How many sets of 8 3-digit consecutive even numbers are possible such that product when divided by 5 gives perfect cube?

Question

The sum of eight three-digit consecutive even number is S.When S is divided by 5, it results in a perfect cube.How many sets of such eight numbers are possible?

Answer 1

The sum of eight consecutive even numbers starting with $a$ is $8(a+7)$. So $\frac{8(a+7)}5 = b^3$, which solves to $$a=\frac58 b^3 - 7$$ For which $b$ is this an even integer between $100$ and $999-14$?

Answer 2

If the fifth of the eight consecutive numbers is $m,$ then the sum $S$ is equal to $8m - 8.$ Let $n$ be an integer. Then we have that $8m - 8 = 5n^{3}.$ Factoring, we receive $8(m - 1) = 5n^{3},$ or $m - 1 = \frac{5}{8}n^{3}.$ Since both $m$ and $n$ are integers, we know from that equation that $n$ is divisible by $2.$ In addition, because $m$ must be even, $\frac{5}{8}n^{3}$ must be odd, so $n$ cannot be divisible by $4.$ Since we also know that $108 \leq m \leq 992,$ as all of the addends must have three digits, rearranging the equation, we have $108 \leq \frac{5}{8}n^{3} + 1 \leq 992.$ Simplifying this, we have that $171.2 \leq n^{3} \leq 1585.6,$ or $5.553 \leq n \leq 11.661.$ Since $n$ is an integer, we see that there are $6$ integer solutions possible. But we said that $n$ must be divisible by $2$ but not by $4.$ So there are only $\boxed{2}$ solutions for $n.$