Numbers Q12

12.	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

Answer – (D)

Solution:

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