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