AMC 10 · 2018 · #22

Grade 11 algebra
stars-and-barsfunction-evaluationsymmetry-argumentdigit-constraintscomplementary-counting complementary-countingconvert-to-algebrasymmetry-argument ↑ Prerequisites: stars-and-barsfunction-evaluation
📏 Long solution 💡 4 insights
Problem
A polynomial of degree at most 3 has four coefficients, and each one is a single digit from 0 to 9. Count how many such polynomials give minus nine when minus one is substituted.

Pick an answer.

(A)
110
(B)
143
(C)
165
(D)
220
(E)
286
How to solve
Strategy Change Focus / Count the Complement

Evaluating at x = -1 turns the condition into one linear equation in the four digits, but two of them arrive with minus signs and that blocks every standard counting formula. The fix is to look at those two digits from the other end: replace a by how far it falls short of 9. Since the digits 0 through 9 are symmetric about that flip, the swap loses nothing, and it converts the mixed-sign equation into a plain "nonnegative numbers adding to a fixed total" problem, which stars and bars counts directly.

1STEP 1

Name the four coefficients

Name the four coefficients.

P(x) = ax³ + bx² + cx + d, a, b, c, d ∈ {0, 1, …, 9}
2STEP 2

Evaluate the polynomial at -1

Substituting makes the signs alternate.

P(-1) = -a + b - c + d = -9
3STEP 3

See what is blocking the count

Mixed signs block a direct count.

#{(a, b, c, d) ∈ {0, …, 9}⁴ : -a + b - c + d = -9}
4STEP 4

Flip the two negative digits

Flip the negative digits by subtracting from nine.

(a' - 9) + b + (c' - 9) + d = -9 ⟹ a' + b + c' + d = 9
5STEP 5

Check the swap loses nothing

The swap loses nothing.

a ↦ 9 - a maps {0, …, 9} onto itself and is its own inverse
6STEP 6

Confirm the digit ceiling never binds

The digit ceiling never binds.

a' + b + c' + d = 9 with all terms ≥ 0 ⟹ each term ≤ 9
7STEP 7

Share 9 units among 4 slots

Sharing nine among four slots gives 220.

C(9 + 4 - 1, 4 - 1) = C(12, 3) = (12 · 11 · 10)/(3 · 2 · 1) = 220
Answer
220
There are 10⁴ = 10000 digit quadruples altogether, so 220 of them is about 2% — a sensible share, because -a + b - c + d spreads across the 37 integers from -18 to 18, peaking at 670 solutions for the value 0, and -9 sits well out toward the tail. The answer choices confirm that the arithmetic is the whole point: C(11, 3) = 165 is choice (C) and C(13, 3) = 286 is choice (E), the values you get from mis-setting the total to 8 or 10. The ceiling check in step 6 also earns its place: had the total come out as 10, the raw stars-and-bars value 286 would overcount by 4, since four solutions push a variable up to 10. At total 9 no such solution exists, so C(12, 3) = 220 is exact.
💡Key takeaway

When a sum equation has minus signs in it, measure those variables from the top instead of the bottom — use 9 - a in place of a — and the whole thing becomes a simple problem of sharing out a fixed total.

  • Name the four coefficients
  • Evaluate the polynomial at -1
  • See what is blocking the count
  • Flip the two negative digits
  • Check the swap loses nothing
  • Confirm the digit ceiling never binds
  • Share 9 units among 4 slots