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

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If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 29 Dec 2012, 06:44
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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
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If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 29 Dec 2012, 06:49
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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

Since a, b, and c are consecutive positive integers and a < b < c, then c = a + 2, from which it follows that c - a = 2. So, I is true.

Next, out of 3 consecutive integers at least 1 must be even, thus abc=even. II is true.

Finally, since b = a + 1, and c = a + 2, then (a + b + c)/3 = (a + a + 1 + a + 2)/3 = a + 1 = integer. III is true as well. (Or: the sum of odd number of consecutive integers is ALWAYS divisible by that odd number. )

Answer: E.

More:

• If \(k\) is odd, the sum of \(k\) consecutive integers is always divisible by \(k\). Given \(\{9,10,11\}\), we have \(k=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If \(k\) is even, the sum of \(k\) consecutive integers is never divisible by \(k\). Given \(\{9,10,11,12\}\), we have \(k=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of \(k\) consecutive integers is always divisible by \(k!\), so by \(k\) too. Given \(k=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.

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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 22 Jun 2013, 11:32
1
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!
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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 28 May 2014, 09:54
1
1
If a, b, and c are consecutive positive integers and a < b < c, which of the following must be true?

I. c - a = 2 #True: Since a,b,c are consecutive integers a and c must be 2 integers apart
II. abc is an even integer. #True: For any 3 consecutive integers a,b,c the product has to be divisible by 3! i.e. 6 --> it is even
III. (a + b + c)/3 is an integer. #True: This is a mean of the series. That is this has to be the middle no. b which as per the question stem is an integer

Answer: E
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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 06 Dec 2017, 14:40
1
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:

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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 22 Jun 2013, 11:37
tricialin wrote:
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.
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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 22 Jun 2013, 11:44
awesome. Thanks a lot.
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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 24 Sep 2015, 01:00
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
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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 15 Jul 2016, 04:49
Walkabout wrote:
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


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.
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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 18 May 2017, 11:52
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.
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If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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New post 26 Feb 2018, 07:11
Bunuel wrote:
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

Since a, b, and c are consecutive positive integers and a < b < c, then c = a + 2, from which it follows that c - a = 2. So, I is true.

Next, out of 3 consecutive integers at least 1 must be even, thus abc=even. II is true.

Finally, since b = a + 1, and c = a + 2, then (a + b + c)/3 = (a + a + 1 + a + 2)/3 = a + 1 = integer. III is true as well. (Or: the sum of odd number of consecutive integers is ALWAYS divisible by that odd number. )

Answer: E.

More:

• If \(k\) is odd, the sum of \(k\) consecutive integers is always divisible by \(k\). Given \(\{9,10,11\}\), we have \(k=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If \(k\) is even, the sum of \(k\) consecutive integers is never divisible by \(k\). Given \(\{9,10,11,12\}\), we have \(k=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of \(k\) consecutive integers is always divisible by \(k!\), so by \(k\) too. Given \(k=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.


how can option I be true if i take \(-4<-3<-2\) (c - a = 2) \(-2-4 = -6\):? isnt it must be true question :)


ok I got it :) I must attentively read the question :)
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Re: If a, b, and c are consecutive positive integers and a < b <  [#permalink]

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Re: If a, b, and c are consecutive positive integers and a < b <   [#permalink] 04 Sep 2019, 12:10
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