ardz24 wrote:If x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200, then x must be a multiple of which of the following?

A) 3
B) 6
C) 7
D) 8
E) 10

If y is a multiple of 5, then we know that y = 5k for some integer k.
So, let's take 3x + 4y = 200 and replace y with 5k to get: 3x + 4(5k) = 200
Simplify to get: 3x + 20k = 200 (where k is some integer)
Subtract 20k from both sides to get: 3x = 200 - 20k
Divide both sides by 3 to get: x = (200 - 20k)/3
Now factor the numerator to get: x = (20)(10 - k)/3
Since x must be an integer, and since 20/3 is not an integer, it must be the case that (10 - k)/3 evaluates to be some integer.
In other words, x = (20)(some integer)
So, x must be a multiple of 20. However, 20 is not one of the answer choices.
Then again, we can rewrite x to get: x = (2)(10)(some integer)
This tells us that x must also be a multiple of 10

Answer: E

Cheers,
Brent

BTGmoderatorAT wrote:If x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200, then x must be a multiple of which of the following?

A) 3
B) 6
C) 7
D) 8
E) 10

Is there a strategic approach to this question? Can any experts help?

Since y is a multiple of 5, we can let y = 5k and we have:

3x + 4(5k) = 200

3x = 200 - 20k

3x = 20(10 - k)

x = [20(10 - k)]/3

Since x is an integer, (10 - k) must be divisible by 3. Then, x is the product of 20 times some integer [which is the quotient of (10 - k)/3]. Thus, x must be a multiple of 10. For instance, when k = 1, x = 90, when k = 4, x = 40, and when k = 7, x= 20.

Answer: E