The positive integer $d$ has the property that each of 468,636, and 867 has the same remainder, $r$, when divided by $d$. What is the largest possible value of $d+r$ ?
The remainder when 468 is divided by $d$ is $r$. This means that $0 \leq r<d$ and there is a unique integer $q_{1}$ such that $468=q_{1} d+r$.

Similarly, there are unique integers $q_{2}$ and $q_{3}$ such that $636=q_{2} d+r$ and $867=q_{3} d+r$.

This gives three equations

$$
\begin{aligned}
& 468=q_{1} d+r \\
& 636=q_{2} d+r \\
& 867=q_{3} d+r
\end{aligned}
$$

Subtracting the first equation from the second, the first from the third, and the second from the third gives the three equations

$$
\begin{aligned}
& 168=\left(q_{2}-q_{1}\right) d \\
& 399=\left(q_{3}-q_{1}\right) d \\
& 231=\left(q_{3}-q_{2}\right) d
\end{aligned}
$$

Since $q_{1}, q_{2}$, and $q_{3}$ are integers, so are $q_{2}-q_{1}, q_{3}-q_{1}$, and $q_{3}-q_{2}$. Therefore, $d$ must be a divisor of 168, 399, and 231.

Since $d$ is a divisor of 168 and 231, it is a factor of $231-168=63$.

Since $d$ is a divisor of 63 and 168 , it is a factor of $168-2 \times 63=42$.

Since $d$ is a divisor of 63 and 42 , it is a factor of $63-42=21$.

Now note that $168=8 \times 21,399=19 \times 21$, and $231=11 \times 21$, so 21 is a divisorof all three integers.

We have shown that $d$ must be a divisor of 21 and that 21 is a divisor of all three integers. This implies that the possible values of $d$ are the positive divisors of 21 , which are $1,3,7$, and 21 .

If $d=1$, then the remainder is 0 when each of 468,636, and 867 is divided by $d$. In this case, $d+r=1+0=1$.

If $d=3$, then the remainder is 0 when each of 468,636, and 867 is divided by $d$. In this case, $d+r=3+0=3$.

If $d=7$, then the remainder is 6 when each of 468, 636, and 867 is divided by $d$. In this case, $d+r=7+6=13$.

If $d=21$, then the remainder is 6 when each of 468,636, and 867 is divided by $d$. In this case, $d+r=21+6=27$.

ANSWER: 27

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