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If a + 1 = (20/a) and b is the average of a set of c consecu

Expert replies
by hazelnut01 » Wed Mar 22, 2017 12:01 am
If a + 1 = (20/a) and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. (a^2)(b^2)(c^2) is even.
II. a + b + c is odd.
III. ab[(c^2) + c] is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

Source : Manhattan Prep
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Source: — Problem Solving |

by Jay@ManhattanReview » Wed Mar 22, 2017 12:41 am
ziyuenlau wrote:If a + 1 = (20/a) and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. (a^2)(b^2)(c^2) is even.
II. a + b + c is odd.
III. ab[(c^2) + c] is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

Source : Manhattan Prep
Hi ziyuenlau,

We have a + 1 = (20/a)

=> a^2 + a = 20

=> a^2 + a - 20 = 0

=> a^2 + 5a - 4a - 20 = 0

=> a(a + 5) - 4(a +5) = 0

=> a = -5 or 4 => a can be even or odd

We know that c is odd.

Let us find out the nature of b.

Say there are three consecutive integers, c = 3.

Case 1: Set: {2, 3, 4} => Average = b = 3 (odd)

Case 2: Set: {3, 4, 5} => Average = b = 4 (even)

b can be even or odd.

Let's see each statement one by one.

S1: (a^2)(b^2)(c^2) is even.

Say a and b each is odd

(a^2)(b^2)(c^2) = (Odd^2)(Odd^2)(Odd^2) = Odd. It's not a must be true statement.

S2: a + b + c is odd.

Say a and b each is odd.

=> a + b + c = Odd + Odd + Odd = Odd. It's not a must be true statement.

S3: ab[(c^2) + c] is even.

ab[(c^2) + c] = ab[Odd^2 + Odd] = ab[Odd + Odd] = ab[Even] = Even.

=> Whatever be the nature of a and b, ab[(c^2) + c] is always even.

Statement III must be true.

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Math Essentials Guide

-Jay
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