When dealing with remainders and equations, it is often helpful to translate the words into formulas.

"When x is divided by 2, remainder is 1" --> This means that x is 1 more than a multiple of 2 --> \(x = 2a+1\), where \(a\) is an integer.
"y is divided by 8, remainder is 2" --> This means that y is 2 more than a multiple of 8 --> \(y = 8b+2\), where \(b\) is an integer.

Now if we plug those expressions into 2x + y we get:

\(2x + y = [2(2a+1)] + [8b+2]\)
\(= 4a+2 + 8b+2\)
\(= 4(a+2b+1)\)

So 2x + y = 4*(some integer), and is therefore a multiple of 4. The only answer choice that is a multiple of 4 is C) 12.

Answer: C
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Use some quick logic to eliminate the options:

"When x is divided by 2, remainder is 1" - means x is odd

"y is divided by 8, remainder is 2" - y is not divisible by 8. It could be 2, 10, 18... (all even numbers)

2x + y -> x is odd but 2x will be even. y is even. So 2x+y will be even. Options (B) and (E) are out.

Consider (A) 10 - y can be 2 but then x will be 4 (even); y can be 10 but then x will be 0(even) - Not possible
Consider (C) 12 - y can be 2 then x will be 5 (odd) - Works

Answer (C)
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Hi,

Apart from the method stated above..
lesser error prone way would be
1) if x is divided by 2, the remainder is 1, that is x is odd..
if the same x multiplied by 2, ( 2x), and divided by 4 will give a remainder 2..

now if y divided by 8 leaves a remainder 2, y when divided by 4 will also give a remainder of 2..
so if 2x+y is divided by 4, it should give us a remainder of 2+2=4, meaning 2x+y should be div by 4..
only C fits in
C
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Hi Karishma,

We know 2 is not divisible by 8. But it does not satisfy the condition stated in the question "y is divided by 8 remainder is 2"?
I could not connect why you have chosen y as 2. Can you please help me in understanding it?

Hi Hari,

2 is a valid choice for y because when you divide 2 by 8, the answer is 0 with remainder 2.

In general, any time you divide a smaller integer by a larger integer, the result will be 0 with the remainder equal to the smaller number. If integer \(a\) is smaller than integer \(b\), then \(\frac{a}{b}\) = 0 with remainder \(a\).

I hope that helps!

Cheers,
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I got it in 1 min by:
X=2a+1..i
Y=8b+2...ii

we want 2X+Y
Multiply eqi by 2
2X=4a+2 And we have Y=8b+2
add both
2X+Y=4(a+2b)+4
only c option is divisible by 4..hence the answer.
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I had to go back to the good old number testing method for this
\(2x + y\)
pick numbers for x and y that fulfils the conditions given.
2(3) + 2
6 + 2
8. Not in the option!

raise x to the next level
2(5) + 2
10 + 2
12
That's in the option!
Defnily their CANNOT be more than one possible value for \(2x + y\) in the options.
C.

Hi chetan2u,
Can you explain me property for the highlited portion. I get that when 18/8 reminder is 2 and 18/ 4 reminder is 2 but can you explain me logic behind the same.

Probus
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next to least value of x=5
least value of y=2
2*5+2=12
C

1) If x is divided by 2, the remainder is 1 so X=2k+1..
Then 2x=2(2k+1)=4k+2
So when 4k+2 is divided by 4, remainder is 2

2)if y divided by 8 leaves a remainder 2.
So y = 8k+2..
Thus dividing by 4, (8k+2)/4
8k is divisible by 4, therefore remainder is 2

Thus 2x+4 will leave the same remainder as Remainder when 2x divided by 4 + Remainder when y is divided by 4...=> 2+2=4 that is it remainder is div by 4..

In choices only 12 is div by 4, hence C
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