ACT math prep question

Question

35.

Given that x and (7-x)/2 are integers, which of the following statements about x MUST be true?

A. x is odd

B. x is even

C. x is prime

D. x is positive

E. x is negative

My answer is "x is odd" b/c when I "plug in" 1, 3, 5 or 7, I get an integer. I don't have a better explanation for why, other than that all the other choices B. - E. can easily be made to not produce integers.

Can someone give me a better/more certain explanation (if I'm even correct)?

Answer 1

Hello John,

1. Short Answer: You have it right. A bit more generally speaking, for integers, an odd minus an odd is always even, and an even is always divisible by 2.

1. Long Answer: If you're really interested in the mechanics of it, keep reading. :-)
2. The below is also instructive as there are any number of questions the ACT Math can ask about even and odd integers.

Let k be an integer. That is, let

k ∈ {... -3, -2, -1, 0, 1, 2, 3 ...}. Then doubling any value chosen for k yields,

2k ∈ {...-6, -4, -2, 0, 2, 4, 6...}, which creates a list of EVEN integers. Now, if you add 1 to 2k, we get

2k + 1 ∈ {... -5, -3, -1, 1, 3, 5, 7 ...}, a list of ODD integers.

So, let's pick two integers, m and n.

Then 2m + 1 and 2n + 1 are odd integers ( in our case 7 and x, where we are letting x be an odd integer).

Subtracting them, we have

(2m + 1) - (2n + 1)

= 2m + 1 - 2n -1

= 2m - 2n

= 2 (m - n).

Since m and n are integers, and the set of integers is closed under subtraction,

m - n is an integer,

→ 2 (m - n) is an even integer,

which will always be divisible by 2.