x1 and x2 are each positive integers. When x1

First thought when reading the question:

x1 could be 1, 4, 7, 10...x1 could be even/odd

x2 could be 4, 16, 28, 40... -> x2 is always even

what can we coclude about y = 2*x1 + x2 -> first term is always even as it is multiplied by an even number (in a multiplication, if one term is even, then the result will be even regardless of the other terms) / second term is always even as stated above

THEREFORE y will ALWAYS be EVEN -> I is true, II is not

Regarding III, we know that neither x1 nor x2 are divisible by 3. Therefore, the sum COULD BE OR NOT divisible by 3, as the general rule is:

term div by 3 + term div by 3 -> Result div by 3 (ALWAYS)
term div by 3 + term NOT div by 3 -> Result NOT div by 3 (ALWAYS)
term NOT div by 3 + term NOT div by 3 -> Result could be div by 3 or not -> this is the case we have

Therefore, we need to test it with two numbers

For instance,

x1=4
x2=16

y = 2 * 4 + 16 = 24 -> DIV by 3

since this result is div by 3, we know that y will always be divisible by 3, as the increases in x1 and x2 are always multiple of 3 (increases of 3 for x1 and increases of 12 (4*3) for x2) -> III is true

ANSWER D

X1=3n+1
X2=12m+4

y=2*(3n+1)+(12m+4)
=6n+2+12m+4=6(n+1+2m)
Means Y is Even and is divisible by 6

Correct option is D

\(x_1\) is divided by 3, the remainder is 1 => \(x_1\) = 3 p + 1

\(x_2\) is divided by 12, the remainder is 4 => \(x_2\) = 12 q + 4

y = 2*\(x_1\) + \(x_2\)

=> y = 2(3 p + 1) + 12 q + 4
=> y = 6p + 2 + 12 q + 4
=> y = 6p + 12q + 6
=> y = 6( p + 2q + 1)

Multiple of 6 will always be even and will be divisible by 3.

The condition I and Condition III only.

Answer D

x1 and x2 are each positive integers. When x1 is divided by 3, the remainder is 1, and when x2 is divided by 12, the remainder is 4. Let y = 2*x1 + x2. What can you conclude about y?

x1 = 3p + 1 (p is a non-negative integer)
x2 = 12m + 4 ( m is a non-negative integer)

=>y = 2*x1 + x2
=>y = 2*(3p + 1) + 12m + 4
=>y = 6p + 2 + 12m + 4
=> y = 6p + 12m + 6 = 6 (p+2m+1) = 6k

Therefore I & III
Option D

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So i need help in figuring out what I have done wrong. I know the right solution but want to understand where i went wrong. Knowing wrong patterns is important hence asking.

I did this in 2 wrong ways
x1 = 3q + 1
x2 = 12y + 4

So
x2 = 3*4y + 4 = 3m + 4 - Wrong method 1
x2 = 3*4y + 3 + 1
= 3(4y+1)+1
= 3m + 1 - wrong method 2

Using method 2
Now y = 2(3q + 1) + 3m + 1
= 3(2q+m+2)

Using Method 1
y = 2(3q+1) + 3m + 4
= 3(6q+3+2)

In both methods cant prove that y us even

I believe your mistake was using y as your variable above (ie, x2 = 12y + 4), since we are told in the question stem that y = 2*x1 + x2.
Had you used a different variable, like p, to get x2 = 12p + 4, then you likely would have reached the correct answer.

I will arrive at same error despite not taking y in x2. Doesn't still help for me to understand what u did wrong.

Another issue occurs here:
x2 = 12y + 4
So
x2 = 3*4y + 4 = 3m + 4 - Wrong method 1

To get 3m + 4, you divided 3*4y + 4 by 4, but in an equation, you must divide BOTH sides by 4.
We get: x2/4 = (12y + 4)/4 which simplifies to: x2/4 = 3y + 1 (note: +1 not +4)

I actually didnt divide by 4 - I just rewrote 12y as 3*4y implying that from initial quotient of y we not have 4y as quotient. However the glaring mistake in my method which now became apparent to me is that in method remainder 4 is more than divisor 3 - hence the remainder will be 1 - this makes method 1 wrong.

Is the method 2 wrong because quotient cant be of the form 4q+1 the way I have written?

Here's part of your method 2:

Using method 2
Now y = 2(3q + 1) + 3m + 1
= 3(2q+m+2)

We can't say x2 = 3m + 1
we CAN say x2 = 12p + 4, but x2 = 3m + 1 is incorrect.