Competition · AMC preparation · step 4 of 4

AMC 8 · 2005 · #8

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

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

📏 Short solution 💡 2 insights

Problem

m and n are positive odd integers. Which of the five expressions m+3n, 3m-n, 3m²+3n², (nm+3)², 3mn 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

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Use Parity or Modular Arithmetic

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 ± odd = even, odd × 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.

1STEP 1

Set the parity rules

Two parity rules settle every choice: odd ± odd = even, and odd × odd = odd. With m, n odd, 3m and 3n are odd too.

odd ± odd = even, odd × odd = odd

2STEP 2

Test choice A

(A) m + 3n is odd + odd, so it is always even — eliminate it.

m + 3n = odd + odd = even

3STEP 3

Test choice B

(B) 3m - n is odd - odd, which is even — eliminate.

3m - n = odd - odd = even

4STEP 4

Test choice C

3m² + 3n² = odd + odd = even

5STEP 5

Test choice D

(D) (nm + 3)²: inside is odd + odd = even, and even squared is even — eliminate.

(nm + 3)² = (odd + odd)² = (even)² = even

6STEP 6

Test choice E

(E) 3mn is odd × odd × odd = odd — the only choice that must be odd.

3mn = odd × odd × odd = odd → (E)

Answer

3mn

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 · 1 · 1 = 3 odd. Try m = 3, n = 5 too: (E) 3 · 3 · 5 = 45 odd. The parity argument predicted exactly this — only (E) lands on odd. The four eliminated choices each produce "odd ± odd" or "even squared," which the parity rules say must be even.

💡Key takeaway

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 ± odd = even, odd × odd = odd) decide all five choices in this AMC 8 problem at once.

Set the parity rules
Test choice A
Test choice B
Test choice C
Test choice D
Test choice E

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