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2020 AMC 10A Problem 18

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2020 AMC 10A Problem 18AMC 10 Step by Step
Name:                             Date: September 13, 2026
Number Properties · difficulty 3 of 5 · about 4 min · position P16-20

Let (a,b,c,d)(a,b,c,d) be an ordered quadruple of not necessarily distinct integers, each one of them in the set {0,1,2,3}.\{0,1,2,3\}. For how many such quadruples is it true that a⋅d−b⋅ca\cdot d-b\cdot c is odd? (For example, (0,3,1,1)(0,3,1,1) is one such quadruple, because 0⋅1−3⋅1=−30\cdot 1-3\cdot 1 = -3 is odd.)

  1. A)

    4848

  2. B)

    6464

  3. C)

    9696

  4. D)

    128128

  5. E)

    192192

Answer

C

Key insight

ad - bc is odd exactly when one product is odd and the other even; a product is odd for 4 of 16 pairs, so 2(4)(12) = 96.

Solution

A difference is odd exactly when its two terms have opposite parity, so we need one of adad, bcbc odd and the other even.

Count ordered pairs (a,d)(a, d) from {0,1,2,3}2\{0,1,2,3\}^2 (1616 pairs) by the parity of adad. The product is odd only when both factors are odd, and each has 22 odd choices: 2⋅2=42 \cdot 2 = 4 pairs. The remaining 16−4=1216 - 4 = 12 pairs give an even product. The same counts hold for (b,c)(b, c).

Quadruples with adad odd and bcbc even: 4⋅12=484 \cdot 12 = 48. Quadruples with adad even and bcbc odd: 12⋅4=4812 \cdot 4 = 48. Total
48+48=96.48 + 48 = 96.

The answer is (C) 96\boxed{\textbf{(C)}\ 96}.

Why this works

Only parity matters, and the set {0,1,2,3}\{0,1,2,3\} contains two evens and two odds, so parity behaves like a fair coin for each entry. "Odd minus even" and "even minus odd" are symmetric, which halves the work. Recognizing that oddness of a product requires all factors odd is the whole computation.

Alternative approach

Probability view: each of adad, bcbc is odd with probability 14\frac14, independently. The difference is odd with probability 2⋅14⋅34=382 \cdot \frac14 \cdot \frac34 = \frac38, and 38⋅256=96\frac38 \cdot 256 = 96.

The trap

Counting only the case ad odd and bc even (48) and forgetting the symmetric case, or working with parities alone without multiplying by 2 for each digit.

Common mistakes

  • Counting only the case ad odd and bc even (48) and forgetting the symmetric case, or working with parities alone without multiplying by 2 for each digit.
  • Believing adad is odd half the time (giving 12⋅256=128\frac12 \cdot 256 = 128).

Techniques

Use an invariant, parity, or coloring argument · Exploit symmetry to reduce work or pair up objects