Problem 659

Suppose the integer in the centre circle is $a$.

Then each integer in a circle connected to the centre circle is either
$d$ more than $a$, which is $a+d$, or it is $d$ less than $a$, which is $a-d$.

The integers in the two circles connected to the centre could both be
$a+d$, as in Figure 1 below, or they
could both be $a-d$, as in Figure 2,
or one could be $a-d$ and one could
be $a+d$, as in Figure 3.

We note that in the last case (Figure 3), swapping locations of the
$a-d$ and the $a+d$ does not change the integers in the
final two empty circles (since they still depend on $a+d$ and $a-d$), and thus does not change the sum of
the five integers.

[[IMAGE0]]
Figure 1

[[IMAGE1]]
Figure 2

[[IMAGE2]]
Figure 3

Next, we explain why it is possible to place integers into the
remaining two circles (in each of the three cases above) so that the
positive difference between each pair of integers in connected circles
is $d$.

For the case that began in Figure 1, the integers in the two empty
circles are either $d$ more than
$a+d$, which is $a+2d$, or they are $d$ less than $a+d$, which is $a$.

These integers could both be $a+2d$,
as in Figure 1a below, or they could both be $a$, as in Figure 1b, or one could be $a+2d$ and one could be $a$, as in Figure 1c.

We note that in the last case (Figure 1c), swapping locations of the
final two integers, $a+2d$ and $a$, does not change the sum of the five
integers.

[[IMAGE3]]
Figure 1a

[[IMAGE4]]
Figure 1b

[[IMAGE5]]
Figure 1c

For the case that began in Figure 2, the integers in the two empty
circles are either $d$ more than
$a-d$, which is $a$, or they are $d$ less than $a-d$, which is $a-2d$.

These integers could both be $a$, as
in Figure 2a below, or they could both be $a-2d$, as in Figure 2b, or one could be
$a$ and one could be $a-2d$, as in Figure 2c.

We again note that in the last case (Figure 2c), swapping locations of
the final two integers, $a$ and $a-2d$, does not change the sum of the five
integers.

[[IMAGE6]]
Figure 2a

[[IMAGE7]]
Figure 2b

[[IMAGE8]]
Figure 2c

Finally, for the case that began in Figure 3, each integer in an
empty circle must have a positive difference of $d$ with both $a-d$ and $a+d$.

The integer $d$ more than $a-d$ is $a$, and the integer $d$ less than $a-d$ is $a-2d$.

The integer $d$ more than $a+d$ is $a+2d$, and the integer $d$ less than $a+d$ is $a$.

Thus, $a$ is the only integer that
has a positive difference of $d$ with
both $a-d$ and $a+d$, and so the integers in the two empty
circles must each be equal to $a$, as
shown in Figure 3a.

[[IMAGE9]]
Figure 3a

Suppose that the sum of the five integers in the circles is $S$.

For Case 1a (which corresponds to Figure 1a), adding the five integers
in the figure, we get$S=a+(a+d)+(a+2d)+(a+d)+(a+2d)=5a+6d$.

In the table below, we determine the value of $S$ for each of the $7$ cases.

Case 1a
Case 1b
Case 1c
Case 2a
Case 2b
Case 2c
Case 3a

$S=5a+6d$
$S=5a+2d$
$S=5a+4d$
$S=5a-2d$
$S=5a-6d$
$S=5a-4d$
$S=5a$

We must determine the number of different integers $d$, between $1$ and $20$ inclusive, for which at least one of
the seven expressions for $S$ is
equal to $54$ and $a$ is an integer.

Consider Case 1a, from which we get $5a+6d=54$.

Since both $a$ and $d$ are integers, and $d$ is between $1$ and $20$ inclusive, we can systematically
substitute values of $d$ into this
equation, and then solve for $a$ to
determine if $a$ is an integer.

For example if $d=1$, we get $5a+6\times1=54$ or $5a=48$.

However, there is no integer $a$ for
which $5a=48$ and so $d=1$ is not a possible value of $d$ in Case 1a. Substituting $d=2$ and $d=3$ similarly give non-integer values of
$a$.

When $d=4$, we get $5a+6\times4=54$ and so $5a=30$ or $a=6$.

In this case, the pair of integers $d=4$ and $a=6$ satisfy the equation $5a+6d=54$.

