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Bunuel
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
Try to disprove each statement.

\(\frac{p}{9}\) = a + 1, so p = 9a + 1

p = 10, 19, 28, 37, 46, 55 ... (leaving out 1 on purpose)

From list, neither I nor II is true. 19, odd, and 10, even, disprove each respectively. At this point you could choose (C) and move on, because all other answers contain I and/or II

III. p = 3z + 1 for some integer z

10 = 3(3) + 1.
19 = 3(6) + 1.
55 = 3(18) + 1.
III is true. Answer C
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Bunuel
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 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
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P CAN BE 1,10,19....

hence I and II can be true or false

III is true for all cases

Option C
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p=3z+1 should be true where 1 is the remainder as per problem statement.
Since it is divisible by 9 z is quotient and the above expression cabe written as multiple of 3.
Option C.
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