AMC 8 · 2005 · #8
Grade 6 arithmeticProblem
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Pick an answer.
AMC 8 2005 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: $m$ and $n$ are positive odd integers. Which of the five expressions $m+3n$, $3m-n$, $3m^2+3n^2$, $(nm+3)^2$, $3mn$ must also be an odd integer?
Givens: $m$ is a positive odd integer; $n$ is a positive odd integer; Answer choices: (A) $m+3n$, (B) $3m-n$, (C) $3m^2+3n^2$, (D) $(nm+3)^2$, (E) $3mn$
Unknowns: Which expression is guaranteed to be odd for every choice of positive odd $m$ and $n$
Understand
Restated: $m$ and $n$ are positive odd integers. Which of the five expressions $m+3n$, $3m-n$, $3m^2+3n^2$, $(nm+3)^2$, $3mn$ must also be an odd integer?
Givens: $m$ is a positive odd integer; $n$ is a positive odd integer; Answer choices: (A) $m+3n$, (B) $3m-n$, (C) $3m^2+3n^2$, (D) $(nm+3)^2$, (E) $3mn$
Plan
Primary tool: #12 Use Parity or Modular Arithmetic
Secondary: #3 Eliminate Possibilities
The actual values of $m$ and $n$ are not given — only their parity. Tool #12 (Use Parity or Modular Arithmetic) is built for exactly this: track whether each piece is odd or even using two short rules (odd $\pm$ odd $=$ even, odd $\times$ odd $=$ odd) and ignore the numerical size. Once the parity of each choice is fixed, Tool #3 (Eliminate Possibilities) sweeps away any choice whose parity comes out even. Only one survives.
Execute — Answer: E
2.OA.C.3 Step 1 - Set the two parity rules we will use.
- With $m$ and $n$ both odd: $3$ is odd, so $3n$ is odd $\times$ odd $=$ odd, and similarly $3m$ is odd.
- Adding or subtracting two odd numbers gives an even number.
- Multiplying any list of odd numbers gives an odd number.
💡 These are the only two parity facts the problem needs — everything else is rule-chasing.
2.OA.C.3 Step 2 - Choice (A): $m + 3n$.
- Both $m$ and $3n$ are odd, and odd $+$ odd $=$ even.
- So (A) is always even.
- Eliminate it.
💡 Two odds combine to an even — same way two odd handfuls of coins pair up perfectly.
2.OA.C.3 Step 3 - Choice (B): $3m - n$.
- Both $3m$ and $n$ are odd, and odd $-$ odd $=$ even.
- Eliminate.
💡 Subtraction obeys the same parity rule as addition.
3.OA.B.5 Step 4 - Choice (C): $3m^2 + 3n^2$.
- Squaring an odd gives odd ($\text{odd} \times \text{odd}$), and multiplying by the odd number $3$ keeps it odd.
- So $3m^2$ and $3n^2$ are both odd, and odd $+$ odd $=$ even.
- Eliminate.
💡 The exponent and the leading $3$ do not change parity — both pieces stay odd, and the sum is still even.
3.OA.B.5 Step 5 - Choice (D): $(nm + 3)^2$.
- Inside the parentheses, $nm$ is odd $\times$ odd $=$ odd, then odd $+$ odd $=$ even.
- Squaring an even gives an even.
- Eliminate.
💡 Squaring never rescues parity — even squared is still even.
6.EE.A.2 Step 6 - Choice (E): $3mn$.
- This is a product of three odd numbers ($3$, $m$, $n$), and odd $\times$ odd $\times$ odd is odd.
- This is the only choice that must be odd.
💡 A product of odd numbers stays odd no matter how many you multiply — so (E) is forced.
2.OA.C.3 Set the two parity rules we will use. With $m$ and $n$ both odd: $3$ is odd, so 2.OA.C.3 Choice (A): $m + 3n$. Both $m$ and $3n$ are odd, and odd $+$ odd $=$ even. So (A 2.OA.C.3 Choice (B): $3m - n$. Both $3m$ and $n$ are odd, and odd $-$ odd $=$ even. Elimi 3.OA.B.5 Choice (C): $3m^2 + 3n^2$. Squaring an odd gives odd ($\text{odd} \times \text{o 3.OA.B.5 Choice (D): $(nm + 3)^2$. Inside the parentheses, $nm$ is odd $\times$ odd $=$ o 6.EE.A.2 Choice (E): $3mn$. This is a product of three odd numbers ($3$, $m$, $n$), and o Review
Reasonableness: Plug in the simplest case $m = n = 1$ to double-check: (A) $1 + 3 = 4$ even, (B) $3 - 1 = 2$ even, (C) $3 + 3 = 6$ even, (D) $(1 + 3)^2 = 16$ even, (E) $3 \cdot 1 \cdot 1 = 3$ odd. Try $m = 3, n = 5$ too: (E) $3 \cdot 3 \cdot 5 = 45$ odd. The parity argument predicted exactly this — only (E) lands on odd. The four eliminated choices each produce "odd $\pm$ odd" or "even squared," which the parity rules say must be even.
Alternative: Tool #9 (Solve an Easier Related Problem): replace the general odd $m, n$ with the smallest case $m = n = 1$. Each choice becomes a single number: $4, 2, 6, 16, 3$. Only $3$ is odd, so the answer must be (E). Because the problem promises one of the choices is odd for every valid $(m, n)$, a single counterexample rules a choice out, and one survivor pins down the answer.
CCSS standards used (min grade 6)
2.OA.C.3Determine whether a group of objects has an odd or even number of members (Naming the two core parity facts — odd $\pm$ odd $=$ even and odd $\times$ odd $=$ odd — and applying them to choices (A) and (B).)3.OA.B.5Apply properties of operations as strategies to multiply and divide (Extending the parity rules through squaring and the constant multiplier $3$ to rule out (C) and (D).)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Reading each algebraic expression $m+3n$, $3m-n$, $3m^2+3n^2$, $(nm+3)^2$, $3mn$ as a recipe in $m$ and $n$ so the parity rules can be applied symbolically.)
⭐ When a problem only tells you the parity of the inputs, throw away the actual numbers and track "odd" or "even" through each step. Two short rules (odd $\pm$ odd $=$ even, odd $\times$ odd $=$ odd) decide all five choices in this AMC 8 problem at once.
⭐ When a problem only tells you the parity of the inputs, throw away the actual numbers and track "odd" or "even" through each step. Two short rules (odd $\pm$ odd $=$ even, odd $\times$ odd $=$ odd) decide all five choices in this AMC 8 problem at once.
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