Problem 654

The smallest five positive integers, each having a divisor of
$d$, are $d$, $2d$, $3d$, $4d$, and $5d$.

Thus, the smallest possible sum of five different positive integers
whose greatest common divisor is $d$
is $d+2d+3d+4d+5d=15d$.

We know that the sum is at least $15d$ and is equal to $264$, which means that $15d\leq264$, and so $d\leq\frac{264}{15}$ or $d\leq17.6$.

Each of the five integers is divisible by $d$, and so the sum of the five integers,
$264$, is divisible by $d$.

Thus, we want the largest possible divisor of $264$ that is less than or equal to $17$.

Since $264=2^3\times3\times11$, the
divisors of $264$ that are less than
or equal to $17$ are: $1$, $2$, $3$, $4$, $6$, $8$, $11$, and $12$, and so the largest possible value of
$d$ is $12$.

The sum of the digits of the largest possible value of $d$ is $1+2=3$.

(We note that $12$, $24$, $36$, $48$, and $144$ are five such integers whose greatest
common divisor is $12$ and whose sum
is $264$.)