x1 and x2 are each positive integers. When x1

Question

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?

I. y is even
II. y is odd
III. y is divisible by 3

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only

nick1816 · Answer

x_1 = 3a+1 
 x_2= 12b+4

y= 6a+2+12b+4= 6(a+2b+1)

D

globaldesi · Answer

x1 = 6k1+2
x2 = 12k2+4
2x1+x2 = 6k1+12k2+6
for variable it will be divisible by 3
since very part is divisible by 3
also all area multiple of even numbers of a even number itself
the sum will be even.

Thus D

garcmillan · Answer

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

Loser94 · Answer

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

MathRevolution · Answer

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

Paras96 · Answer

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

kpop1234567890 · Answer

KarishmaB 
 GMATNinja 
 MartyMurray 
 GMATCoachBen 
 ScottTargetTestPrep 
 MartyTargetTestPrep 
 bb 
 Bunuel

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

ScottTargetTestPrep · Answer

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.

kpop1234567890 · 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.

ScottTargetTestPrep · Answer

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)

kpop1234567890 · Answer

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?

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

ScottTargetTestPrep · Answer

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.

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x1 and x2 are each positive integers. When x1

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