theachiever wrote:If positive integer x is a multiple of 6 and positive integer y is a multiple of 14,is xy a multiple of 105?

1.x is a multiple of 9.

2.y is a multiple of 25.

Target question: Is xy a multiple of 105?

Important stuff:
First, If N is a multiple of k, then N is divisible by k.

Second, 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 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7


Since 105 = (3)(5)(7), then we can rewrite the target question as . . .
Rephrased target question: Is there a 3, a 5 and a 7 hiding in the prime factorization of xy?

Given: x is a multiple of 6
In other words, x = (2)(3)(other possible prime numbers)

Given: y is a multiple of 14
In other words,y = (2)(7)(other possible prime numbers)

Combine both of the above to see that xy = (2)(2)(3)(7)(other possible prime numbers)

So, the given information tells us that we ALREADY have a 3 and a 7 hiding in the prime factorization of xy. The only piece missing is the 5.

So, we can rephrase our target question one last time. . .

Rephrased target question: Is there a 5 hiding in the prime factorization of xy?

Statement 1: x is a multiple of 9.
Since 9 = (3)(3), all this tells us is that there are two 3's hiding in the prime factorization of xy.
So, there may or may not be a 5 hiding in the prime factorization of xy.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 1: y is a multiple of 25.
Since 25 = (5)(5), this tells us is that there is definitely a 5 hiding in the prime factorization of xy.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent