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Is b/3 an integer?

This topic has 1 expert reply and 0 member replies

Is b/3 an integer?

Post Tue Sep 12, 2017 10:30 am
Is b/3 an integer?

(1) (b^2-9)/3 is an integer.
(2) p and p + 1 are prime factors of b.

OA is B.

Statement 2 confuses me. Can any expert help me.

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Top Reply
Post Tue Sep 12, 2017 12:29 pm
Vincen wrote:
Is b/3 an integer?

(1) (b² - 9)/3 is an integer.
(2) p and p + 1 are prime factors of b.
Target question: Is b/3 an integer?

Statement 1: (b² - 9)/3 is an integer
There are several values of b that satisfy statement 1. Here are two:
Case a: b = 6. Here, (b² - 9)/3 = (6² - 9)/3 = (36 - 9)/3 = 27/3 = 9, and 9 is an integer. In this case, b/3 = 6/3 = 2. So, b/3 IS an integer
Case b: b = √18. Here, (b² - 9)/3 = (√18² - 9)/3 = (18 - 9)/3 = 9/3 = 3, and 3 is an integer. In this case, b/3 = (√18)/3 = (3√2)/3 = √2 (. So, b/3 is NOT an integer
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: p and p + 1 are prime factors of b
Notice that p and p+1 are CONSECUTIVE integers
Since p and p+1 are CONSECUTIVE integers, we know that one of the numbers must be odd and one must be even.
Since 2 is the ONLY even integer, we can conclude that p = 2 and p+1 = 3
Since p and p+1 are factors of b, we can see that b is divisible by 2 and by 3.
If b is divisible by 3, then b/3 must be an integer
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

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Brent Hanneson – Founder of GMATPrepNow.com
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Thanked by: Vincen
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