Given that The remainder when m is divided by 5 is 2, and the remainder when m is divided by 3 is 2 and m > 2 and we need to find the least possible value of m

Let's solve the problem with two methods:

Method 1: Substitution

Let's take each option choice and see which one satisfies the question

(A) 5

=> 5 when divided by 5 gives 0 remainder. But we needed 2 as remainder => NOT POSSIBLE

(B) 7

=> 7 when divided by 5 gives 2 remainder. => TRUE
=> 7 when divided by 3 gives 1 remainder. But we needed 2 as remainder => NOT POSSIBLE

(C) 15

=> 15 when divided by 5 gives 0 remainder. But we needed 2 as remainder => NOT POSSIBLE

(D) 17

=> 17 when divided by 5 gives 2 remainder. => TRUE
=> 17 when divided by 3 gives 2 remainder. POSSIBLE
In Test, we don't need to solve further, but I am solving to complete the solution

(E) 27

=> 27 when divided by 5 gives 2 remainder. => TRUE
=> 27 when divided by 3 gives 0 remainder. But we needed 2 as remainder => NOT POSSIBLE

So, Answer will be D

Method 2: Algebra

The remainder when m is divided by 5 is 2

Theory: Dividend = Divisor*Quotient + Remainder

m -> Dividend
5 -> Divisor
a -> Quotient (Assume)
2 -> Remainders
=> m = 5*a + 2 = 5a + 2 ...(1)

The remainder when m is divided by 3 is 2

=> Let quotient be b
=> m = 3b + 2 ..(2)

From 1 and 2 we get that when divided by 5 and 3 gives 2 as remainder
=> m = a multiple of LCM(5,3) + 2
=> m = 15c + 2 (where c is an integer)

Option choice 17 satisfies this condition as 17 = 15 + 2

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Remainders