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Re: If x and y are different positive integers, which of the [#permalink]
I don't see why B is wrong.
X = 3
Y = 8

2x = 6
6/8 = 3/4 ===> Reminder 3.
Makes the second condition correct.

GreenlightTestPrep wrote:
GreenlightTestPrep wrote:
If x and y are different positive integers, which of the following COULD be true:
Select all that apply.

A) When x is divided by y, the remainder is x
B) When 2x is divided by y, the remainder is x
C) When x+y is divided by x , the remainder is x-y


A) When x is divided by y, the remainder is x
This occurs any time x < y
For example, if x = 5 and y = 7, then statement A becomes: When 5 is divided by 7, the remainder is 5
So TRUE!


-----------------------------------------------------
B) When 2x is divided by y, the remainder is x
Nice rule: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

So, from statement B, we can say: some possible values of 2x are: x, x + y, x + 2y, x + 3y, . . . etc
Let's examine the first option: 2x = x. Solve to get x = 0, but we're told x is POSITIVE No good.

Check the second option: 2x = x + y. Solve to get x = y. This means the remainder is y (aka x), but the remainder CANNOT be greater than the divisor. See the rule below:

When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0

Check the third option: 2x = x + 2y. Solve to get x = 2y.
This means the remainder = 2y, which means the remainder is greater than the divisor (see rule above). No good.

In fact, we can see that, with all of the possible values of 2x, the remainder will be greater than the divisor.
So, statement B is NOT TRUE

-----------------------------------------------------
C) When x+y is divided by x, the remainder is x-y
Some possible values of x+y are: (x-y), (x-y)+x, (x-y)+2x, (x-y)+3x, . . . etc
Let's examine the first option: x+y = x-y
Solve to get y = 0. No good.

Check the second option: x+y = (x-y)+x
Simplify: x+y = 2x - y
Solve to get: x = 2y

So, one possible case is: x = 6 and y = 3
Statement C becomes: When (6 + 3) is divided by 6, the remainder is 3
So TRUE!
------------------------------

Answer: A, C

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Re: If x and y are different positive integers, which of the [#permalink]
2
Paul121 wrote:
I don't see why B is wrong.
X = 3
Y = 8

2x = 6
6/8 = 3/4 ===> Reminder 3.
Makes the second condition correct.


B) When 2x is divided by y, the remainder is x
With your numbers, 2x = 6 and y = 8
So, 2x divided by 8 becomes 6 divided by 8, with equals 0 with remainder 6 (not remainder 3)
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Re: If x and y are different positive integers, which of the [#permalink]
GreenlightTestPrep wrote:
Paul121 wrote:
I don't see why B is wrong.
X = 3
Y = 8

2x = 6
6/8 = 3/4 ===> Reminder 3.
Makes the second condition correct.


B) When 2x is divided by y, the remainder is x
With your numbers, 2x = 6 and y = 8
So, 2x divided by 8 becomes 6 divided by 8, with equals 0 with remainder 6 (not remainder 3)


What if X = 5 and Y= 2
5 * 2 = 10; 10/2 = 5.
Does this not make B true under these conditions?
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Re: If x and y are different positive integers, which of the [#permalink]
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