which is a clean common factor. Tool #7 (Subproblems) breaks the prime question into two: 'find a factorization' and 'check both factors are $>1$'. If we can show every term factors into two pieces both bigger than $1$, no term is prime.

Execute — Answer: A

#9 Solve an Easier Related Problem 4.OA.B.4 Step 1

Try the smallest case.
$a_1 = 121 = 110 + 11 = 11 \cdot 10 + 11 = 11 \cdot (10 + 1) = 11 \cdot 11$.
So $a_1 = 121$ already factors as $11 \times 11$ — composite.

$$a_1 = 121 = 11 \cdot 11$$

💡 Finding a factor pair is exactly the Grade 4 way to show a number is not prime.

#9 Solve an Easier Related Problem 4.NBT.B.5 Step 2

Try $n=2$.
$a_2 = 11211 = 11100 + 111 = 111 \cdot 100 + 111 = 111 \cdot (100 + 1) = 111 \cdot 101$.
Same shape: a string of ones times one-zero-zero-…-one.

$$a_2 = 11211 = 111 \cdot 101$$

💡 Splitting the digits at the $2$ lines up with a clean multiplication by a power of $10$.

#5 Look for a Pattern 4.OA.C.5 Step 3

Try $n=3$ to confirm the pattern.
$a_3 = 1112111 = 1111000 + 1111 = 1111 \cdot 1000 + 1111 = 1111 \cdot 1001$.
The same structure appears for the third time.

$$a_3 = 1112111 = 1111 \cdot 1001$$

💡 Three matching cases is enough to trust the rule and write it in general form.

#5 Look for a Pattern 5.NBT.A.2 Step 4

Pattern in plain words: $a_n$ equals (a string of $n+1$ ones) times (a $1$, then $n-1$ zeros, then a $1$).
Call the first factor $R_{n+1}$ and the second factor $10^n + 1$.
So $a_n = R_{n+1} \cdot (10^n + 1)$.

$$a_n = R_{n+1} \cdot (10^n + 1)$$

💡 Multiplying by $10^n$ shifts a number $n$ places left, which is what the splitting did each time.

#7 Identify Subproblems 5.OA.A.2 Step 5

Now check the two parts for every $n$ from $1$ to $10$.
The first factor $R_{n+1}$ has at least two ones ($n+1 \ge 2$), so $R_{n+1} \ge 11$, which is bigger than $1$.
The second factor $10^n + 1$ is at least $10 + 1 = 11$, also bigger than $1$.

$$R_{n+1} \ge 11 > 1\;\text{and}\;10^n + 1 \ge 11 > 1$$

💡 If both factors are bigger than $1$, the product has divisors other than $1$ and itself.

#5 Look for a Pattern 4.OA.B.4 Step 6

Conclusion: every term $a_n$ for $n = 1, 2, \ldots, 10$ is the product of two whole numbers each bigger than $1$, so every one of the first ten terms is composite.
None are prime — count = $0$.
Answer (A).

$$\text{prime count} = 0 \;\Rightarrow\; \textbf{(A)}$$

💡 A factorization with two factors $>1$ rules out primality, by definition.

[1] #9 4.OA.B.4 Try the smallest case. $a_1 = 121 = 110 + 11 = 11 \cdot 10 + 11 = 11 \cdot (10 +

[2] #9 4.NBT.B.5 Try $n=2$. $a_2 = 11211 = 11100 + 111 = 111 \cdot 100 + 111 = 111 \cdot (100 + 1

[3] #5 4.OA.C.5 Try $n=3$ to confirm the pattern. $a_3 = 1112111 = 1111000 + 1111 = 1111 \cdot 1

[4] #5 5.NBT.A.2 Pattern in plain words: $a_n$ equals (a string of $n+1$ ones) times (a $1$, then

[5] #7 5.OA.A.2 Now check the two parts for every $n$ from $1$ to $10$. The first factor $R_{n+1

[6] #5 4.OA.B.4 Conclusion: every term $a_n$ for $n = 1, 2, \ldots, 10$ is the product of two wh

Review

Reasonableness: Spot-check $n=1$: $11 \cdot 11 = 121$ ✓. Spot-check $n=2$: $111 \cdot 101 = 11211$ ✓. Spot-check $n=3$: $1111 \cdot 1001 = 1112111$ ✓. The factorization is verified on three independent cases, and the inequality $R_{n+1} \ge 11, 10^n + 1 \ge 11$ holds for all $n \ge 1$, so the conclusion covers $n = 1$ through $n = 10$ without exception. The other answer choices $1, 2, 3, 4$ would each require pointing to a specific $n$ where the factorization fails — which it never does.

Alternative: Tool #6 (Guess and Check) on small primality: compute $a_1 = 121 = 11^2$ (composite), then notice $a_2 = 11211$ is divisible by $3$ (digit sum $1+1+2+1+1 = 6$), and $a_3 = 1112111$ — try $1112111 / 11 = 101101$ ✓ (so also composite). After three composite hits a sharper student suspects all of them are composite and looks for a general factorization, which is exactly the path above. Pure case-by-case checking is slower and gives no guarantee about $a_{10}$ without the pattern.

CCSS standards used (min grade 5)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Using the definition of prime/composite — a product of two integers $>1$ is composite.)
4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number (Verifying small-case products like $111 \cdot 101 = 11211$ by multi-digit multiplication.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Confirming the splitting rule on three small cases before generalizing.)
5.NBT.A.2 Explain patterns in number of zeros and placement of decimal point (Reading the shift 'string of ones times $10^n$' as place-value movement of $n$ digits.)
5.OA.A.2 Write simple expressions that record calculations with numbers (Writing the general factorization $a_n = R_{n+1} \cdot (10^n + 1)$ as a compact expression.)

⭐ This AMC 10 problem only needs Grade 5 place value and factor pairs you already know — split each number at the $2$, peel out the common string of ones, and every single one of the ten terms is a product of two whole numbers bigger than $1$, so none are prime.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 5)

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