A three-digit positive integer $n$ has digits $a b c$. (That is, $a$ is the hundreds digit of $n, b$ is the tens digit of $n$, and $c$ is the ones (units) digit of $n$.) Determine the largest possible value of $n$ for which

- $a$ is divisible by 2 ,
- the two-digit integer $a b$ (that, $a$ is the tens digit and $b$ is the ones (units) digit) is divisible by 3 but is not divisible by 6 , and
- $n$ is divisible by 5 but is not divisible by 7 .
To make $n$ as large as possible, we make each of the digits $a, b, c$ as large as possible, starting with $a$.

Since $a$ is divisible by 2 , its largest possible value is $a=8$, so we try $a=8$.

Consider the two digit integer $8 b$.

This integer is a multiple of 3 exactly when $b=1,4,7$.

We note that 84 is divisible by 6 , but 81 and 87 are not.

To make $n$ as large possible, we try $b=7$, which makes $n=87 c$.

For $n=87 c$ to be divisible by 5 , it must be the case that $c=0$ or $c=5$.

But $875=7 \cdot 125$ so 875 is divisible by 7 .

Therefore, for $n$ to be divisible by 5 and not by 7 , we choose $c=0$.

Thus, the largest integer $n$ that satisfies the given conditions is $n=870$.

ANSWER: 870