A number when divided by 7 leaves remainder 3 then which of the following is exactly divisible by 49? ("^" means power)

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.

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