We want to decompose each of the numbers 19, 28, and 37 into the sum of three numbers, such that the resulting nine numbers, when arranged in an appropriate order, have the same difference between any two consecutive numbers. How many ways can this be done, if we do not consider the order of the addends in the decompositions? - Solve the problem starting from the numbers $6, 72, 48$ as well.
I. Let the members of the questioned resolutions form a list (number sequence) that meets the requirement, with the middle fifth member being $a$ and the difference between adjacent members being $d$ (meaning that we always subtract the preceding member from the one that follows). Then the members of the list are

$$
a-4 d, \quad a-3 d, \quad a-2 d, \quad a-d, \quad a, \quad a+d, \quad a+2 d, \quad a+3 d, \quad a+4 d
$$

Let's call the coefficient of $d$ in the index of the respective member.

The sum of the 9 members is $9a$, and this is equal to the sum of the numbers to be resolved, $19+28+37=84$, from which $a=28/3$. Let the sum of the 3 indices resulting from the resolution of 19, 28, 37 be $s_{1}, s_{2}, s_{3}$, respectively, so the sums of these 3 members are $3a+s_{1}d, 3a+s_{2}d$, and $3a+s_{3}d$, respectively, so

$$
\begin{gathered}
28+s_{1} d=19, \quad 28+s_{2} d=28, \quad 28+s_{3} d=37, \quad \text { from which } \\
s_{2}=0, \quad s_{1}=-s_{3} \neq 0, \text { and } \\
d=9 / s_{3}
\end{gathered}
$$

According to these, if the 9 indices are arranged into 3 groups of 3, such that the sum of the 3 indices in one group is 0, and in another group it is non-zero - when of course the sum of the remaining group is the negative of the latter sum, since the sum of the 9 indices is 0, - then taking the non-zero sum as $s_{3}$ and calculating the common difference from (4), and then the members of (1), the 3 members of the list corresponding to the index triplet with the sum $s_{3}$ according to (2) give 37, the 3 members corresponding to the index triplet with the sum 0 give 28, and the remaining 3 members give 19. For example, the

$$
-3,0,3 \quad\left(s_{2}\right), \quad 1,2,4 \quad\left(s_{3}\right), \quad-4,-2,-1 \quad\left(s_{1}\right)
$$

index arrangement meets (3), $d=9/(1+2+4)=9/7$, and the members of the list (brought to a common denominator), and the decompositions of our numbers are:

$$
\begin{gathered}
\frac{88}{21}, \quad \frac{115}{21}, \quad \frac{142}{21}, \quad \frac{169}{21}, \quad \frac{196}{21}, \quad \frac{223}{21} \quad \frac{250}{21}, \quad \frac{277}{21}, \quad \frac{304}{21} \\
19=\frac{88}{21}+\frac{142}{21}+\frac{169}{21}, \quad 28=\frac{115}{21}+\frac{196}{21}+\frac{277}{21}, \quad 37=\frac{223}{21}+\frac{250}{21}+\frac{304}{21}
\end{gathered}
$$

If the last group in (5) is taken as $s_{3}$, $d=-9/7$, and the members of the list and decompositions in (6) are obtained in reverse order, but this should not be considered a new solution. (6) is also led to by the following grouping of the indices:

$$
-3,1,2 ; \quad 0,3,4 ; \quad-4,-2,-1 \quad \text { and } \quad-2,-1,3 ; 1,2,4 ; \quad-4,-3,0
$$

II. To determine the number of resolutions, it is not necessary to write down every possible arrangement of the 9 indices, it is enough to generate the index triplets with a sum of 0, to be used as $s_{2}$. It is easy to show that these are the following:

$$
\begin{array}{llll}
(-4,0,4), & (-3,-0,3), & (-2,0,2), & (-1,0,1), \\
(-4,1,3), & (-3,-1,4), & (-3,1,2), & (-2,-1,3),
\end{array}
$$

their number is 8. Indeed, starting from any of them, the second, $s_{3}$, index triplet - as we will show below - can be assembled from the remaining 6 indices in 20 ways, which also determines the arrangement, the members of $s_{1}$ will be the indices not used. Of the $8 \cdot 20=160$ groupings thus conceivable, those do not give a (1) shaped list, which have $s_{3}=0$, that is, $s_{1}=s_{2}=s_{3}=0$. From the (8) triplets, essentially two such groupings can be formed:

$$
\begin{aligned}
& (-4,0,4), \quad(-3,1,2), \quad(-2,-1,3) \\
& \text { and }(-2,0,2), \quad(-4,1,3), \quad(-3,-1,4) \text {. }
\end{aligned}
$$

(Due to the 0 index, one of the triplets in the first row of (8) must be present, but the $(-3,0,3)$ and $(-1,0,1)$ triplets cannot appear in an $s_{1}=s_{2}=s_{3}$ type arrangement because they have a common index with each of the other triplets.) Each of the (9) arrangements appears in the above 160 groupings in 6 different orders according to which of the 3 triplets gets the role of $s_{2}$, and $s_{3}$: $s_{2}$ can be chosen in 3 ways, $s_{3}$ in 2 ways.

Thus, in $160-2 \cdot 6=148$ arrangements of the 9 indices, the (3) condition is met. These - as in the example shown - give the same decomposition of the numbers 19, 