AMC 10 · 2020 · #18
Grade 7 arithmeticProblem
Let be an ordered quadruple of not necessarily distinct integers, each one of them in the set For how many such quadruples is it true that is odd? (For example, is one such quadruple, because is odd.)
Pick an answer.
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 $\textbf{(C)}$.