AMC 10 · 2025 · #3

Grade 11 algebra
complex-numbersdifference-of-squarespolynomial-factoring identify-subproblems ↑ Prerequisites: complex-numbers
📏 Short solution 💡 2 insights
Problem
Multiply the four complex factors i, i-1, i-2, and i-3 together, where i is the imaginary unit, and report the single number the product equals.

Pick an answer.

(A)
6-5i
(B)
-10i
(C)
10i
(D)
-10
(E)
10
How to solve
Strategy Organize Information in More Ways

Multiplying the four factors left to right works, but it drags a messy complex number through three separate multiplications and gives three chances to lose a sign. So I rearrange the factors before touching them. The four factors are i shifted down by 0, 1, 2, and 3, and if I pair the outer two together and the inner two together, each pair's shifts add to the same total 3 — which forces both pairs into the same shape. That regrouping turns one long product into two short independent ones, and their matching shape then collapses in a single move. Tracking whether any i can survive to the end also crosses off most of the choices before the arithmetic is even finished.

1STEP 1

Pair the outer and inner factors

Pair the outer two and the inner two.

i(i-1)(i-2)(i-3) = [i(i-3)]·[(i-1)(i-2)]
2STEP 2

Multiply each pair on its own

They become negative one minus three i and one minus three i.

i(i-3) = i² - 3i = -1 - 3i, (i-1)(i-2) = i² - 3i + 2 = 1 - 3i
3STEP 3

Reveal a difference of squares

Pulling out a sign reveals a difference of squares.

(-1-3i)(1-3i) = -(1+3i)(1-3i) = -(1² - (3i)²) = -(1 - 9i²)
4STEP 4

Check that no stray i survives

Every imaginary part cancels, leaving a real number.

i² = -1 → -(1 - 9i²) ∈ R
5STEP 5

Substitute and finish

Substituting gives -10.

-(1 - 9(-1)) = -(1+9) = -10
Answer
-10
Redo the product in the plain left-to-right order as an independent check: i(i-1) = i² - i = -1 - i; then (-1-i)(i-2) = -i + 2 - i² + 2i = -i + 2 + 1 + 2i = 3 + i; then (3+i)(i-3) = 3i - 9 + i² - 3i = -9 - 1 = -10. Both routes land on -10. The size can be checked separately without any expanding, because absolute values multiply: |i| = 1, |i-1| = √(2), |i-2| = √(5), and |i-3| = √(10), so the product has absolute value 1·√(2)·√(5)·√(10) = √(100) = 10. A value of -10 has absolute value 10, exactly as required, and the structural argument in step 4 already ruled out anything with an i in it. This matches choice (D).
💡Key takeaway

Before multiplying a long chain of factors, hunt for a pairing that makes the pieces match — matching pieces cancel each other's messy parts and leave a clean number behind.

  • Pair the outer and inner factors
  • Multiply each pair on its own
  • Reveal a difference of squares
  • Check that no stray i survives
  • Substitute and finish