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Difficulty: 555-605 Levelx   Algebrax   Divisibility/Multiples/Factorsx                        
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]
Either write y in terms of x: y = (200-3x)/(20) then we see that denominator is a multiple of BOTH 2 and 5 = so only (E)10 will fit here,

Or, work backwards and try different numbers and see if the end result is divisible by 5.

Example: if x=6, then we have 2y=91 which will not be divisible by 5...
Eventually, only x=10 will work. Option (E) again
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]
EducationAisle is the following process correct?

3x = 200 - 4y
3x = 4(50-y)/3

So we can infer:
1. 50-y is a multiple of 3
2. 50-y is a multiple of 5
hence from 1. and 2. 50-y is a multiple of 15

So x = 4 x Multiple of 15. This means that x will be composed of at least a 3, 4, and 5. So x will be a multiple of 10 (E)
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]
Expert Reply
Hoozan wrote:
2. 50-y is a multiple of 5

How did you get this?
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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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EducationAisle wrote:
Hoozan wrote:
2. 50-y is a multiple of 5

How did you get this?


50 is a multiply of 5. Y is a multiple of 5. So Multiple of 5 (50) - another multiple of 5 (y) = some multiple of 5
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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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Since y is a multiple of 5 then 4y is also a multiple of 5 and an even multiple of 5 as it has a 4 in it. Thus the unit digit of 4y is 0.

3x = 200-4x

=>3x = 20(0) - ...(0) [Representing the unit digit under the brackets]

=......(0)

=> x is a multiple of 10 as 3 itself is not.

E is the only answer possible.
(option e)

D.S
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]
Why cannot x be a multiple of 3?
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]
Expert Reply
x could be multiple of 3 too.

but ques is asking MUST be true.

x must be multiple of 20 and hence 10.

Akshay004 wrote:
Why cannot x be a multiple of 3?
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]
my doubt is that y can be 0 also.... as 0 is multiple of very number .......in this case no option is correct... please help where iM wrong
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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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kakakakaak wrote:
my doubt is that y can be 0 also.... as 0 is multiple of very number .......in this case no option is correct... please help where iM wrong

The question mentions that x and y are positive integers.
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]
I've always found the plug and chug method to work best for me with similar problems.

In this case, we know that y can be 5, 10, 15... all the way to 50. I'll then just work through the equation as follows with the numbers in the parenthesis representing my "y":

3x + 4(5) = 200 --- x = 60
3x + 4(10) = 200 --- x is not an integer
3x + 4(15) = 200 --- x is not an integer
3x + 4(20) = 200 --- x = 40

Looking at 40 and 60, each are not individually divisible by 3, 6, 7, and 8. However, they are each individually divisible by 10. Answer choice E.
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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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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: If x and y are positive integers such that y is a multiple of 5 and 3x [#permalink]
For 3x+4y=200 given that y is already a multiple of 5, 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]
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