retroville 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 = 2x2x2x
3
70 is divisible by
5 <--> 70 = 2x
5x7
330 is divisible by
6 <--> 330 =
2x
3x5x11
56 is divisible by
8 <--> 56 =
2x
2x
2x7
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?
Now we can check the statements.
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 2: 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
