AMC 10 · 2014 · #19

Grade 8 algebranumber-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

quadratic-equationspolynomial-rootsinterval-arithmetic convert-to-algebrabound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: quadratic-equationspolynomial-roots

📏 Medium solution 💡 3 insights

Problem

A quadratic with a bounded rational coefficient must have a whole-number root. Count the coefficients.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Change Focus / Count the Complement

Searching directly over k is hopeless: there are infinitely many rationals with |k| ＜ 200, and for each one you would have to test whether a root happens to be an integer. So change focus (Tool #16): name the integer root instead (Tool #4) and let k be whatever that root forces it to be. The unknown then runs over integers, which can be listed. Two questions remain, and each has its own tool: how far can the integer root go before |k| reaches 200 (Tool #14, the boundary case), and does every root give a different k (needed before the list of roots may be counted as a list of k values, Tool #2).

1STEP 1

Name the integer root

Naming the root gives the coefficient directly.

5n²+kn+12=0, n ≠ 0

2STEP 2

Flip the search onto n

The search moves onto the root instead.

k=-(5n²+12)/n=-5n-12/n

3STEP 3

Push n to the boundary

The bound bites at a clean boundary.

n ≥ 40: |k|=5n+12/n ＞ 200; 3 ≤ n ≤ 39: |k| ≤ 195+4=199 ＜ 200

4STEP 4

Mirror the negative roots

Negative roots mirror the positive ones.

k(-n)=5n+12/n=-k(n), |k(-n)|=|k(n)|

5STEP 5

Check for repeated k values

No two roots give the same coefficient.

5=12/mn → mn=12/5 ∉Z

6STEP 6

Tally the k values

The tally is 78, choice (E).

N=39+39=78 → (E)

Answer

Spot-check the ends of the list. For n=1, k=-17, and 5x²-17x+12=(5x-12)(x-1) really does have the integer root x=1. For n=-39, k=195+12/39=2539/13≈ 195.31, which is under 200; the next one, n=-40, gives k=200.3, which is over. So 39 is the right cutoff. The smaller answer choices are the traps: demanding that k itself be an integer forces n to divide 12, leaving only n=± 1,± 2,± 3,± 4,± 6,± 12 and the answer 12. The problem says rational, not integer, so the count is much larger, and 78 is the choice that reflects that.

💡Key takeaway

Do not hunt for k — pick the integer root first, and k=-5n-12/n has no choice but to follow.

Name the integer root
Flip the search onto n
Push n to the boundary
Mirror the negative roots
Check for repeated k values
Tally the k values