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If x and y are positive integers such that y is a multiple of 5 and 3x

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Intern
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B
Joined: 29 Jan 2017
Posts: 46
Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]

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New post 09 Feb 2018, 16:09
Can someone pls verify if this method will work for all similar question types? This is the way I approached as well

wvu wrote:
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.
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Joined: 31 Jul 2017
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]

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New post 10 Feb 2018, 00:37
mrdlee23 wrote:
Can someone pls verify if this method will work for all similar question types? This is the way I approached as well

wvu wrote:
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.


We know that -
\(3x + 4y = 200, y = 5m\)
\(3x + 20m = 200\)
\(x = \frac{10(20 - m)}{3}\)

So, x has to be multiple of 10.
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]

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New post 12 Feb 2018, 03:38
mrdlee23 wrote:
Can someone pls verify if this method will work for all similar question types? This is the way I approached as well

wvu wrote:
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.


Since y is a factor of 5, 4y will end in 0.
So get the sum of 200, 3x should end in 0 as well. So x should be a multiple of 10.
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x   [#permalink] 12 Feb 2018, 03:38

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