# Easy Number System Solved QuestionAptitude Discussion

 Q. If  $n$  is a positive integer, which one of the following numbers must have a remainder of 3 when divided by any of the numbers 4, 5 and 6?
 ✖ A. $12n + 3$ ✖ B. $24n + 3$ ✖ C. $90n + 2$ ✔ D. $120n + 3$

Solution:
Option(D) is correct

Let m be a number that has a remainder of 3 when divided by any of the numbers 4, 5 and 6.
Then $m - 3$ must be exactly divisible by all three numbers.

Hence, $m - 3$ must be a multiple of the Least Common Multiple of the numbers 4, 5 and 6.
The LCM is $3×4×5 = 60$.

Hence, we can suppose $m - 3 =60p$, where $p$ is a positive integer.

Replacing $p$ with $n$, we get $m – 3 = 60n$.
So, $m = 60n + 3$.

Choice (D) is in the same format 120n + 3 $= 60(2n) + 3$