Competition · AMC preparation · step 4 of 4

AMC 10 · 2020A · #18

Grade 7 arithmetic
Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split
paritycaseworkcombinations-basicsystematic-enumeration caseworkidentify-subproblemscomplementary-counting ↑ Prerequisites: paritycombinations-basic
📏 Medium solution 💡 3 insights
Problem
Pick four numbers a, b, c, d — each from {0, 1, 2, 3}, repeats allowed, order matters. Count the ordered quadruples (a, b, c, d) for which a · d - b · c is odd.

Pick an answer.

(A)
48
(B)
64
(C)
96
(D)
128
(E)
192

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Tool #7 (Subproblems): the question only cares about parity (odd vs even), not the actual values. Break it into two sub-questions: "how many (a, d) make ad odd?" and "how many (b, c) make bc odd?" Then ad - bc is odd exactly when one product is odd and the other is even. Tool #2 (Systematic List): in {0, 1, 2, 3} count odd numbers (1, 3) and even numbers (0, 2) — two of each. Tool #9 (Easier Problem): replacing the 4-value set with just "odd / even" turns this into a simple parity-counting puzzle.

1STEP 1

Count odd ad pairs

Odd needs both factors odd; {0, 1, 2, 3} has 2 odd values, so 4 of the 16 (a, d) pairs make a · d odd, the other 12 even.

#{(a,d): ad odd} = 2 · 2 = 4, #{(a,d): ad even} = 12
2STEP 2

Repeat for b and c

By the same reasoning, (b, c) mirrors (a, d): 4 pairs give b · c odd and 12 give b · c even.

#{(b,c): bc odd} = 4, #{(b,c): bc even} = 12
3STEP 3

Find when the difference is odd

a · d - b · c is odd only when the two products have opposite parity — one odd, one even; two exclusive cases.

odd - even = odd, even - odd = odd
4STEP 4

Count case 1

Case 1 — a · d odd, b · c even: the two pairs are independent, so multiply to get 48.

Case 1 = 4 · 12 = 48
5STEP 5

Count case 2

Case 2 — a · d even, b · c odd: by symmetry this also gives 48.

Case 2 = 12 · 4 = 48
6STEP 6

Add the two cases

The two cases are disjoint — a product can't be both odd and even — so add: 96, matching choice (C).

48 + 48 = 96 → (C)
Answer
96
Out of 4⁴ = 256 total quadruples, our answer 96 is exactly 96256\frac{96}{256} = 38\frac{3}{8}. That makes sense: each product is odd with probability 14\frac{1}{4} (two odd choices out of four, squared), so the chance that the parities disagree is 2 · 14\frac{1}{4} · 34\frac{3}{4} = 616\frac{6}{16} = 38\frac{3}{8}, giving 38\frac{3}{8} · 256 = 96. The answer choices 48 and 192 are exactly half and double of 96 — common one-case-only or symmetry-mistake traps.
💡Key takeaway

This AMC 10 problem only needs Grade 7 parity-counting you already know — a product is odd only when both factors are odd, so among 16 pairs only 4 give an odd product. Mixing one odd-product pair with one even-product pair gives 4 · 12 + 12 · 4 = 96. The answer is (C).

  • Count odd ad pairs
  • Repeat for b and c
  • Find when the difference is odd
  • Count case 1
  • Count case 2
  • Add the two cases

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