Problem 578

Since the mean of the first three rolls is $3$, then the sum of the first three rolls
is $3\times3=9$. The results of the
4th, 5th and 6th rolls are $a$, $b$ and $c$, and so the mean of all six rolls is
$\dfrac{9+a+b+c}{6}$. For this mean
to be an integer, $9+a+b+c$ must be a
multiple of $6$.

Each roll is a positive integer between $1$ and $6$ inclusive, and so $a+b+c$ is at least $1+1+1=3$ and at most $6+6+6=18$.

Therefore, $9+a+b+c$ is at least
$9+3=12$ and at most $9+18=27$.

The multiples of $6$ between $12$ and $27$ inclusive are $12$, $18$ and $24$, and so the mean is an integer exactly
when $a+b+c=3$ or $a+b+c=9$ or $a+b+c=15$.

Next, we count the number of ordered triples $(a,b,c)$ for each of these $3$ cases.

Case 1: $a+b+c=3$.

There is exactly $1$ ordered
triple, $(a,b,c)=(1,1,1)$, in this
case.

Case 2: $a+b+c=9$.

If $a=6$, then $b+c=3$ and so $(b,c)=(1,2)$ or $(b,c)=(2,1)$.

If $a=5$, then $b+c=4$ and so $(b,c)=(1,3)$ or $(b,c)=(2,2)$ or $(b,c)=(3,1)$.

We continue in this way and summarize the results in the table that
follows.

Value of $a$
Value of $b+c$
Possible ordered pairs $(a,b)$
Number of ordered triples

$6$
$3$
$(1,2)$, $(2,1)$
$2$

$5$
$4$
$(1,3)$, $(2,2)$, $(3,1)$
$3$

$4$
$5$
$(1,4)$, $(2,3)$, $(3,2)$, $(4,1)$
$4$

$3$
$6$
$(1,5)$, $(2,4)$, $(3,3)$, $(4,2)$, $(5,1)$
$5$

$2$
$7$
$(1,6)$, $(2,5)$, $(3,4)$, $(4,3)$, $(5,2)$, $(6,1)$
$6$

$1$
$8$
$(2,6)$, $(3,5)$, $(4,4)$, $(5,3)$, $(6,2)$
$5$

In total, there are $2+3+4+5+6+5=25$ such ordered triples in
this case.

Case 3: $a+b+c=15$.

We count in a manner similar to that in Case 2, and summarize the
results in the table that follows.

Value of $a$
Value of $b+c$
Possible ordered pairs $(a,b)$
Number of ordered triples

$6$
$9$
$(3,6)$, $(4,5)$, $(5,4)$, $(6,3)$
$4$

$5$
$10$
$(4,6)$, $(5,5)$, $(6,4)$
$3$

$4$
$11$
$(5,6)$, $(6,5)$
$2$

$3$
$12$
$(6,6)$
$1$

If $a<3$, then $b+c>12$ which is not possible.

In total, there are $4+3+2+1=10$ such
ordered triples in this case.

The number of ordered triples $(a,b,c)$ for which the mean of all six
rolls is an integer is $1+25+10=36$.