If positive integer p divided by 9 leaves a remainder of 1

swerve wrote:If positive integer p divided by 9 leaves a remainder of 1, which of the following must be true?

I. p is even.
II. p is odd.
III. p = 3Z+1 for some integer z.

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

We see that p can be values such as 1, 10, 19, 28, ….

Therefore, neither Roman numeral I nor Roman numeral II is true. Furthermore, there is no answer choice stating that none of the Roman numerals is true.

Thus, the Roman numeral that must be true is that p = 3z+1 for some integer z.

Alternate Solution:

Since positive integer p divided by 9 leaves a remainder of 1, we have:

p/9 = Q + 1/9

p = 9Q + 1

Thus, p can be values such as 1, 10, 19, 28, ….

Thus, p can be either even or odd.

Now we can look at Roman numeral III:

p = 3z + 1 for some integer z.

Recall that we have expressed that p = 9Q + 1, so p = 3(3Q) + 1. We see that if we let z = 3Q, we can express p as 3z + 1. So Roman numeral III is true.

Answer: C