If a, b, and c are consecutive positive integers and a < b <

Hey Bunuel, I know I'm a few years late to this but I have a general question about consecutive integers.
According to your explanation, consecutive integers are always 1 apart. However in Sackmann's Total GMAT Math, he defines 'consecutive integers' as any set of integers that are EVENLY SPACED. I'm a little confused here. What's the correct way to think about them?
Thanks!

When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1: ... x-3, x-2, x-1, x, x+1, x+2, ....

For example:

-7, -6, -5 are consecutive integers.

2, 4, 6 ARE NOT consecutive integers, they are consecutive even integers.

3, 5, 7 ARE NOT consecutive integers, they are consecutive odd integers.

So, not all evenly spaced sets represent consecutive integers.

awesome. Thanks a lot.

Assume the following variables to make the calculations simpler:
3 consecutive integers: x-1, x, x+1
3 consecutive even/odd integers: x-2, x, x+2

Although, this question can be solved without the information, still you should keep it in mind if you encounter questions involving consecutive integers/even integers/odd integers etc.

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

I. c - a = 2. By our assumption, (a, b, c) are x-1, 1, x+1.
c - a = (x+1) - (x-1) = 2
Correct

II. abc is an even integer.
2 or more consecutive integers will always be even as every alternate number is even
Correct

III. (a + b + c)/3 is an integer.
((x+1) + x + (x-1))/3 = 3x/3 =x
And x is an integer.
Correct.

Hence option E: I, II, and III

The easiest way to solve the problem is to plug in some real 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.

The answer is E.

Here we have a, b, and c are consecutive integers, a<b<c =>

n, n+1, n+2 or
a, b = a+1, c = b +1, c = a+1 + 1 = a + 2 .

1. c=a+2 => sufficient

2. a*b*c => we have 2 variations here: - odd * even * odd or even *odd *even => the result is always going to be even, since we have even number in multiplication => sufficient


3. (a + b + c)/3 = (a + a + 1 + a + 3)/3 = a + 1 – always an integer => sufficient.

The answer is E.

Hi All,

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:

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

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

always take x-a , x , x+a
in solving consecutive integers problem
let see now
a = x-a
b = x
c = x+a

now 1. c-a=2
x+a - x + a = 2a correct

2. abc = even integer

as it consecutive integer so one number has to be a even number

3. x-a+x+x+a /3
3x/3
hence e answer

here in the option 2 abc is an even integer so it means that abc are multiplying right?

Right. abc there means a*b*c.

a, b, c are consecutive integers and a<b<c
=> We can express them as a=n, b= n+1 and c = n+2 where n is the value of the first integer

Now considering the options:

I. c- a = (n+2)-n = 2 so must be true for all cases

II. abc = n*(n+1)*(n+2)
at least one of the three integers will be even irrespective of the value of n. For example - 2*3*4 for n=1 ; (-3)*(-2)*(-1) for n=-4 etc
(Rule -For 3 consecutive integers - product = EVEN always)
So abc must be an EVEN integer

III. (a+b+c)/3 = (n + n+1+ n+3)/3 = (3n+3)/3 = n+1 is an INTEGER
Must be true.
(another rule for consecutive integers - product of three consecutive integers must always be divisible by 3)

So the solution is E.