AbeNeedsAnswers wrote:If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?
I. 8
II. 12
III 18
A) II only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III
B
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
Consider these examples:
24 is divisible by
3 because 24 = (2)(2)(2)
(3)
Likewise, 70 is divisible by
5 because 70 = (2)
(5)(7)
And 112 is divisible by
8 because 112 = (2)
(2)(2)(2)(7)
And 630 is divisible by
15 because 630 = (2)(3)
(3)(5)(7)
-----------ONTO THE QUESTION-------------------------------
positive integer x is a multiple of 4
In other words, x is divisible by
4, which means
4 is hiding in the prime factorization of x
So, we can write: x =
(2)(2)(?)(?)(?)(?)....
[ASIDE: The ?'s represents other primes that COULD be in the prime factorization. However, the only part of the prime factorization that we are certain of is the two 2's]
positive integer y is a multiple of 6
In other words, y is divisible by
6, which means
6 is hiding in the prime factorization of y
So, we can write: y =
(2)(3)(?)(?)(?)(?)....
This means xy =
(2)(2)(2)(3)(?)(?)(?)(?)....
This tells us that xy is divisible by all products formed by any combination of
(2)(2)(2)(3)
So, xy must be divisible by 2, 3, 4, 6, 8, 12, 24
Answer:
B
Cheers,
Brent
