Factors

kop wrote:Is P divisible by 168??

1) p is divisible by 14
2) p is divisible by 12

Target question: Is P divisible by 168?

A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = (2)(2)(2)(3)
70 is divisible by 5 <--> 70 = (2)(5)(7)
330 is divisible by 6 <--> 330 = (2)(3)(5)(11)
56 is divisible by 8 <--> 56 = (2)(2)(2)(7)

So, for P to be divisible by 168, 168 must be hiding in the prime factorization of P.

Since 168 = (2)(2)(2)(3)(7), we need to determine whether there are AT LEAST three 2's, one 3 and one 7 are hiding in the prime factorization of P. So, let's rephrase our target question . . .

REPHRASED target question: Are three 2's, one 3 and one 7 "hiding" in the prime factorization of P?

Statement 1: p is divisible by 14
Since 14 = (2)(7), we can conclude that one 2 and one 7 are "hiding" in the prime factorization of P.
So, we can't be certain whether or not three 2's, one 3 and one 7 are hiding in the prime factorization of P
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: p is divisible by 12
Since 12 = (2)(2)(3), we can conclude that two 2's and one 3 are "hiding" in the prime factorization of P.
So, we can't be certain whether or not three 2's, one 3 and one 7 are hiding in the prime factorization of P
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that one 2 and one 7 are "hiding" in the prime factorization of P.
Statement 2 tells us that two 2's and one 3 are "hiding" in the prime factorization of P.
So, we can conclude that there are at least TWO 2's, one 3 and one 7 "hiding" in the prime factorization of P.
HOWEVER, our goal is to determine whether there are THREE 2's, one 3 and one 7 "hiding" in the prime factorization of P.
So, we can't be certain whether or not three 2's, one 3 and one 7 are hiding in the prime factorization of P
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E

Cheers,
Brent

kop wrote:Is P divisible by 168??

1) p is divisible by 14
2) p is divisible by 12

Another approach is to look for counterexamples...

Target question: Is P divisible by 168?
NOTE: 168 = (2)(2)(2)(3)(7)

Statement 1: p is divisible by 14
There are several values of P that satisfy this condition. Here are two:
Case a: P = 168, in which case P is divisible by 168
Case b: P = 84, in which case P is NOT divisible by 168
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: p is divisible by 12
There are several values of P that satisfy this condition. Here are two:
Case a: P = 168, in which case P is divisible by 168
Case b: P = 84, in which case P is NOT divisible by 168
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
There are several values of P that satisfy this condition. Here are two:
Case a: P = 168, in which case P is divisible by 168
Case b: P = 84, in which case P is NOT divisible by 168
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E

Cheers,
Brent