AMC 10 · 2024 · #17

Grade 11 probabilityalgebra
vieta-formulaspolynomial-rootsprobability-basicsystematic-enumeration work-backwardssystematic-enumerationbound-inequality-then-enumerate ↑ Prerequisites: polynomial-factoringprobability-basicfactors
📏 Long solution 💡 4 insights
Problem
There are 21 integers whose absolute value is at most 10, namely -10 through 10. Pick a and then b from that set without replacement, so a and b are different. Find the probability that the cubic x³+ax²+bx+6 has three distinct integer roots.

Pick an answer.

(A)
$\frac{1}{240}$
(B)
$\frac{1}{221}$
(C)
$\frac{1}{105}$
(D)
$\frac{1}{84}$
(E)
$\frac{1}{63}$
How to solve
Strategy Work Backwards

Testing all 420 pairs (a,b) one at a time is hopeless by hand, so Tool #11 (Work Backwards) turns the problem around: instead of choosing coefficients and hunting for roots, choose the roots first and let them tell you the coefficients. Tool #4 (Introduce a Variable) makes that turn possible — name the roots p,q,r, write the cubic as (x-p)(x-q)(x-r), and expand. The constant term then forces pqr=-6, which is a tiny condition, so Tool #2 (Make a Systematic List) can write down every triple of distinct integers with product -6. Finally Tool #3 (Eliminate Possibilities) throws out the triples whose resulting a or b escapes the pool [-10,10].

1STEP 1

Count every equally likely pair

Ordered draws give 420 pairs.

21 × 20 = 420
2STEP 2

Name the roots and factor

Write it as a product of three factors.

x³+ax²+bx+6=(x-p)(x-q)(x-r)
3STEP 3

Expand and reverse the arrow

The roots must multiply to negative six.

(x-p)(x-q)(x-r)=x³-(p+q+r)x²+(pq+qr+rp)x-pqr → a=-(p+q+r), b=pq+qr+rp, pqr=-6
4STEP 4

List every triple with product -6

There are five distinct triples.

{1,-1,6}, {-1,2,3}, {1,-2,3}, {1,2,-3}, {-1,-2,-3}
5STEP 5

Read off each pair

Each triple fixes one a and one b.

(-6,-1), (-4,1), (-2,-5), (0,-7), (6,11)
6STEP 6

Drop the pairs the pool forbids

Removing the out-of-range one leaves four.

(6,11): |11| > 10 → discard; 4 pairs remain
7STEP 7

Divide favorable by total

Four over 420 is 1 over 105.

4/420=1/105 → (C)
Answer
1/105
The probability 1/105 is under one percent, which fits: a random cubic almost never factors over the integers, so only a handful of the 420 draws can work. Checking all 420 ordered pairs directly confirms exactly four winners, (a,b)=(-6,-1),(-4,1),(-2,-5),(0,-7), with roots {-1,1,6},{-1,2,3},{-2,1,3},{-3,1,2}. The wrong answers are the shapes of the two easy mistakes. Forgetting that b=11 is outside the pool keeps five triples and gives 5/420=1/84, which is choice (D). Treating the draw as an unordered pair uses C(21, 2)=210 instead of 420 and also miscounts the favorable side. Both traps are avoided by checking the pool membership of a and b separately and by remembering that a and b sit in different slots.
💡Key takeaway

When the coefficients are hard to search, build the polynomial from its roots instead, and let the constant term tell you which roots are even possible.

  • Count every equally likely pair
  • Name the roots and factor
  • Expand and reverse the arrow
  • List every triple with product -6
  • Read off each pair (a,b)
  • Drop the pairs the pool forbids
  • Divide favorable by total