What is the remainder when the positive integer x is divided by 6?

(1) This says that x is a multiple of 3 but not a multiple of 2.
All even multiples of 3 must be divisible by 6, and all odd multiples must be of the form 6n+3. (this is easy to see, consider the multiples - 3,6,9,12,15,18,21,... - they are alternatively even and odd, the even ones are multiples of 6 and the others sit in the middle of two multiples, hence leave remainder 3 each time divided by 6).
Sufficient

(2) x = 12k + 3 = 6(2k) + 3
Hence remainder when divided by 6 is 3
Sufficient

Answer is (d)

Stmnt 1: When x is divided by 2, the remainder is 1; - It means x is odd.
and when x is divided by 3, the remainder is 0. - It means x is divisible by 3

So basically, stmnt 1 tells us that x is an odd multiple of 3 e.g. 3, 9, 15 etc. When you divide these numbers by 6, you will always get 3 as remainder.
Deduce it logically - x i.e. an odd multiple of 3 will always be 3 more than an even multiple of 3 since odd and even multiples of 3 will alternate (e.g. 3 (O), 6(E), 9(O), 12(E), 15(O), 18(E) etc)
Every even multiple of 3 is divisible by 6 so we can say that odd multiples of 3 are always 3 more than multiples of 6. Hence the remainder when you divide x by 6 will always be 3.

Stmnt 2: x is 3 more than a multiple of 12. Since 12 is divisible by 6, x is 3 more than a multiple of 6 too. Hence, remainder is always 3.

Answer (D)

Is there an algebraic approach to this problem? i am always confused whether to take an algebraic approach or number testing approach esp on remainder prblems. I don't want to make this decision in the exam hall. If I want to go in the exam hall with one approach which one it should be for remainder problems? Thanks

Hi,
each Q may have different method to be tackled efficiently ..
But the statements generally give you sufficient info to find the answer..
Algebric way may be better if you are to find the numeric value of remainder, may be in PS..
and working on the info avail in terms of putting values etc in case we are to find if there would be any remainder, but value is not required..

here the info is very straightforward..

What is the remainder when the positive integer x is divided by 6?
1). When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0
it is not div by 2, so will not be div by 6... suff

2). When x is divided by 12, the remainder is 3.
since div by 12 leaves an odd remainder, x is an odd number but 6 is an even number.. again suff

thanks for your reply chetan.
however the question does not ask if x is divisible by 6? it asks what is the remainder when x is divided by 6. St2 is quite straight forward. we can clearly see how the remainder should be 3. But st1 is not as intuitively clear. any thoughts there?

Hi,
sorry , i did not read the Q properly..
Statement 1 is slightly complex and we require to again play with the propperties of number..

three things..
1) 3 has alternate odd and even multiple..
2) we know x is an odd number, since it is not div by 2..
3) from 1 and 2 above x is an odd multiple of 3..
4) Also all even multiple of 3 will be div of 3, so the diff in x and lower even multiple of 3 will be 3, ..
hence remainder is 3

Hello,

I still don't understand why statement (1) is sufficient.

The number 3 fulfills all the requirements of statement (1). Hence, when it is divided by 6, there will be no remainder.
Like already mentioned before all the other numbers e.g. 9, 15, ... will result in a remainder of 3 when it is divided by 6.
My conclusion would be that there will be two possible solutions, either 0 or 3 --> Insufficient

Thank you very much for your help!

3 divided by 6 gives the remainder of 3: 3 = 0*6 + 3.

Reviving an old thread, but I do have a question
Summoning Bunuel and KarishmaB

I know that statement (2) is sufficient since 6 is a factor of 12
Using the format of
\(number = quotient * some multiplier + remainder\):

I know that all of the other numbers satisfy all the 2 constraints from statement (1) and the constraint from the question stem ->
\( x = [9, 15, 21, ......] \)
all of which point to
\(x = 6 * multiplier + 3\),
thus we can conclude that the remainder is 3 for all of these numbers

Everyone agrees that statement (1) is also sufficient, but when I looked at statement 1, however, I'm having doubts especially considering one extreme value; 3.
The number three ticks both of the constraints of the statement (1) ie
\(3 = 2 *1 + 1\) and
\(3 = 3 * 1 + 0\)

Is the remainder of \(\frac{3}{6}\) is also 3? If so, why?

Or maybe generally speaking, how do you calculate the remainder when the numerator is smaller than the denominator ie a fraction?

thanks!

Let me ask you a question: if you had 1 apple and wanted to distribute it evenly in 9 baskets, each basket would get 0 apples and 1 apple would be leftover (remainder).

When a divisor is larger than the dividend, then the remainder is the same as the dividend. For example:

3 divided by 4 yields a remainder of 3: 3 = 4*0 + 3.
9 divided by 14 yields a remainder of 9: 9 = 14*0 + 9.
1 divided by 9 yields a remainder of 1: 1 = 9*0 + 1.

Let's analyze each statement one by one.

Statement 1: When x is divided by 2, the remainder is 1, and when x is divided by 3, the remainder is 0.
From this statement, we know that x is an odd multiple of 2 because the remainder when divided by 2 is 1. Also, x is a multiple of 3 because the remainder when divided by 3 is 0. The smallest positive integer that satisfies these conditions is 3. When 3 is divided by 6, the remainder is 3. Therefore, statement 1 alone is sufficient to answer the question.

Now, let's consider statement 2: When x is divided by 12, the remainder is 3.
Statement 2 tells us that when x is divided by 12, the remainder is 3, telling us directly that the remainder of x when divided by 6 is 3.
In conclusion, statements 1 & 2 are sufficient to answer the question; the right answer is D.

in a way we need to know what is the remainder when X divided by (3*2)
so statement 1 gives information of remainder when x is divided by 3 and 2 ie (3*2) so suffi

statement 2 since 6 is factor of 12, st 2 suffi

D