the sum of eight three digit consequtive even numbers is S.When Sis divided by 5,it results in a perfect cube

Question

how many sets of such eight numbers are possible? 1,2,3,4,5 (choose among this)

which of the following can be one of those eight numbers 644, 328, 108, 126, 140

Answer 1

Let the eight numbers be: $x-7,x-5,x-3,x-1,x+1,x+3,x+5,x+7$. (as each number must be even, that implies that $x$ is odd)

The sum of those eight numbers is then $8x$.

Since $8x/5$ is a perfect cube that tells us that $x$ is divisible by five. Further, as $x$ is odd, that implies that the last digit of $x$ is five, implying that the eight numbers are of the form $\star\star8,\star\star 0, \star\star2,\star\star4,\star\star6,\star\star8,\star\star0,\star\star 2$

We ask now, what values of $115\leq x\leq 985$ when multiplied by $8$ and divided by $5$ result in a perfect cube. (the limits on the inequalities arise because we know that $x$ must end in a five and we must have $x-7$ and $x+7$ a three digit number)

Equivalently, by letting $y=\frac{x}{5}$, what values of $23\leq y\leq 197$ when multiplied by $8$ yield a perfect cube. Well, if $y\cdot 8$ is a perfect cube, that implies that $y$ is a perfect cube. We ask now, what values of $y$ in that range actually are perfect cubes.

The sequence of perfect cubes is: $1,8,27,64,125,216,343,\dots$, so valid choices for $y$ seem to be $27,64,125$.

These would imply $x$ values of $135, 320, 625$ respectively, however we had determined that $x$ needed to be odd to make the eight numbers in our sum even, so we can rule out $320$ as a possibility.

There are then two possible sets of such numbers:

$$128,130,132,134,136,138,140,142$$

and

$$618,620,622,624,626,628,630,632$$