AMC 10 · 2002 · #14
Grade 6 number-theoryBoth roots of the quadratic equation x2−63x+k=0 are prime numbers. The number of possible values of k is
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: The quadratic $x^2 - 63x + k = 0$ has two roots, and both of them are prime numbers. Count how many different values of $k$ can make that happen.
Givens: The equation $x^2 - 63x + k = 0$; Both of its roots are prime numbers; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $4$, (E) more than $4$
Unknowns: How many different values of $k$ are possible
Understand
Restated: The quadratic $x^2 - 63x + k = 0$ has two roots, and both of them are prime numbers. Count how many different values of $k$ can make that happen.
Givens: The equation $x^2 - 63x + k = 0$; Both of its roots are prime numbers; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $4$, (E) more than $4$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #3 Eliminate Possibilities, #6 Guess and Check
Name the two roots $p$ and $q$ (tool #4). A monic quadratic factors as $x^2-63x+k=(x-p)(x-q)$, and expanding that shows $p+q=63$ and $pq=k$. So the whole problem becomes: which prime pairs add up to $63$? Tool #3 (Eliminate Possibilities) does the heavy lifting through parity — $63$ is odd, and two odd numbers always add to an even number, so one root is forced to be the only even prime, $2$. That pins the other root to $63-2=61$; tool #6 (Guess and Check) just confirms $61$ is prime. Each surviving pair gives one $k$, so counting pairs counts values of $k$.
Execute — Answer: B
6.EE.A.3 Step 1 Name the roots, read off sum and product
- Call the two roots $p$ and $q$.
- A quadratic whose leading coefficient is $1$ can be written from its roots as $(x-p)(x-q)$.
- Expanding, $(x-p)(x-q)=x^2-(p+q)x+pq$.
- Matching this with $x^2-63x+k$ term by term gives $p+q=63$ and $pq=k$.
- So the two prime roots must add to $63$, and their product is the value of $k$.
💡 Building the quadratic back from its roots lets you read the sum and product straight off the coefficients.
2.OA.C.3 Step 2 Use parity to force one root to be 2
- The two prime roots must add up to $63$, which is odd.
- Every prime except $2$ is odd, and an odd number plus an odd number is always even.
- So two odd primes would add to an even total, never $63$.
- The only way to reach an odd sum is for one of the roots to be even — and the only even prime is $2$.
- So one root must be $2$.
💡 An odd total can't come from two odd numbers, so the even prime $2$ has to be in the mix.
4.OA.B.4 Step 3 Find the other root and check it is prime
- If one root is $2$, the other must be $63-2=61$.
- Now check whether $61$ is prime: it is not divisible by $2$, $3$, $5$, or $7$, and $7^2=49$ while $8^2=64>61$, so testing primes up to $7$ is enough.
- Since none divide it, $61$ is prime.
- So the roots $2$ and $61$ are both prime and are the only pair that works.
💡 You only need to test prime divisors up to the square root before declaring a number prime.
4.OA.B.4 Step 4 Count the values of k
- The only prime pair is $2$ and $61$, which gives $k=pq=2\times 61=122$.
- There is exactly one such value of $k$.
- So the number of possible values of $k$ is $1$, which is choice (B).
💡 One valid pair of roots means one product, so one value of $k$.
6.EE.A.3 Call the two roots $p$ and $q$. A quadratic whose leading coefficient is $1$ can 2.OA.C.3 The two prime roots must add up to $63$, which is odd. Every prime except $2$ is 4.OA.B.4 If one root is $2$, the other must be $63-2=61$. Now check whether $61$ is prime 4.OA.B.4 The only prime pair is $2$ and $61$, which gives $k=pq=2\times 61=122$. There is Review
Reasonableness: Check the pair directly: $2+61=63$ and $2\times 61=122$, so $x^2-63x+122=(x-2)(x-61)$ really does have the prime roots $2$ and $61$. The parity argument rules out every other pair, because any other prime pair would be two odd primes summing to an even number, never $63$. So exactly one $k=122$ works, matching choice (B). Choice (A) $0$ is the trap for anyone who forgets $2$ is prime and concludes no odd-sum pair exists; choice (E) is the trap for treating $(2,61)$ and $(61,2)$ as different, but swapping the roots gives the same $k$.
Alternative: Skip Vieta's and think of $63$ as a target to hit with a Goldbach-style search: list primes below $63$ and ask which have a prime partner summing to $63$. Because $63$ is odd, only $2$ can pair with an odd prime; $63-2=61$ is prime, so $\{2,61\}$ is the single pair, and every other candidate (like $63-4$, $63-6$) starts from a non-prime. Same conclusion: one value, $k=122$, choice (B).
CCSS standards used (min grade 6)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(x-p)(x-q)=x^2-(p+q)x+pq$ to read off $p+q=63$ and $pq=k$.)2.OA.C.3Determine whether a group of objects has an odd or even number (The parity argument that odd + odd is even, forcing one root to be the even prime $2$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Checking that $61$ is prime and that $\{2,61\}$ is the only prime pair, so there is exactly one $k$.)
⭐ Two primes can only add to an odd number if one of them is 2, so an odd target like 63 leaves a single prime pair — and a single value of k.
⭐ Two primes can only add to an odd number if one of them is 2, so an odd target like 63 leaves a single prime pair — and a single value of k.
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