AMC 10 · 2002 · #14

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
vieta-formulasprime-numbersparity casework ↑ Prerequisites: prime-numbersvieta-formulas
📏 Medium solution 💡 2 insights
Problem
The quadratic x² - 63x + k = 0 has two roots, and both of them are prime numbers. Count how many different values of k can make that happen.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
4
(E)
more than 4

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Name the two roots p and q (tool #4). A monic quadratic factors as x²-63x+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.

1STEP 1

Name the roots, read off sum and product

Call the roots p and q. Expanding (x-p)(x-q)=x²-(p+q)x+pq and matching term by term gives p+q=63 and pq=k.

x²-63x+k=(x-p)(x-q)=x²-(p+q)x+pq → p+q=63, pq=k
2STEP 2

Use parity to force one root to be 2

63 is odd and odd plus odd is even, so one root must be even — and the only even prime is 2.

odd+odd=even ≠ 63 → one root is even → that root = 2
3STEP 3

Find the other root and check it is prime

The other root is 61, since 63-2=61, and no prime up to 7 divides it while 8²=64 already exceeds it.

63-2=61; 61 ÷ 2,3,5,7 all leave a remainder, 8²=64 ＞ 61 → 61 is prime
4STEP 4

Count the values of k

The only prime pair is 2 and 61, so k=2×61=122 — exactly one possible value of k, choice (B).

k=2 × 61=122 → exactly 1 value → (B)
Answer
1
Check the pair directly: 2+61=63 and 2 × 61=122, so x²-63x+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.
💡Key takeaway

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.

  • Name the roots, read off sum and product
  • Use parity to force one root to be 2
  • Find the other root and check it is prime
  • Count the values of k

More like this

Same archetype — closest grade level first.