Substituting $d=4$ and $a=6$ into Figure 1a, we get the following
diagram.

[[IMAGE10]]

We can confirm that the positive difference between each pair of
integers in connected circles is $4$
(an integer between $1$ and $20$ inclusive), and the sum of the five
integers in the circles is $54$, as
required. Thus $d=4$ is a possible
value satisfying the given conditions.

We can systematically continue to substitute $d=5,6,7,\dots, 20$ into $5a+6d=54$ and solve the equation to
determine which values of $d$ give
integer values of $a$.

The next smallest value of $d$ for
which $a$ is an integer is $d=9$. In this case, we get $5a+6\times9=54$ and so $5a=0$ or $a=0$.

We could continue in this systematic way, however since there are $20$ possible values of $d$ and $7$ cases to check, this would take a while
to complete. Instead, we might recognize that $d=4$, $a=6$ and $d=9$, $a=0$ are both solutions to $5a+6d=54$.

Notice that from the first solution to the second, the value of $d$ increases by $5$, and the value of $a$ decreases by $6$.

Can you see why increasing $d$ by
$5$ and decreasing $a$ by $6$ gives the next possible pair of
integers for which $5a+6d=54$? (Hint:
Take a close look at the left side of the equation.)

If we increase $d$ by $5$ again, and decrease $a$ by $6$, we get $d=9+5=14$ and $a=0-6=-6$, and since $5a+6d=5\times(-6)+6\times14=-30+84=54$,
then $d=14$ and $a=-6$ is a solution to the equation (and
in fact, this is the next smallest value of $d$ that works).

The final integer value of $d$
between $1$ and $20$ inclusive for which $5a+6d=54$ is $d=14+5=19$, and in this case $a=-6-6=-12$ or $(d,a)=(19, -12)$

Therefore, Case 1a gives $d=4,9,14$,
and $19$, or $4$ values of $d$ which satisfy the given conditions.

We continue in this way for each of the first four cases, and
summarize all possible integer solutions for those cases in the table
below.

Case 1a
$5a+6d=54$
$(d,a)=(4,6),
(9,0), (14, -6), (19, -12)$
$d=4,9,14,19$

Case 1b
$5a+2d=54$
$(d,a)=(2,10),
(7,8), (12,6), (17,4)$
$d=2,7,12,17$

Case 1c
$5a+4d=54$
$(d,a)=(1,10),
(6,6), (11, 2), (16, -2)$
$d=1, 6, 11,
16$

Case 2a
$5a-2d=54$
$(d,a)=(3,12),
(8,14), (13, 16), (18,18)$
$d=3, 8, 13,
18$

Notice that after the first four cases shown above, all possible
values of $d$ from $1$ to $20$ inclusive satisfy the given conditions
with the exception of $d=5,10,15$,
and $20$.

In Case 3a we get, $5a=54$ and so
$a$ is not an integer. Next, consider
Case 2b, $5a-6d=54$.

Each value of $d$ left to check
($d=5,10,15, 20$) is a multiple of
$5$, and so $6d$ is a multiple of $5$ for each of these possible values of
$d$.

Since $5a$ is also a multiple of
$5$ for all possible integers $a$, then $5a-6d$ is the difference between two
multiples of $5$, and thus is a
multiple of $5$.

However, the right side of the equation $5a-6d=54$ is not a multiple of $5$ and so $d$ cannot be equal to a multiple of $5$.

In the final case, $5a-4d=54$, it is
similarly not possible for $d$ to be
equal to a multiple of $5$.

Thus $d$ can be equal to each of the
first $20$ positive integers with the
exception of $5,10,15$, and $20$, and so there are $20-4=16$ different possible values of
$d$.

It is worth noting that there are many different ways to find the
integer solutions to each of the $7$
equations (cases) above. For example, the value of each of the terms
$6d$, $2d$, $4d$, $-2d$, $-6d$, $-4d$, $0d$ is even for all integers $d$, and the right side of each equation,
$54$, is also even. This means that
in each equation, the value of $5a$
must be even, and so $a$ is even.
Further, when $a$ is even, the units
digit of $5a$ is $0$. Since the units digit of $54$ is $4$, what do we now know about the units
digit of each term containing a $d$,
and in each case, what does that tell us about the possible values of
$d$?