AMC 10 · 2022 · #6
Grade 5 number-theoryProblem
How many of the first ten numbers of the sequence 121,11211,1112111,… are prime numbers?
Pick an answer.
AMC 10 2022 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.
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
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) $4$
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
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.
💡 Finding a factor pair is exactly the Grade 4 way to show a number is not prime.
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.
💡 Splitting the digits at the $2$ lines up with a clean multiplication by a power of $10$.
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.
💡 Three matching cases is enough to trust the rule and write it in general form.
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 $a_n = R_{n+1} \cdot (10^n + 1)$.
💡 Multiplying by $10^n$ shifts a number $n$ places left, which is what the splitting did each time.
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 $1$.
💡 If both factors are bigger than $1$, the product has divisors other than $1$ and itself.
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).
💡 A factorization with two factors $>1$ rules out primality, by definition.
4.OA.B.4 Try the smallest case. $a_1 = 121 = 110 + 11 = 11 \cdot 10 + 11 = 11 \cdot (10 + 4.NBT.B.5 Try $n=2$. $a_2 = 11211 = 11100 + 111 = 111 \cdot 100 + 111 = 111 \cdot (100 + 1 4.OA.C.5 Try $n=3$ to confirm the pattern. $a_3 = 1112111 = 1111000 + 1111 = 1111 \cdot 1 5.NBT.A.2 Pattern in plain words: $a_n$ equals (a string of $n+1$ ones) times (a $1$, then 5.OA.A.2 Now check the two parts for every $n$ from $1$ to $10$. The first factor $R_{n+1 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.4Find 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.5Multiply 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.5Generate a number or shape pattern following a given rule (Confirming the splitting rule on three small cases before generalizing.)5.NBT.A.2Explain 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.2Write 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.
⭐ 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.
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