, 2025

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$?