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divisibility

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by kanha81 » Tue May 26, 2009 3:55 pm
When the integer k is divided 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

How to attack such problems?

The way I try to resolve is: q- quotient, R-remainder
k = 12q + R = 12q + 3 ..........(1)

and then I get lost :oops:
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Re: divisibility

by dtweah » Tue May 26, 2009 4:08 pm
kanha81 wrote:When the integer k is divided 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

How to attack such problems?

The way I try to resolve is: q- quotient, R-remainder
k = 12q + R = 12q + 3 ..........(1)

and then I get lost :oops:
You are on to something. What is 2K?

2 (12q+3)=24q+6, remainder 6 when divided by 12

6(12q+3)= 6 x 12q + 18 remainder 6 when divided by 12

4 (12q+3) +6= 4 (12q) +18, again remaider 6

Choose E.
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Re: divisibility

by kanha81 » Tue May 26, 2009 4:43 pm
dtweah wrote:
You are on to something. What is 2K?
You must be right. How could I not see something that simple? :evil:
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