When the integer k is divid by 12, the remainder is 3. Which of the following, when divided by 12, will have a remainder of 6?
I. 2k
II. 6k
III. 4k+6

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

Answer with explaination will be highly appreciated.
OA E

Plug in values dude....

What do you plug in? Try k=15 ( because 15/12 gives you a remainder 3)

I. 2k=30 and remainder is 6 (when divided by 12)
II. 6k=90 and remainder is 6 (when divided by 12)
III. 4k+6=66 and remainder is 6(when divided by 12)

Hence all the three conditions return the same remainder 6.(E)

If k is divided by 12 and leaves remainder as 3,then we can assume values of k to be 3,15,27 and so on..
So,we will take value of k=3..

1.)2k = 6 divided by 12 will always leave remainder as 6.
2.) 6k = 18 when divided by 12 will always leave remainder as 6
3.) 4k+6 = 18 when divided by 12 will always leave remainder as 6

So All above conditions will give the answers.

Hence correct answer is E

the above solutions aren't good enough, actually; in these kinds of problems, you can't just pick one value and watch what happens. in order to be convinced that the pattern is genuine, you would have to pick a succession of numbers that satisfies each of the statements, and make sure that you keep getting the same remainder every time.

so, you should try at least a few values -- 3, 15, 27, 39, etc. -- until you are satisfied that the pattern is going to continue indefinitely.

--

here's a theory-based approach:

if k leaves a remainder of 3 when it is divided by 12, then you can write k = 12n + 3 (where n is an integer).

(i) 2k = 24n + 6.
since 24n is a multiple of 12, this will have an overall remainder of 6 when you divide it by 12.

(ii) 6k = 72n + 18.
= 72n + 12 + 6.
since 72n and 12 are multiples of 12, this will have an overall remainder of 6 when you divide it by 12.

(iii) 4k + 6 = (48n + 12) + 6 = 48n + 18.
= 48n + 12 + 6.
since 48n and 12 are multiples of 12, this will have an overall remainder of 6 when you divide it by 12.

so, all three of them.

I totally understand the point you are making here Ron. But the question does not necessarily specify which of the following options must leave a remainder 6( In which case we need to check for multiple cases) ; rather the stem asks which of the following leaves a remainder 6( Even if 1 plug in satisfies the condition, the stem is satisfied.) In other words, the question is not a must be true type( for all cases).

What do you think?

yeah ok, i see what you're saying here -- the language of this problem does pretty much imply that the remainder will be the same every time.

i don't think that this is actually an official problem, since i don't think the official problems would be this cavalier with their language; i'm pretty sure that any official problem would contain either "must" or "could" in the problem statement.

technically, though, if you DO see a problem like this, you should treat it as a "must" problem.