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555-605 Level|   Algebra|   Multiples and Factors|                        
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If x and y are positive integers such that y is a multiple of 5 and 3x 12 May 2021, 00:51
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AbdurRakib
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

OG 2017 New Question
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Answer: Option E

Video solution by GMATinsight

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If x and y are positive integers such that y is a multiple of 5 and 3x 15 Dec 2021, 04:52
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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)

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If x and y are positive integers such that y is a multiple of 5 and 3x 28 Jul 2022, 14:46
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Video solution from Quant Reasoning:
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If x and y are positive integers such that y is a multiple of 5 and 3x 01 Mar 2025, 06:25
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KarishmaB , based on the laws of even and odd, we can conclude that x must be even. But how are you ruling out 6 and 8 immediately?
KarishmaB
AbdurRakib
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

OG 2017 New Question

If y is a multiple of 5, 4y must be a multiple of 5 too.

3x = 200 - 4y = Multiple of 5 - Multiple of 5 = Multiple of 5

Since 3 is not a multiple of 5, x MUST be a multiple of 5.

Also, 4y is even. So 3x = Even - Even = Even
Since 3 is not even, x must be even.

Hence x must be a multiple of 10.

Answer (E)

Check out this discussion on numbers and their factors: https://youtu.be/DxIH8rjhpKY
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If x and y are positive integers such that y is a multiple of 5 and 3x 01 Mar 2025, 06:53
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INprimesItrust
KarishmaB , based on the laws of even and odd, we can conclude that x must be even. But how are you ruling out 6 and 8 immediately?
KarishmaB
AbdurRakib
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

OG 2017 New Question

If y is a multiple of 5, 4y must be a multiple of 5 too.

3x = 200 - 4y = Multiple of 5 - Multiple of 5 = Multiple of 5

Since 3 is not a multiple of 5, x MUST be a multiple of 5.

Also, 4y is even. So 3x = Even - Even = Even
Since 3 is not even, x must be even.

Hence x must be a multiple of 10.

Answer (E)

Check out this discussion on numbers and their factors: https://youtu.be/DxIH8rjhpKY
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Since x MUST be a multiple of 5, and 6 and 8 aren't.
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If x and y are positive integers such that y is a multiple of 5 and 3x 03 Mar 2025, 00:39
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INprimesItrust
KarishmaB , based on the laws of even and odd, we can conclude that x must be even. But how are you ruling out 6 and 8 immediately?
KarishmaB
AbdurRakib
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

OG 2017 New Question

If y is a multiple of 5, 4y must be a multiple of 5 too.

3x = 200 - 4y = Multiple of 5 - Multiple of 5 = Multiple of 5

Since 3 is not a multiple of 5, x MUST be a multiple of 5.

Also, 4y is even. So 3x = Even - Even = Even
Since 3 is not even, x must be even.

Hence x must be a multiple of 10.

Answer (E)

Check out this discussion on numbers and their factors: https://youtu.be/DxIH8rjhpKY
Show more

Look at it this way:

Given: 3x + 4y = 200
y is a multiple of 5 so let's say y = 5a (where a is an integer)

We get: 3x + 4*5a = 200
3x = 200 - 20a
3x = 20(10 - a)

Now, from where do we get this factor of 20? It has to come from x since 3 does not have 20 as a factor. This means x MUST be a multiple of 20 and hence 10 must be a factor of x.

Nothing can be said about the other options.

Answer (E)
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If x and y are positive integers such that y is a multiple of 5 and 3x 05 Mar 2025, 01:30
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You don't even need math. Simply put, a multiple of 5 will end with a 5 or a 0, lets says it ends with a 0, 3x (given the options) will no way get you to the unit digit of zero unless x = is a multiple of 10, hence. E
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If x and y are positive integers such that y is a multiple of 5 and 3x 31 Mar 2025, 05:42
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Bunuel, karishma

after arriving at x = 20 (10-a) / 3

For x to be an integer ---> 10-a has to be divisible by 3.

Is it necessary to plug in values for a and find the pattern? I did for a = 0,1,2,3 etc.

only a= 1 and a= 4 fit the criterion : 20*9/3 = 60 , 20*6/3 = 40

so only 10 fits.

this is cumbersome, thanks!
KarishmaB
AbdurRakib
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

OG 2017 New Question

If y is a multiple of 5, 4y must be a multiple of 5 too.

3x = 200 - 4y = Multiple of 5 - Multiple of 5 = Multiple of 5

Since 3 is not a multiple of 5, x MUST be a multiple of 5.

Also, 4y is even. So 3x = Even - Even = Even
Since 3 is not even, x must be even.

Hence x must be a multiple of 10.

Answer (E)

Check out this discussion on numbers and their factors: https://youtu.be/DxIH8rjhpKY

Method 2:

Given: 3x + 4y = 200
y is a multiple of 5 so let's say y = 5a (where a is an integer)

We get: 3x + 4*5a = 200
3x = 200 - 20a
3x = 20(10 - a)

Now, from where do we get this factor of 20? It has to come from x since 3 does not have 20 as a factor. This means x MUST be a multiple of 20 and hence 10 must be a factor of x.

Nothing can be said about the other options.

Answer (E)
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If x and y are positive integers such that y is a multiple of 5 and 3x 31 Mar 2025, 07:16
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INprimesItrust

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

Bunuel, karishma

after arriving at x = 20 (10-a) / 3

For x to be an integer ---> 10-a has to be divisible by 3.

Is it necessary to plug in values for a and find the pattern? I did for a = 0,1,2,3 etc.

only a= 1 and a= 4 fit the criterion : 20*9/3 = 60 , 20*6/3 = 40

so only 10 fits.

this is cumbersome, thanks!
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You don't really need any of that. We have that y = 5a, so we'd get 3x + 20a = 200. Both 20a and 200 are multiples of 10, so 3x must also be a multiple of 10, which makes x a multiple of 10. Only E fits.

Answer: A.
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