To solve this question, we need to find the number which is exactly divisible by 49.
Given that a number when divided by 7 leaves a remainder of 3, we can express this as:
[n \equiv 3 \pmod{7}]
This means that the number can be written in the form:
[n = 7k + 3]
where k is an integer.
Now, let's substitute this expression for n into each of the given options and check which one is exactly divisible by 49.
Option A) ((n^2)-2)
Substituting (n = 7k + 3) into this option, we get:
((7k + 3)^2 - 2)
Simplifying this expression, we get:
(49k^2 + 42k + 7)
This expression is not divisible by 49.
Option B) ((n^2)-3)
Substituting (n = 7k + 3) into this option, we get:
((7k + 3)^2 - 3)
Simplifying this expression, we get:
(49k^2 + 42k + 6)
This expression is not divisible by 49.
Option C) ((n-2)^2)
Substituting (n = 7k + 3) into this option, we get:
((7k + 3 - 2)^2)
Simplifying this expression, we get:
((7k + 1)^2)
This expression is not divisible by 49.
Option D) ((n-3)^2)
Substituting (n = 7k + 3) into this option, we get:
((7k + 3 - 3)^2)
Simplifying this expression, we get:
((7k)^2)
This expression is divisible by 49.
Option E) ((n-4)^2)
Substituting (n = 7k + 3) into this option, we get:
((7k + 3 - 4)^2)
Simplifying this expression, we get:
((7k - 1)^2)
This expression is not divisible by 49.
Therefore, the correct answer is Option D) ((n-3)^2) because it is exactly divisible by 49.