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If a, b, and c are consecutive positive integers and a < b <
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18 Aug 2020, 09:36
18 Aug 2020, 09:36
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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
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
Re: If a, b, and c are consecutive positive integers and a < b <
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18 Aug 2020, 09:36
18 Aug 2020, 09:36
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Re: If a, b, and c are consecutive positive integers and a < b <
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18 Aug 2020, 12:08
18 Aug 2020, 12:08
a, b and c are three consecutive numbers. Let's say they are of the form
a=k, b=k+1 and c=k+2, k being any integer we want.
Then we look at the three assertions:
I - c-a=k+2-k=2 (this makes sense because a,b and c are consecutive so the gap between each one of them is 1)
II - abc=k*(k+1)*(k+2). Now we know that at least one of the three consecutive numbers is even (1,2,3 satisfy this example for instance but it also works if we take 2,3,4 which has two even numbers). So abc = 2 * some integer. abc is thus even because 2 divides abc.
III - a+b+c = 3k+3 =3 * (k+1) and thus 3 divides (a+b+c) so (a+b+c)/3 is an integer.
Thus all three assertions are true.
a=k, b=k+1 and c=k+2, k being any integer we want.
Then we look at the three assertions:
I - c-a=k+2-k=2 (this makes sense because a,b and c are consecutive so the gap between each one of them is 1)
II - abc=k*(k+1)*(k+2). Now we know that at least one of the three consecutive numbers is even (1,2,3 satisfy this example for instance but it also works if we take 2,3,4 which has two even numbers). So abc = 2 * some integer. abc is thus even because 2 divides abc.
III - a+b+c = 3k+3 =3 * (k+1) and thus 3 divides (a+b+c) so (a+b+c)/3 is an integer.
Thus all three assertions are true.
