AMC 8 · 2005 · #8

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

parityformula-substitutionsystematic-enumeration systematic-enumerationguess-and-check ↑ Prerequisites: parity

📏 Short solution 💡 2 insights

Problem

Suppose m and n are positive odd integers. Which of the following must also be an odd integer?

Pick an answer.

(A)

m+3n

(B)

3m-n

(C)

$3m^2 + 3n^2$

(D)

$(nm + 3)^2$

(E)

3mn

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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

#12 Use Parity or Modular Arithmetic 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.

$$\text{odd} \pm \text{odd} = \text{even}, \qquad \text{odd} \times \text{odd} = \text{odd}$$

💡 These are the only two parity facts the problem needs — everything else is rule-chasing.

#3 Eliminate Possibilities 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.

$$m + 3n = \text{odd} + \text{odd} = \text{even}$$

💡 Two odds combine to an even — same way two odd handfuls of coins pair up perfectly.

#3 Eliminate Possibilities 2.OA.C.3 Step 3

Choice (B): $3m - n$.
Both $3m$ and $n$ are odd, and odd $-$ odd $=$ even.
Eliminate.

$$3m - n = \text{odd} - \text{odd} = \text{even}$$

💡 Subtraction obeys the same parity rule as addition.

#3 Eliminate Possibilities 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.

$$3m^2 + 3n^2 = \text{odd} + \text{odd} = \text{even}$$

💡 The exponent and the leading $3$ do not change parity — both pieces stay odd, and the sum is still even.

#3 Eliminate Possibilities 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.

$$(nm + 3)^2 = (\text{odd} + \text{odd})^2 = (\text{even})^2 = \text{even}$$

💡 Squaring never rescues parity — even squared is still even.

#12 Use Parity or Modular Arithmetic 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.

$$3mn = \text{odd} \times \text{odd} \times \text{odd} = \text{odd} \;\Rightarrow\; \textbf{(E)}$$

💡 A product of odd numbers stays odd no matter how many you multiply — so (E) is forced.

[1] #12 2.OA.C.3 Set the two parity rules we will use. With $m$ and $n$ both odd: $3$ is odd, so

[2] #3 2.OA.C.3 Choice (A): $m + 3n$. Both $m$ and $3n$ are odd, and odd $+$ odd $=$ even. So (A

[3] #3 2.OA.C.3 Choice (B): $3m - n$. Both $3m$ and $n$ are odd, and odd $-$ odd $=$ even. Elimi

[4] #3 3.OA.B.5 Choice (C): $3m^2 + 3n^2$. Squaring an odd gives odd ($\text{odd} \times \text{o

[5] #3 3.OA.B.5 Choice (D): $(nm + 3)^2$. Inside the parentheses, $nm$ is odd $\times$ odd $=$ o

[6] #12 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.3 Determine 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.5 Apply 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.2 Write, 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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