Target question: Is P divisible by 168?kop wrote:Is P divisible by 168??
1) p is divisible by 14
2) p is divisible by 12
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
