If $4^n+15 n+P$ is divisible by 9 for all $n \in N$, then the least negative integral value of $P$ is

Let $f(n) = 4^n + 15n + P$. We are given that $f(n)$ is divisible by 9 for all $n \in N$.

For $n=1$, $f(1) = 4^1 + 15(1) + P = 19 + P$. Since $f(1)$ is divisible by 9, $19 + P = 9k$ for some integer $k$.

For $n=2$, $f(2) = 4^2 + 15(2) + P = 16 + 30 + P = 46 + P$. Since $f(2)$ is divisible by 9, $46 + P = 9m$ for some integer $m$.

Subtracting the two expressions: $(46 + P) - (19 + P) = 27$, which is divisible by 9. This confirms the condition is consistent for $n=1$ and $n=2$.

From $19 + P = 9k$, we have $P = 9k - 19$.

To find the least negative integral value of $P$, we test values of $k$:
If $k=1$, $P = 9 - 19 = -10$.
If $k=2$, $P = 18 - 19 = -1$.

Checking $P = -1$: $f(n) = 4^n + 15n - 1$.
Using modular arithmetic, $4^n + 15n - 1 \equiv 1^n + 6n - 1 \pmod{3}$ is not sufficient. Let us check modulo 9.
$4^n + 15n - 1 \equiv 4^n + 6n - 1 \pmod{9}$.
For $n=1$, $4+6-1 = 9 \equiv 0 \pmod{9}$.
For $n=2$, $16+12-1 = 27 \equiv 0 \pmod{9}$.
For $n=3$, $64+18-1 = 81 \equiv 0 \pmod{9}$.
Since $P = -1$ satisfies the condition, and it is greater than $-10$, $-1$ is the least negative integer satisfying the condition.

Answer: -1