2005 AMC 8 Problem 8

Question

Suppose

m

and

n

are positive odd integers. Which of the following must also be an odd integer?

m + 3n

3m - n

3m^2 + 3n^2

(nm + 3)^2

3mn

Concepts:parity

Test the choices using parity rules for odd numbers.

The product of odd integers is odd.

Recall the four following rules:

• odd plus odd and even plus even is even

• even plus odd is odd

• even times anything is even

• odd times odd is odd

These rules can be easily verified by representing arbitrary odd numbers as

2m+1

and arbitrary even numbers as

2n

respectively, for integers

m,

n.

With this in mind, let’s examine each answer choice individually:

A:

Note that

3

is odd. Modulo

2

, this gives us

1 + 1 \cdot 1 \equiv 0 \pmod 2.

From our above rules, we know that this is even.

B:

1 \cdot 1 - 1 \equiv 0 \pmod 2.

Once again, this is even.

C:

1 \cdot 1^2 + 1 \cdot 1^2 \equiv 0 \pmod 2.

This is also even.

D:

(1 \cdot 1 + 1)^2 \equiv 0 \pmod 2.

Unfortunately, this is also even.

E:

1 \cdot 1 \cdot 1 \equiv 1 \pmod 2.

This is odd.

Therefore, E is the only answer choice that is an odd integer.

Thus, E is the correct answer.

Problem 8 in other years