Given that x is a positive integer such that 19 divided by x has a remainder of 3 and we need to find the sum of all the possible values of x

Theory: Any number when divided by n gives remainder from 0 to n-1

Since we have a remainder of 3 so the divisor should be more than 3
=> divisor ≥ 4

Let's solve the problem using two methods

Method 1: Algebra

19 divided by x gives 3 remainder

Theory: Dividend = Divisor*Quotient + Remainder

19 -> Dividend
x -> Divisor
a -> Quotient (Assume. Also a is an integer)
5 -> Remainders
=> 19 = x*a + 3 = xa + 3
=> ax = 19-3 = 16

We can take values of x greater than 3 which satisfies this and gives a as an integer
x = 4, a = 4 (possible)
x = 8, a = 2 (possible)
x = 16, a = 1 (possible)

Sum of possible values of x = 4 + 8 + 16 = 28

So, Answer will be D

Method 2: Taking Numbers

Let's take numbers from 4 to 19 and see which one gives 3 remainder

19 divided by 4 gives 3 remainder => POSSBILE
19 divided by 5 gives 4 remainder => NOT POSSIBLE
19 divided by 6 gives 1 remainder => NOT POSSIBLE
19 divided by 7 gives 5 remainder => NOT POSSIBLE
19 divided by 8 gives 3 remainder => POSSBILE
19 divided by 9 gives 1 remainder => NOT POSSIBLE
19 divided by 10 gives 9 remainder => NOT POSSIBLE
19 divided by 11 gives 8 remainder => NOT POSSIBLE
19 divided by 12 gives 7 remainder => NOT POSSIBLE
19 divided by 13 gives 6 remainder => NOT POSSIBLE
19 divided by 14 gives 5 remainder => NOT POSSIBLE
19 divided by 15 gives 4 remainder => NOT POSSIBLE
19 divided by 16 gives 3 remainder => POSSBILE
19 divided by 17 gives 2 remainder => NOT POSSIBLE
19 divided by 18 gives 1 remainder => NOT POSSIBLE
19 divided by 19 gives 0 remainder => NOT POSSIBLE

So, 4 + 8 + 16 = 28

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Remainders

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