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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

Re: If x and y are positive integers such that y is a multiple of 5 and 3x
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14 Jun 2016, 22:25

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its given y is multiple of 5 therefore 4y is even and will end in zero as even multiple of 5 ends in zero(zero at units place) now equation 3x + 4y =200, as we know 4y ends in zero (zero at units place ) we need to add something that ends in zero (zero at units place) to get to 200,3x can only end in zero if x is multiple of 10

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28 Nov 2016, 10:32

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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

---------------------- I drowned in these numbers and failed on my first attempt at this question, but I've got it now.

The most simple approach I can think of is to plug in values of y that are multiples of 5.

Re: If x and y are positive integers such that y is a multiple of 5 and 3x
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02 Dec 2016, 10:45

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AbdurRakib 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

We are given that x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200. We need to determine which answer choice MUST be a multiple of x.

Because y is a multiple of 5, we can represent y as 5k, in which k is an integer, and we have:

3x + 4(5k) = 200

3x + 20k = 200

Since the sum of 3x and 20k is 200 (or a number that has a units digit of zero), and since 20k has a units digit of zero, 3x must also contain a units digit of zero. The only way for 3x to contain a units digit of zero is if it’s multiplied by a multiple of 10. Thus, x must be a multiple of 10.

Re: If x and y are positive integers such that y is a multiple of 5 and 3x
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25 Mar 2017, 06:08

We know: 1. x and y are positive integers 2. y is a multiple of 5 3. 3x + 4y = 200

3x + 4y = 200

=> 3x = 200 - 4y

It's known that if we add or subtract multiples of N, the result is a multiple of N. Here, both 200 and 4y are multiple of 10(20 actually). So 3x is a multiple of 10. And therefore x is a multiple of 10.

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28 Apr 2017, 10:49

I approached this in a different style I think than the other posters - is this way correct as well?

First, 200 factors down to \(2^3 * 5^2\)

Looking at the information we know that from the equation \(3x + 4y\), where y is a multiple of 5, we have the factors of \(2^2\) (from the 4), 3, and 5. Given this, we can then determine that we need one more factor of both 2 and 5. E - 10 is the only answer choice that provides those two factors.

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

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14 Sep 2017, 23:07

3x + 4y = 200 y is a multiple of 5 so go for the trail and error 1. 3x+4*5 = 200 3x = 180, it is mutiple of both 3 and 10 2. 3x+4*(5*2) = 200 3x = 160, 3 can be eliminated so answer should be 10 E
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x
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22 Oct 2017, 17:59

I think the trick here is to use unit digit concept. 3X+20K=200 means 3X must have 0 as unit digit. from the answer choices, only E works.

Thanks
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