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

[swerve]

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

The OA is C.

Please, can anyone explain this PS question? I don't understand why C is the correct answer. I need help. Thanks.

[Shahrukh_mbabreakspace]

P leaves 1 as remainder when divided by 9, it means p is of 9x+1 form where x is a non-negative integer
It can be odd or even depending on value of x, so I and II can't be true always.

We can write P as 3*3x +1
This clearly Shows that III is always true and Z is of 3x form.

________________________________________

Shahrukh Moin Khan
QA Mentor at Breakspace
https://www.mbabreakspace.com
PGDM: IIM Calcutta
B.Tech IIT Roorkee

[GMATGuruNY]

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

Since dividing p by 9 yields a remainder of 1, p is ONE MORE THAN MULTIPLE OF 9:
p = 9a + 1, where a is a nonnegative integer.
Options for p
1, 10, 19, 28...

If p=1, then Statement I is not true.
Eliminate A, D and E.
If p=10, then Statement II is not true.
Eliminate B.

The correct answer is C.

The expression in blue indicates that Statement III must be true.

[Scott@TargetTestPrep]

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