Hi lionsshare,

This Roman Numeral question can be solved by either TESTing VALUES or using Number Properties. Here are the various Number Properties involved in this prompt:

We're told that A, B and C are CONSECUTIVE, POSITIVE INTEGERS and that A < B < C. We're asked which of the following MUST be true.

Since the numbers are consecutive, positive integers and A < B < C, we can 'rewrite' the three variables as..
A
B = A+1
C = A+2

I. C - A = 2

Since C = A+2....
C - A =
(A+2) - A =
2
Roman Numeral 1 is always true.
Eliminate Answers B and D.

II. ABC is an EVEN integer.

When dealing with 3 consecutive integers, we're guaranteed to have at least one even integer. The options would be:
(even)(odd)(even)
(odd)(even)(odd)

When multiplying ANY integer by an EVEN number, the product is ALWAYS EVEN. Thus Roman Numeral II is always true.
Eliminate Answer A.

III. (A+B+C)/3 is an integer.

Using the 'rewritten' versions of B and C above, we know that...
(A+B+C) = (A+A+1+A+2) = 3A+3
Since A is an integer, we know that 3A will always be a multiple of 3. Adding a multiple of 3 to 3 (which is also clearly a multiple of 3), we will end up with a sum that is ALWAYS a multiple of 3. Finally, dividing a multiple of 3 by 3 will always give you an integer. Thus, Roman Numeral III is always true.
Eliminate Answer C.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

lionsshare wrote: ↑

Sat Sep 16, 2017 1:51 am

If a, b, and c are consecutive positive integers and a < b < c, which of the following must be true?

I. c - a = 2
II. abc is an even integer.
III. (a + b + c)/3 is an integer.

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III

OA: E

Hello, Experts. Please share the solution to this problem.

The easiest way to solve the problem is to plug in some actual numbers for a, b, and c. Since we know they are consecutive and we know that a < b < c, we can say:

a = 1

b = 2

c = 3

Or

a = 2

b = 3

c = 4

It is good to test two cases because in our first case we start with an odd integer and in the second case we start with an even integer.

Let’s use these values in each Roman numeral answer choice. Remember we need to determine which answer must be true, meaning in all circumstances.

I. c – a = 2

Case #1

3 – 1 = 2

Case #2

4 - 2 = 2

I must be true.

II. abc is an even integer.

Case #1

1 x 2 x 3 = 6

Case #2

2 x 3 x 4 = 24

II must be true.

III. (a + b + c)/3 is an integer.

Case #1

(1 + 2 + 3)/3 = 6/3 = 2

Case #2

(2 + 3 + 4)/3 = 9/3 = 3

III must be true.

I, II, and III are all true.

Answer: E