find out what it means to be a multiple of 105 by breaking down 105 to its prime factors:

105 = 5*3*7

Thus, in order to be a multiple of 105, xy have to be multiples of 3, 5 and 7 between them. Think of 3,5,7 as the building blocks needed to "make" a 105 - do x and y have what it takes?

From the Q stem, x is a multiple of 6, which makes it a multiple of 3. So xy will also be a multiple of 3
By the same reasoning, y is a multiple of 14, which means that it's a multiple of 7. So xy will also be a multiple of 7.

It just remains to be seen whether the xy has the last needed building block of 5:

Stat. (1) - x is a multiple of 9, but you don't know whether it's also a multiple of 5 (for example, 6*9*5), or not (for example, x=6*9). The the statement is insufficient: whether xy is a multiple of 105 depends on whether x brings the additional remaining 5, which we are unable to determine.

Stat. (2) - if y is a multiple of 25, the it is also a multiple of 5: This closes the list, as xy includes all of the building blocks needed: 5, 3, and 7. Thus, the statement is sufficient, and the answer is B.

Another problem dealing with number properties is

If k,m, and t are positive integers and k/6 + m/4 = t/12 , do t and 12 have a common factor greater than 1?

1. k is a multiple of 3
2. m is a multiple of 3.

shoot4greatness wrote:Another problem dealing with number properties is

If k,m, and t are positive integers and k/6 + m/4 = t/12 , do t and 12 have a common factor greater than 1?

1. k is a multiple of 3
2. m is a multiple of 3.

Multiply by 12 on both sides to get rid of these nasty fractions and isolate t:

12k/6 + 12m/4 = 12t/12
reduce:

2k+3m=t

Now, what could t be? your goal is to show that t can have a common factor with 12, but doesn't have to. primes are good for the 'no' example, as a prime will not share any factor other than 1.

(1): let k=3, and m=1, so that t=2*3+3*1=9, which shares 3 with as a common factor with 12.
Let k=3, m=2 to get t=11, which is prime and shares no common factor.

IS

(2) let m=3 and k=1, which gives t=2+9=11, giving us the no answer from earlier.
Can we find a yes anwer? keep m=3, and try out ks: k=3 yields t=15, which shares a 3 with 12.

IS

combined: if both m and k are multiples of 3, then t will also be a multiple of 3, so it will always have 3 as a common factor and the answer is a definite yes - sufficient --> C.

shoot4greatness wrote:Another problem dealing with number properties is

If k,m, and t are positive integers and k/6 + m/4 = t/12 , do t and 12 have a common factor greater than 1?

1. k is a multiple of 3
2. m is a multiple of 3.

I would approach this problem just as Geva did, but the solution above has a slight miscalculation. When we evaluate statement 1, if k=3 and m=2, then t = 2*3 + 3*2 = 12. Thus t and 12 will share the factors 2, 3, 4, 6, and 12.

The correct answer is A. Here's why:

We're given the equation 2k + 3m = t.

Statement 1: k is multiple of 3
Then 2k and 3m are each a multiple of 3.
This tells us that t is a multiple of 3. (If each term of a sum is a multiple of x, then the sum is a multiple of x.)
Thus t and 12 will share 3 as a factor.

If we plug in values, no matter what we plug in, t and 12 will always share 3 as a factor.
Thus, sufficient.