AMC 10 · 2002 · #6
Grade 6 number-theoryFor how many positive integers n is n2−3n+2 a prime number?
Pick an answer.
AMC 10 2002 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 expression $n^2 - 3n + 2$. Feed in the positive integers $n = 1, 2, 3, \ldots$ one at a time and see how often the result is a prime number. The task is to count how many positive integers $n$ make $n^2 - 3n + 2$ prime, and match that count to one of the five descriptions.
Givens: The expression is $n^2 - 3n + 2$; $n$ ranges over the positive integers $1, 2, 3, \ldots$; A number counts only if $n^2 - 3n + 2$ is prime for that $n$; Answer choices describe a count: (A) none, (B) one, (C) two, (D) more than two but finitely many, (E) infinitely many
Unknowns: How many positive integers $n$ make $n^2 - 3n + 2$ a prime number
Understand
Restated: Look at the expression $n^2 - 3n + 2$. Feed in the positive integers $n = 1, 2, 3, \ldots$ one at a time and see how often the result is a prime number. The task is to count how many positive integers $n$ make $n^2 - 3n + 2$ prime, and match that count to one of the five descriptions.
Givens: The expression is $n^2 - 3n + 2$; $n$ ranges over the positive integers $1, 2, 3, \ldots$; A number counts only if $n^2 - 3n + 2$ is prime for that $n$; Answer choices describe a count: (A) none, (B) one, (C) two, (D) more than two but finitely many, (E) infinitely many
Plan
Primary tool: #7 Identify Subproblems
Secondary: #6 Guess and Check, #3 Eliminate Possibilities
Plugging in numbers forever cannot prove a count of 'none,' 'one,' or 'infinitely many,' so raw checking is not enough by itself. Tool #7 (Identify Subproblems) supplies the key: the quadratic factors as $(n-1)(n-2)$, which turns 'is this prime?' into a question about a product of two integers. A prime can only be a product of two whole numbers when one of them is $1$, so the search collapses to a handful of cases. Tool #6 (Guess and Check) then tests just those few $n$ values, and Tool #3 (Eliminate Possibilities) reads the final count off against the five choices.
Execute — Answer: B
6.EE.A.3 Step 1 Factor the expression
- The expression $n^2 - 3n + 2$ is a product in disguise.
- Look for two numbers that multiply to $2$ and add to $-3$; those are $-1$ and $-2$, giving $n^2 - 3n + 2 = (n-1)(n-2)$.
- Check by expanding with the distributive property: $(n-1)(n-2) = n^2 - 2n - n + 2 = n^2 - 3n + 2$, which matches.
- So instead of a mysterious quadratic, we have a clean product of two pieces, $n-1$ and $n-2$.
💡 Rewriting the quadratic as a product turns a hard 'is it prime?' question into a question about its two factors.
4.OA.B.4 Step 2 Ask what makes a product prime
- A prime number has exactly two factors: $1$ and itself.
- So if a product of two whole numbers is prime, one of those factors is forced to be $1$ and the other must be the prime.
- If instead both factors are $2$ or larger, the product has extra factors and is composite; and if either factor is $0$, the product is $0$, which is not prime.
- For the value $(n-1)(n-2)$ to be prime, then, the smaller factor $n-2$ has to equal $1$ — any larger and both factors exceed $1$, making the product composite.
💡 A prime cannot be split into two factors that are both bigger than one, so one factor is pinned to $1$.
6.EE.A.2 Step 3 Check the small cases
- Test the positive integers where a factor is small.
- At $n = 1$: $(1-1)(1-2) = 0 \cdot (-1) = 0$, not prime.
- At $n = 2$: $(2-1)(2-2) = 1 \cdot 0 = 0$, not prime.
- At $n = 3$: $(3-1)(3-2) = 2 \cdot 1 = 2$, and $2$ is prime — this one works.
- This is exactly the case $n-2 = 1$ found in the previous step.
💡 The only place a factor can equal $1$ for a positive $n$ is $n=3$, and it gives the prime $2$.
4.OA.B.4 Step 4 Rule out every larger n and count
- For any $n \ge 4$ both factors are at least $2$: $n-2 \ge 2$ and $n-1 \ge 3$.
- Their product is then a composite number with factors other than $1$ and itself, so it is never prime.
- That leaves $n = 3$ as the single positive integer that works — $n=1$ and $n=2$ give $0$, and everything from $n=4$ upward is composite.
- Exactly one value succeeds, so the count is one: the answer is $\textbf{(B)}$.
- This rules out (A) none, (C) two, (D) more than two, and (E) infinitely many.
💡 Once both factors pass $1$, the product always splits further, so no large $n$ can ever be prime.
6.EE.A.3 The expression $n^2 - 3n + 2$ is a product in disguise. Look for two numbers tha 4.OA.B.4 A prime number has exactly two factors: $1$ and itself. So if a product of two w 6.EE.A.2 Test the positive integers where a factor is small. At $n = 1$: $(1-1)(1-2) = 0 4.OA.B.4 For any $n \ge 4$ both factors are at least $2$: $n-2 \ge 2$ and $n-1 \ge 3$. Th Review
Reasonableness: Spot-check a few values directly against the original expression: $n=3$ gives $9-9+2 = 2$ (prime), $n=4$ gives $16-12+2 = 6 = 2\cdot3$ (composite), $n=5$ gives $25-15+2 = 12$ (composite), and $n=1,2$ give $0$. Every check agrees with the factored analysis, and the values only grow past $n=4$, so nothing prime hides further out. Exactly one positive integer, $n=3$, produces a prime, confirming a count of one.
Alternative: Use parity instead of size. Among any two consecutive integers $n-1$ and $n-2$, one is even, so the product $(n-1)(n-2)$ is always even. The only even prime is $2$, so a prime value forces $(n-1)(n-2) = 2$, i.e. $2 \cdot 1$, which happens only at $n = 3$. Same single answer, reached through even/odd reasoning rather than comparing the factors to $1$.
CCSS standards used (min grade 6)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring $n^2 - 3n + 2$ into $(n-1)(n-2)$ and confirming it by expanding with the distributive property.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Using the definition of prime to argue a product is prime only when one factor is $1$, and that both factors above $1$ force a composite.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating $(n-1)(n-2)$ at $n = 1, 2, 3$ to test the small cases.)
⭐ Factor it into $(n-1)(n-2)$: a product is prime only when one piece is $1$, and that happens for just one positive $n$, namely $n=3$, giving the prime $2$.
⭐ Factor it into $(n-1)(n-2)$: a product is prime only when one piece is $1$, and that happens for just one positive $n$, namely $n=3$, giving the prime $2$.
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