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

Since our target expression $29-12\sqrt{5}$ contains a negative coefficient with the square root term, we need to be strategic about our approach.

Key Insight: If we try $(a+b\sqrt{n})^2$ with positive $b$, we'd get a positive coefficient for the $\sqrt{n}$ term. But we need a negative coefficient!

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Let's expand $(p-q\sqrt{5})^2$:

$(p-q\sqrt{5})^2 = p^2 + (q\sqrt{5})^2 - 2 \cdot p \cdot q\sqrt{5}$

$= p^2 + q^2 \cdot 5 - 2pq\sqrt{5}$

$= p^2 + 5q^2 - 2pq\sqrt{5}$

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We need this to equal $29-12\sqrt{5}$:

$p^2 + 5q^2 - 2pq\sqrt{5} = 29 - 12\sqrt{5}$

Comparing the rational parts: $p^2 + 5q^2 = 29$

Comparing the irrational parts: $-2pq = -12$, so $pq = 6$

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From $pq = 6$, the possible positive integer pairs are:

$(p,q) = (1,6), (2,3), (3,2), (6,1)$

Let's check which satisfies $p^2 + 5q^2 = 29$:

$(1,6)$: $1 + 5(36) = 1 + 180 = 181 \neq 29$

$(2,3)$: $4 + 5(9) = 4 + 45 = 49 \neq 29$

$(3,2)$: $9 + 5(4) = 9 + 20 = 29$ This works!

$(6,1)$: $36 + 5(1) = 36 + 5 = 41 \neq 29$

Only $(p,q) = (3,2)$ works!

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We found that $(3-2\sqrt{5})^2 = 29-12\sqrt{5}$

But let's check if $3-2\sqrt{5}$ is positive or negative:

Since $\sqrt{5} \approx 2.236$, we have $2\sqrt{5} \approx 4.472 > 3$

So $3-2\sqrt{5} < 0$, which means it's negative!

Since we need the positive square root, we take:

$\sqrt{29-12\sqrt{5}} = -(3-2\sqrt{5}) = 2\sqrt{5}-3$

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We need to express $2\sqrt{5}-3$ in the form $a+b\sqrt{n}$:

$2\sqrt{5}-3 = -3+2\sqrt{5}$

This gives us: $a = -3$, $b = 2$, $n = 5$

So $a+b+n = -3+2+5 = 4$

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The question asks for the maximum possible value, so let's see if there are other valid representations.

Notice that $2\sqrt{5} = \sqrt{4 \cdot 5} = \sqrt{20}$

So we can also write: $2\sqrt{5}-3 = -3+\sqrt{20}$

This gives us: $a = -3$, $b = 1$, $n = 20$

So $a+b+n = -3+1+20 = 18$

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Comparing our two valid forms:

$a = -3, b = 2, n = 5$: $a+b+n = 4$

$a = -3, b = 1, n = 20$: $a+b+n = 18$

Therefore, the maximum possible value of $(a+b+n)$ is 18.