If $$ (a + b\sqrt{n}) $$ is the positive square root of $$ (29 - 12\sqrt{5}) $$, where $$ a $$ and $$ b $$ are integers, and $$ n $$ is a natural number, then the maximum possible value of $$ (a + b + n) $$ is :

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Accepted Answer

Given the square root of $29 - 12\sqrt{5}$ is $a + b\sqrt{n}$.

$$
\Rightarrow (a + b\sqrt{n})^2 = 29 - 12\sqrt{5}
$$

$$
\Rightarrow a^2 + nb^2 = 29
$$

$$
2ab\sqrt{n} = -12\sqrt{5}
$$

Equating on both sides, we get $n = 5$ and $ab = -6$

$$
a^2 + 5b^2 = 29
$$

$$
9 + 5(4) = 29
$$

The values that will satisfy the above equation, $(a = 3 	ext{ and } b = -2)$ or $(a = -3 	ext{ and } b = 2)$

* Positive root is $-3 + 2\sqrt{5}$
* Negative root is $3 - 2\sqrt{5}$

Now maximum possible value of $(a + b + n)$ in $-3 + 2\sqrt{5} = -3 + 1\sqrt{20}$ (modified to get maximum value of $a + b + n$)

$$
a = -3 \
b = 1 \
n = 20
$$

Now the maximum possible value of $(a + b + n) = -3 + 1 + 20 = 18$

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### Hence, the answer is: **'18'**

**Choice C is the correct answer.**

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