AMC 10 · 2020 · #18
Grade 7 arithmeticProblem
Let (a,b,c,d) be an ordered quadruple of not necessarily distinct integers, each one of them in the set {0,1,2,3}. For how many such quadruples is it true that a⋅d−b⋅c is odd? (For example, (0,3,1,1) is one such quadruple, because 0⋅1−3⋅1=−3 is odd.)
Pick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $a \cdot d - b \cdot c$ is odd.
Givens: Each of $a, b, c, d$ is independently chosen from $\{0, 1, 2, 3\}$; Total quadruples: $4^4 = 256$; In $\{0, 1, 2, 3\}$: even values are $\{0, 2\}$ (2 of them), odd values are $\{1, 3\}$ (2 of them); Choices: (A) $48$, (B) $64$, (C) $96$, (D) $128$, (E) $192$
Unknowns: Number of quadruples making $a \cdot d - b \cdot c$ odd
Understand
Restated: 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 $a \cdot d - b \cdot c$ is odd.
Givens: Each of $a, b, c, d$ is independently chosen from $\{0, 1, 2, 3\}$; Total quadruples: $4^4 = 256$; In $\{0, 1, 2, 3\}$: even values are $\{0, 2\}$ (2 of them), odd values are $\{1, 3\}$ (2 of them); Choices: (A) $48$, (B) $64$, (C) $96$, (D) $128$, (E) $192$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem
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.
Execute — Answer: C
4.OA.B.4 Step 1 - Count how many ways the product $a \cdot d$ is odd.
- A product of two integers is odd only when both are odd.
- In $\{0, 1, 2, 3\}$ the odd numbers are $\{1, 3\}$ — there are 2 of them.
- So $a$ must be odd (2 choices) and $d$ must be odd (2 choices), giving $2 \cdot 2 = 4$ odd products.
- The remaining $16 - 4 = 12$ pairs $(a, d)$ give an even product.
💡 Odd $\times$ odd $=$ odd — both factors must be odd.
4.OA.B.4 Step 2 By the same reasoning for $(b, c)$, there are also $4$ pairs giving $bc$ odd and $12$ pairs giving $bc$ even — the pair $(b, c)$ behaves identically to $(a, d)$.
💡 Same logic as above — the second pair mirrors the first.
2.OA.C.3 Step 3 - $a \cdot d - b \cdot c$ is odd exactly when the two products have opposite parity: one is odd and the other is even (so their difference is odd $\pm$ even $=$ odd).
- Two mutually exclusive cases.
💡 Odd minus even (or even minus odd) is odd — the parities must disagree.
7.SP.C.8 Step 4 - Case 1: $ad$ odd and $bc$ even.
- The $(a, d)$ choices and $(b, c)$ choices are independent, so multiply: $4 \cdot 12 = 48$.
💡 Independent pieces — multiply the counts.
7.SP.C.8 Step 5 - Case 2: $ad$ even and $bc$ odd.
- By symmetry between $(a, d)$ and $(b, c)$, this is also $12 \cdot 4 = 48$.
💡 Symmetric to Case 1 — same count.
2.OA.C.3 Step 6 - The two cases never overlap (a product can't be both odd and even), so add: $48 + 48 = 96$.
- That matches choice (C).
💡 Mutually exclusive cases — just add.
4.OA.B.4 Count how many ways the product $a \cdot d$ is odd. A product of two integers is 4.OA.B.4 By the same reasoning for $(b, c)$, there are also $4$ pairs giving $bc$ odd and 2.OA.C.3 $a \cdot d - b \cdot c$ is odd exactly when the two products have opposite parit 7.SP.C.8 Case 1: $ad$ odd and $bc$ even. The $(a, d)$ choices and $(b, c)$ choices are in 7.SP.C.8 Case 2: $ad$ even and $bc$ odd. By symmetry between $(a, d)$ and $(b, c)$, this 2.OA.C.3 The two cases never overlap (a product can't be both odd and even), so add: $48 Review
Reasonableness: Out of $4^4 = 256$ total quadruples, our answer $96$ is exactly $\tfrac{96}{256} = \tfrac{3}{8}$. That makes sense: each product is odd with probability $\tfrac{1}{4}$ (two odd choices out of four, squared), so the chance that the parities disagree is $2 \cdot \tfrac{1}{4} \cdot \tfrac{3}{4} = \tfrac{6}{16} = \tfrac{3}{8}$, giving $\tfrac{3}{8} \cdot 256 = 96$. The answer choices $48$ and $192$ are exactly half and double of $96$ — common one-case-only or symmetry-mistake traps.
Alternative: Tool #16 (Complement): count even differences instead. $ad - bc$ is even when both products share parity. Even-even: $12 \cdot 12 = 144$. Odd-odd: $4 \cdot 4 = 16$. So even differences total $144 + 16 = 160$, and odd differences are $256 - 160 = 96$. Same answer.
CCSS standards used (min grade 7)
2.OA.C.3Determine whether a group of objects has an odd or even number (Reasoning about odd/even sums and differences (odd $-$ even $=$ odd).)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that a product is odd iff both factors are odd — counting odd-odd pairs in $\{0,1,2,3\}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Multiplying independent counts $4 \cdot 12 = 48$ for each parity case and summing the two cases.)
⭐ 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 $4 \cdot 12 + 12 \cdot 4 = 96$. The answer is $\textbf{(C)}$.
⭐ 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 $4 \cdot 12 + 12 \cdot 4 = 96$. The answer is $\textbf{(C)}$.
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